Matrix Methods for Linear Systems of Differential Equations
|
|
- Darren Gilmore
- 7 years ago
- Views:
Transcription
1 Matrix Methods for Linear Systems of Differential Equations We now present an application of matrix methods to linear systems of differential equations. We shall follow the development given in Chapter 9 of Fundamentals of Differential Equations and Boundary Value Problems by Nagle, Saff, Snider, third edition. Calculus of Matrices If we allow the entries a ij t in an n n matrix At to be functions of the variable t, thenat is a matrix function of t. Similarly if the entries x i t of a vector xt are functions of t, thenxt is a vector function of t. A matrix At is said to be continuous at t if each a ij t is continuous at t. At is differentiable at t if each a ij t is differentiable at t and we write Also da dt t A t a ij t nn a b Atdt a b aij tdt nn We have the following differentiation formulas for matrices d CA C da dt dt, C a constant matrix d A B da dt dt db dt d AB A db dt dt B da dt In the last formula the order in which the matrices are written is important, since matrix multiplication need not be commutative. Linear Systems in Normal Form Asystemofn linear differential equations is in normal form if it is expressed as x t Atxt ft where xt and ft are n column vectors and At a ij t nn. A system is called homogeneous if ft ; otherwise it is called nonhomogeneous. When the
2 elements of A are constants, the system is said to have constant coefficients. We note that a linear nth order differential equation y n t p n ty n p ty gt can be rewritten as a first order system in normal form using the substitution x t yt, x t y t,..., x n t y n t.5 Then x t y t x t x t y t x 3 t x n t y n t x n t x n t y n t p n ty n p ty gt From.5 we can write this last equation as x n t p tx t p n tx n t gt Thus the differential equation can be put in the form with x xt x,ft x n gt and A p t p t p t p n t p n t The initial value problem for the normal system is the problem of finding a differential vector function xt that satisfies the system on an interval I and also satisfies the initial condition xt x, where t is a given point of I and x is a given constant vector.
3 Example: Convert the initial value problem y 3y y y y 3 into an initial value problem for a system in normal form. Solution: y 3y y. x t yt x t y t x x x 3x x Thus x t 3 x x We also have the initial condition x x x 3 x. Theorem (Existence and Uniqueness) Suppose At and ft are continuous on an open interval I that contains the point t. Then, for any choice of the initial vector x there exists a unique solution xt on the entire interval I to the initial value problem x t Atxt ft, xt x Remark: Just as in Ma we may introduce the Wronskian of n vectors functions and use it to test for linear independence. We have Definition: The Wronskian of the n vector functions x t colx,x,...,x n,...,x n t colx n,x n,...,x nn is defined to be the real-valued function 3
4 Wx,...,x n t x t x t x n t x t x t x n t x n t x n t x nn t One can show that the Wronskian of solutions x,...,x n to x Ax is either identically zero or never zero on and interval I. Also, a set of n solutions x,...,x n to x Ax on I is linearly independent if and only if their Wronskian is never zero on I. Thus the Wronskian provides us with an easy test for linear independence for solutions of x Ax. Theorem (Representation of Solutions - Homogeneous Case) Let x,x,...,x n be n linearly independent solutions to the homogeneous system x t Atxt 3 on the interval I, whereat is an n n matrix function continuous on I. Then every solution of 3 on I can be expressed in the form xt c x t c n x n t 4 where c,...,c n are constants. A set of solutions x,...,x n that are linearly independent on I is called a fundamental solution set for 3. The linear combination 4 is referred to as the general solution of 3. Exercise: e t e t e t Verify that e t e t,, e t e t is a fundamental solution set for the system x t xt 5, eigenvectors:,, 4
5 Consider x 3 t e t e t.then e t e t Ax 3 t e t e t x 3 t e t e t e t Remark: The matrix Xt e t e t e t e t solution of 5 can be written as is a fundamental matrix for the DE 5. The general e t e t e t xt Xtc c e t e t c e t c 3 e t Remark:IfwedefineanoperatorL by Lx x Ax then this operator is linear. Thatis,Lc x c x c Lx c Lx. Thus if x and x are homogeneous solutions of the homogeneous equation x Ax the c x c x is also a solution of this equation. Another consequence of this linearity is the superposition principle for linear systems. It states that if x p and x p are solutions respectively of the nonhomogeneous systems Lx g and Lx g,thenx p x p is a solution of Lx g g. This leads to Theorem 3 (Representation of Solutions - Nonhomogeneous Case) Let x p be a particular solution to the nonhomogeneous system x t Atxt ft 6 on the interval I, andletx,x,...,x n be a fundamental solution set on I for the corresponding homogeneous system x t Atxt. Then every solution to 6 on I can be expressed in the form xt x p t c x t c n x n t 7 where c,...,c n are constants. 5
6 Remark: The linear combination of x p,x,...,x n written in 7 with arbitrary constants c,...,c n is called the general solution of 6. We may express this solution as x x p Xc, wherex is a fundamental matrix for the homogeneous system and c is an arbitrary constant vector. Solving Normal Systems. To determine a general solution to the n n homogeneous system x Ax : a. Find a fundamental solution set x,...,x n that consists of n linearly independent solutions to the homogeneous equation. b. Form the linear combination x Xc c x c n x n where c colc,...,c n is any constant vector and X x,...,x n is the fundamental matrix, to obtain a general solution.. To determine a general solution of to the nonhomogeneous system x Ax f : a. Find a particular solution x p to the nonhomogeneous system. b. Form the sum of the particular solution and the general solution Xc c x c n x n to the corresponding homogeneous system in part, x x p Xc x p c x c n x n to obtain a general solution. Homogeneous Linear Systems with Constant Coefficients Consider now the system x t Axt 8 where A is a (real) constant n n matrix. Theorem 4 Suppose the n n constant matrix A has n linearly independent eigenvectors u,u,...,u n.letr i be the eigenvalue corresponding to the u i.then e r t u,e r t u,...,e r nt u n 9 is a fundamental solution set on, for the homogeneous system x Ax. Hence the general solution of x Ax is xt c e r t u c n e r nt u n where c,...,c n are arbitrary constants. 6
7 Remark: The eigenvalues may be real or complex and need not be distinct. Proof Since Au i r i u i we have d dt er it u i r i e r it u i e r it Au i Ae r it u i so each element of the set 9 is a solution of the system 8. Also the Wronskian of these solutions is Wt dete r t u,...,e r nt u n e r r r nt detu,...,u n since the eigenvectors are linearly independent. Example Find a general solution of x 5 4 x 5 4, eigenvectors:, 4 4 Thus xt c e t c e 4t 4 Thus the solution is x t c e t 4c e 4t x t c e t c e 4t SNB gives the following strange looking result: x 5x 4x x x, Exact solution is: x t 3 C e t 4 3 C e 4t 4 3 C e 4t 4 3 C e t x t 3 C e 4t 3 C e t 4 3 C e t 3 C e 4t This is correct and is equivalent to, ifweletc 3 C 4 3 C and c 3 C 3 C. However, it is a most cumbersome form of the solution. 7
8 Exercise: Nagle and Saff page 535 #3. Find a fundamental matrix for the system x t 3 7 xt Solution: 3, eigenvectors: 3,, 7, Hence the four linearly independent solutions are e t 3,e t,e 7t 8 Therefore a fundamental matrix is,e 3t e t e t e 7t e 3t 3e t e 7t e 7t e 3t 8e 7t We know that if a matrix has n distinct eigenvalues, then the eigenvectors associated with these eigenvalues are linearly independent. Hence Corollary If the n n constant matrix A has n distinct eigenvalues r,...,r n and u i is an eigenvector associated with r i then e r t u,...,e r nt u n is a fundamental solution set for the homogeneous system x Ax. 8
9 Example Solve the initial value problem x t xt x Solution: 4 4 5, eigenvectors:,, c e t c e t c 3 e 3t Thus xt c e t c e t c 3 e 3t c e t c e t c 3 e 3t 4 4 c e t 4c e t 4c 3 e 3t We define xt via this. xt c e t c e t c 3 e 3t c e t c e t c 3 e 3t c e t 4c e t 4c 3 e 3t so that c c c 3 x c c c 3 c 4c 4c 3 Thus we form, row echelon form: 4 4 c,c,c 3. Then the solution is. Hence xt e t e 3t 4 4 Complex Eigenvalues 9
10 We now discuss how one solves the system x t Axt in the case where A is a real matrix and the eigenvalues are complex. We shall show how to obtain two real vector solutions of the system. Recall that if r i is a solution of the equation that determines the eigenvalues, namely, p deta ri then r i is also a solution of this equation, and hence is an eigenvalue. Recall that r is called the complex conjugate of r and r r. Let z a ib, wherea and b are real vectors, be an eigenvector corresponding to r. Then it is not hard to see that z a ib is an eigenvector corresponding to r.since A r Iz then taking the conjugate of this equation and noting that since A and I are real matrices then A A and I I A r Iz A r Iz A r Iz so z is an eigenvector corresponding to r. Therefore the vectors w t e it a ib and w t e it a ib are two linearly independent vector solutions of. However, they are not real. To get real solutions we proceed as follows: Since e it e t cost isint then w t e it a ib e t costa sin tb isinta costb Therefore w t x t ix t where x t e t costa sintb x t e t sin ta costb Since w t is a solution of, then w t Aw t so x t ix t Ax t iax t Equating real and imaginary parts of this last equation leads to the real equations x t Ax t x t Ax t
11 so that x t and x t are real vector solutions of corresponding to the eigenvalues i. Note that we can get the two expressions above for x t and x t by taking the real and imaginary parts of w t Example Find the general solution of the initial value problem x t 3 xt x Solution: We first find the eigenvalues and eigenvectors of the matrix. We want the roots of 3 r r 3 r r r 4r 5 Thus r i The system of equations to determine the eigenvectors is 3 rx x x rx or 3 rx x x rx For r i we have ix x x ix Multiplication of the first equation by i i yields the second equation, since i i. Thus x ix Letting x gives the eigenvector conjugate of the first we have 3, eigenvectors: i i. Since the second eigenvector is the complex i, i i i i Thus,,a,b. The two linearly independent solutions are
12 wt e it i e t e it ie t e it e t cost isin t ie t cost isint e t cost ie t sint e t cost sin t ie t cost sin t e t cost sin t ie t sint cost Therefore x t e t costa sintb e t cost sint x t e t sin ta costb e t sint cost Thus xt c e t cost sint c e t sint cost x c e c e. Therefore so c e and c xt e t cost sin t Example Find a [real] general solution to x x 3 x x. Solution: We first find the eigenvalues and eigenvectors of the matrix. We want the solutions to 3 r det 3 r r r 4r 5 r Thus r The system of equations to determine the eigenvectors is i
13 3 rx x x rx For r i we have ix x x ix or ix x x ix One can see that the first and second equations are the same by multiplying the first equation by i and recalling that i i. Thus x i x We have x i x i x i i i x Letting x we have the eigenvector i Since the eigenvectors are complex conjugates we have that the eigenvalues of the matrix i i are i and i and the corresponding eigenvectors are and. 3 3 The eigenvalues of the matrix i i and. Let r i, so,. Also are i and i and the corresponding eigenvectors are i i so a and b. x t e t costa sintb e t cost sint x t e t sin ta costb e t sin t cost 3
14 xt c e t cost sint c e t sint cost c e t cost sin t e t cost c e t sint cost e t sint Nonhomogeneous Linear Systems The techniques of Undetermined Coefficients and Variation of Parameters that are used to find particular solutions to the nonhomogeneous equation y pxy qxy gx have analogies to nonhomogeneous systems. Thus we now discuss how one solves the nonhomogeneous system x t Atxt ft Undetermined Coefficients Consider the nonhomogeneous constant coefficient system x t Axt ft Before presenting the method for systems we recall the following result for second nonhomogeneous order differential equations. For more on this see Linear Second Order DEs (Hold down the Shift key and click.) A particular solution of where a,b,c are constants is where ay by cy Ke x y p Kex p if p y p Kxex p if p, p y p K p x e x if p p Example pr ar br c y 5y 4y e x 4
15 Homogeneous solution: p , y h c e x c e 4x Now to find a particular solution for e x. p Since p 5 p 5 3 y p kxex p xe x 3 y y h y p c e x c e 4x 3 xex Example Problem 3, page 547 of text. Find the general solution of x t xt e t 4e t e t Solution: We first find the homogeneous solution., eigenvectors: 3,, 3 Since these eigenvectors are linearly independent, then x h t c e 3t We seek a particular solution of the form c e 3t x p t e t a a a 3 c 3 e 3t Then x p t e t a a a 3 Ax p t e t 4e t e t e t a a a 3 e t 4e t e t e t a a a 3 a a a 3 a a a 3 4 Thus 5
16 a a a a 3 a a a a 3 4 a 3 a a a 3, Solution is: a,a,a 3 Therefore x p t e t and xt x h t x p t c e 3t c e 3t c 3 e 3t e t Note: the Method of Undetermined Coefficients works only for constant coefficient systems. Example Problem, Page 547 of text. Find a general solution to x t t xt 4 4t Solution: We first find a general homogeneous solution. r r 4 r r 3 r 3r 4 r Thus the eigenvalues are r,3. The equations that determine the eigenvectors are rx x 4x rx For r 3 we have x x 4x 3x Thus x x and an eigenvector is. For r we have x x 4x x Thus x x and we have the eigenvector. Hence 6
17 x h t c e 3t c e t. To find x p we let x p t a t b a t b since ft is a polynomial. Plugging into the DE we have a a 4 a t b a t b t 4t b t b ta ta 4b 4t b 4ta ta Equating the coefficients of t on both sides we have a a or 4 4a a a a 4a a 4 Therefore a,a. Equating the constant terms on both sides we have a b b a 4b b Using the values for a, a we have b b 4b b Thus b,b. With these values for the constants we have that Finally x p t xt x h t x p t c e 3t t c e t t Example Problem 4, page 547 of text. Find a general solution to 7
18 x t Solution: We first find a homogeneous solution. xt 4cost sin t r r 4 4 4r r r r 4r rr 4 r Thus the eigenvalues are r,4. The equations for the eigenvectors are rx x x rx For r we have x x or x x. Thus we have the eigenvector. Forr 4 we have x x x x or x x Thus we have the eigenvector Hence x h t c e t. c e 4t We assume Hence Plugging into the DE yields x p t x p a cost b sint a cost b sint a sint b cost a sint b cost a sin t b cost a sin t b cost a cost b sint a cost b sint 4cost sin t or a sin t b cost a sin t b cost a cost a cost b sin t b sin t a cost a cost b sin t b sin t 4cost sin t We equate the coefficients of the sint and cost terms. Thus 8
19 b a a 4 b a a or To solve this system we form a b b a b b a a b 4 a a b a b b a b b 4 and row reduce. 4 R R 3 R R R R 3 R R R R 3R R 4 R R 3R R 4 4 R 4R 3 4 R 3 R R 3 R 5 R 4 R R 4 R 9
20 Thus a,a,b,b and Finally a general solution is x p t sint cost sint xt x h t x p t c c e 4t sint cost sint 3 Example The eigenvalues of the matrix i i eigenvectors are and. Find a [real] general solution to x x 3 are i and i and the corresponding x x 5t. Solution: First we find a real general solution to x x 3 x x. Here, and Since then i i a ib. x t e t costa sintb x t e t sin ta costb x t e t cost sint x t e t sin t cost Hence x h t c x t c x t c e t cost sint e t cost c e t sint cost e t sint Or we may expand one of the complex solutions and take the real and imaginary parts.
21 e it i e t cost isin t i e t cost sint i sin t cost cost isint e t cost sint e t cost i e t sint cost e t sint x h c e t cost sin t e t cost c e t sint cost e t sin t e t cost sint e t cost e t sin t cost e t sint c c Next, we find a particular solution to the given non-homogeneous equation. Since the non-homogeneous term is a polynomial of degree one, the solution must be the same. Thus let x p t x x at b ct d We substitute into the system of DEs and find the coefficients. x x 3 x x 5t a c 3 at b ct d 5t a c 3a ct 3b d a ct b d 5t We equate like terms. 3a c 5 a c a 3b d c b d Thus (from the first pair of equations) a 5, c and then b andd 8. Combining homogeneous and particular solutions, we have a general solution.
22 x x c e t cost sint e t cost c e t sin t cost e t sin t 5t t 8 e t cost sin t e t cost e t sint cost e t sint c c 5t t 8 Example a) Find the eigenvalues and eigenvectors of A. Solution: We solve deta ri. deta ri r r r r r i r i So, the eigenvalues are a complex conjugate pair. We find the eigenvector for one and take the complex conjugate to get the other. For r i, we solve A riu i i u u Thus we have the equations iu u u iu The second row is redundant, so iu u oru i u. Hence any multiple of i is an eigenvector for r i. Then an eigenvector corresponding to r i is b) Find the [real] general solution to i. x x x x e t. Solution: The solution is the general solution x h to the homogeneous equation plus one [particular] solution x p to the full non-homogeneous equation. First we ll find x p. It is in the form x p c e t c e t
23 Substituting into the D.E., we obtain x x c e t c e t c e t c e t e t Hence c e t c e t c e t c e t c e t c e t e t c e t c e t c e t c e t e t We can divide by e t (which is never zero) and move the unknowns to the left side to obtain c c x p e t To find a solution to the homogeneous solution we use the eigenvalue i i and the corresponding eigenvector i a ib. i Since x t e t costa sintb x t e t sin ta costb then x t e t cost sint x t e t sint cost Hence x h t c x t c x t c e t cost e t sin t c e t sin t e t cost Or, for the solution to the homogeneous equation, we may use one of the eigenvalues and eigenvectors found in a to write a complex solution and break it into real and imaginary parts. we ll use i. 3
24 x e it i e t cost isint i e t cost ie t sin t e t sint ie t cost x h c e t cost e t sin t c e t sin t e t cost e t cost e t sint e t sint e t cost c c Finally, we add to obtain the desired solution. x e t cost e t sin t e t sin t e t cost c c e t Variation of Parameters (This material is not covered in Ma 7.) Consider now the nonhomogeneous system x t Atxt ft where the entries in At may be any continuous functions of t. LetXt be a fundamental matrix for the homogeneous system x t Atxt The general solution of is Xtc where c is an n constant vector. To find a particular solution of, consider x p t Xtvt where vt is an n vector function of t that we wish to determine. Then x p t Xtv t X tvt so that yields Xtv t X tvt Atx p t ft AtXtvt ft Since Xt is a fundamental matrix for, thenx t AXt, so the last equation becomes Xtv t AtXtvt AtXtvt ft or Xtv t ft Since the columns of Xt are linearly independent, X t exists and v t X tft 4
25 Hence vt X tftdt and Finally the general solution is given by x p t Xtvt XtX tftdt xt x h t x p t Xtc XtX tftdt 3 Example Page 578 Solve the initial value problem x t 3 xt e t, x Solution: We first find a fundamental matrix for the homogeneous solution. 3, eigenvectors:, 3 Xt 3e t e t e t e t Hence 3e t e t e t e t Formula 3 yields e t e t 3 e t e t X xt 3e t e t c e t e t c 3e t e t e t e t e t e t 3 e t e t e t dt 3e t e t c e t e t c 3e t e t e t e t et e t 6 e3t 3 e t 3e t e t c e t e t c 3e t et e t e t 6 e3t 3 e t e t et e t e t 6 e3t 3 e t 5
26 xt x Thus 3e t e t c e t e t c 3c c 3 3 c c 7 3, Solution is: c 5 6,c 3 xt 6 et 6 et 3e t e t 3 e t e t 5 6 7e t 5e 3t 8 8e t 9e t 5e 3t e t 3e t et e t e t 6 e3t 3 e t e t et e t e t 6 e3t 3 e t 3c c 3 3 c c 7 3. Finally 3e t et e t e t 6 e3t 3 e t e t et e t e t 6 e3t 3 e t The Matrix Exponential Function (Not covered in F) Recall that the general solution of the scalar equation x t axt where a is a constant is xt ce at. We will now see that the general solution of the normal system x t Axt where A is a constant n n matrix is xt e At c. We must, of course, define e At. Definition: Let A be a constant n n matrix. The we define e At I At A t! An tn n! n A n tn n! This is an n n matrix. Remark: If D is a diagonal matrix, then the computation of e Dt is straightforward. Example Let D.Then 6
27 D 4, D 3 8,... D n n n Therefore e Dt n D n n! tn n n t n n! n n t n n! e t e t In general if D is an n n diagonal matrix with r,r,...,r n down its main diagonal, then e Dt is the diagonal matrix with e r t,e r t,...,e r nt down its main diagonal. It can be shown that the series converges for all t and has many of the same properties as the scalar exponential e at. Remark: It can be shown that if a matrix A has n linearly independent eigenvectors, then P AP is a diagonal matrix, where P is formed from the n linearly independent eigenvectors of A. Thus P AP D 3 where D is a diagonal matrix. In fact, D has the eigenvalues of A along its diagonal. Let A Thus P Then, inverse: 4 3, P, 4 Hence P AP Now 3 implies that when A has n linearly independent eigenvalues we have A PDP so that 7
28 e At e PDP t I PDP t PDP t PDP t I PDP t PDP PDP t I PDP t PD P t P I Dt Dt P Pe Dt P Example: Let A 5 4,, eigenvectors: 3, P,P and D 3. Show that A P DP and use this to compute e At. A PDP as required. Thus e 5 4 t e t e 3t e 3t e t e 3t e t e t e 3t from SNB. Also Pe Dt P e 3 t so e At Pe Dt P e t e 3t e t e 3t e t e 3t e t e 3t e t e 3t This e At is a fundamental matrix for the system x t 5 4 xt sinceifwelet 8
29 x h t e 5 4 t c c e t e 3t e t e 3t c e t e 3t e t e 3t c c e 3t e t c e t e 3t c e t e 3t c e 3t e t c e 3t e t c e t e 3t c e t e 3t c e 3t e t then x h t c 6e 3t e t c e t 6e 3t c e t 3e 3t c 3e 3t e t and Ax h t 5 4 c e 3t e t c e t e 3t c e t e 3t c e 3t e t 4c e t e 3t 5c e t e 3t 4c e 3t e t 5c e 3t e t c e t e 3t c e t e 3t c e 3t e t c e 3t e t 6c e 3t c e t c e t 6c e 3t c e t c e t 3c e 3t 3c e 3t c 6e 3t e t c e t 6e 3t c e t 3e 3t c 3e 3t e t x h t : Theorem 5 (Properties of the Matrix Exponential Function) Let A and B be n n constant matrices and r,s and t be real (or complex) numbers. Then. a. e A e I b. e Ats e At e As c. e At e At d. e ABt e At e Bt provided that AB BA e. e rit e rt I f. d dt e At Ae At Remark: c. tells us that the matrix e At has an inverse. Proof of f. 9
30 d dt e At d dt I At A t! A A t A 3 t! An tn n! An tn n! Ae At A I At A t! An tn n! Theorem 6 (e At is a Fundamental Matrix) If A is an n n constant matrix, then the columns of the matrix e At form a fundamental solution set for the system x t Axt. Therefore, e At is a fundamental matrix for the system, and a general solution is xt e At c. Lemma (Relationship Between Fundamental Matrices) Let Xt and Yt be two fundamental matrices for the same system x Ax. Then there exists a constant column matrix C such that Yt XtC. Remark: Let Yt e At XtC and set t. Then I XC C X and e At XtX If the nxn matrix A has n linearly independent eigenvectors u i,thene r t u, e r t u,,e r nt is a fundamental matrix for x Ax and e At e rt u, e rt u,,e rnt u,u,,u n Calculating e At for Nilpotent Matrices Definition: An nxn matrix A matrix is nilpotent if for some positive integer k A k. Since e At I At A t! An tn n! n A n tn n! we see that if A is nilpotent, then the infinite series has only a finite number of terms since A k A k and in this case e At I At A t! Ak tk k! This may be taken further. The Cayley-Hamilton Theorem says that a matrix satisfies its own characteristic equation, that is, pa. Therefore, if the characteristic polynomial for A has the form pr n r r n,thatisahas only one multiple eigenvalue r,then pa n A r I n. Hence A r I is nilpotent and e At e r IA r It e rit e A rit e r t I A r It A r In tn n! Example Find the fundamental matrix e At for the system 3
31 x t Axt where A Solution: The characteristic polynomial for A is pr det r r r r 3 3r 3r r 3 Hence r isaneigenvalueofa with multiplicity 3. By the Cayley-Hamilton Theorem A I 3 and e At e t e A It e t I A It A I t! A I and A I Thus e At e t te t e t te t te t te t te t e t te t te t te t te t e t e t 3
r (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + 1 = 1 1 θ(t) 1 9.4.4 Write the given system in matrix form x = Ax + f ( ) sin(t) x y 1 0 5 z = dy cos(t)
Solutions HW 9.4.2 Write the given system in matrix form x = Ax + f r (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + We write this as ( ) r (t) θ (t) = ( ) ( ) 2 r(t) θ(t) + ( ) sin(t) 9.4.4 Write the given system
More informationLS.6 Solution Matrices
LS.6 Solution Matrices In the literature, solutions to linear systems often are expressed using square matrices rather than vectors. You need to get used to the terminology. As before, we state the definitions
More informationSystem of First Order Differential Equations
CHAPTER System of First Order Differential Equations In this chapter, we will discuss system of first order differential equations. There are many applications that involving find several unknown functions
More informationSo far, we have looked at homogeneous equations
Chapter 3.6: equations Non-homogeneous So far, we have looked at homogeneous equations L[y] = y + p(t)y + q(t)y = 0. Homogeneous means that the right side is zero. Linear homogeneous equations satisfy
More informationHW6 Solutions. MATH 20D Fall 2013 Prof: Sun Hui TA: Zezhou Zhang (David) November 14, 2013. Checklist: Section 7.8: 1c, 2, 7, 10, [16]
HW6 Solutions MATH D Fall 3 Prof: Sun Hui TA: Zezhou Zhang David November 4, 3 Checklist: Section 7.8: c,, 7,, [6] Section 7.9:, 3, 7, 9 Section 7.8 In Problems 7.8. thru 4: a Draw a direction field and
More informationSimilarity and Diagonalization. Similar Matrices
MATH022 Linear Algebra Brief lecture notes 48 Similarity and Diagonalization Similar Matrices Let A and B be n n matrices. We say that A is similar to B if there is an invertible n n matrix P such that
More informationBrief Introduction to Vectors and Matrices
CHAPTER 1 Brief Introduction to Vectors and Matrices In this chapter, we will discuss some needed concepts found in introductory course in linear algebra. We will introduce matrix, vector, vector-valued
More informationHigher Order Equations
Higher Order Equations We briefly consider how what we have done with order two equations generalizes to higher order linear equations. Fortunately, the generalization is very straightforward: 1. Theory.
More informationMATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m
MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 1. SYSTEMS OF EQUATIONS AND MATRICES 1.1. Representation of a linear system. The general system of m equations in n unknowns can be written a 11 x 1 + a 12 x 2 +
More informationNotes on Determinant
ENGG2012B Advanced Engineering Mathematics Notes on Determinant Lecturer: Kenneth Shum Lecture 9-18/02/2013 The determinant of a system of linear equations determines whether the solution is unique, without
More informationMatrix Algebra. Some Basic Matrix Laws. Before reading the text or the following notes glance at the following list of basic matrix algebra laws.
Matrix Algebra A. Doerr Before reading the text or the following notes glance at the following list of basic matrix algebra laws. Some Basic Matrix Laws Assume the orders of the matrices are such that
More informationa 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.
Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given
More informationGeneral Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1
A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics General Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1 Dr. John Ehrke Department of Mathematics Fall 2012 Questions
More informationSystems of Linear Equations
Systems of Linear Equations Beifang Chen Systems of linear equations Linear systems A linear equation in variables x, x,, x n is an equation of the form a x + a x + + a n x n = b, where a, a,, a n and
More informationMATRIX ALGEBRA AND SYSTEMS OF EQUATIONS
MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Systems of Equations and Matrices Representation of a linear system The general system of m equations in n unknowns can be written a x + a 2 x 2 + + a n x n b a
More information4.5 Linear Dependence and Linear Independence
4.5 Linear Dependence and Linear Independence 267 32. {v 1, v 2 }, where v 1, v 2 are collinear vectors in R 3. 33. Prove that if S and S are subsets of a vector space V such that S is a subset of S, then
More informationLinear Algebra Notes for Marsden and Tromba Vector Calculus
Linear Algebra Notes for Marsden and Tromba Vector Calculus n-dimensional Euclidean Space and Matrices Definition of n space As was learned in Math b, a point in Euclidean three space can be thought of
More information3. Let A and B be two n n orthogonal matrices. Then prove that AB and BA are both orthogonal matrices. Prove a similar result for unitary matrices.
Exercise 1 1. Let A be an n n orthogonal matrix. Then prove that (a) the rows of A form an orthonormal basis of R n. (b) the columns of A form an orthonormal basis of R n. (c) for any two vectors x,y R
More informationThe Characteristic Polynomial
Physics 116A Winter 2011 The Characteristic Polynomial 1 Coefficients of the characteristic polynomial Consider the eigenvalue problem for an n n matrix A, A v = λ v, v 0 (1) The solution to this problem
More information1 2 3 1 1 2 x = + x 2 + x 4 1 0 1
(d) If the vector b is the sum of the four columns of A, write down the complete solution to Ax = b. 1 2 3 1 1 2 x = + x 2 + x 4 1 0 0 1 0 1 2. (11 points) This problem finds the curve y = C + D 2 t which
More informationby the matrix A results in a vector which is a reflection of the given
Eigenvalues & Eigenvectors Example Suppose Then So, geometrically, multiplying a vector in by the matrix A results in a vector which is a reflection of the given vector about the y-axis We observe that
More informationAu = = = 3u. Aw = = = 2w. so the action of A on u and w is very easy to picture: it simply amounts to a stretching by 3 and 2, respectively.
Chapter 7 Eigenvalues and Eigenvectors In this last chapter of our exploration of Linear Algebra we will revisit eigenvalues and eigenvectors of matrices, concepts that were already introduced in Geometry
More informationUniversity of Lille I PC first year list of exercises n 7. Review
University of Lille I PC first year list of exercises n 7 Review Exercise Solve the following systems in 4 different ways (by substitution, by the Gauss method, by inverting the matrix of coefficients
More informationCONTROLLABILITY. Chapter 2. 2.1 Reachable Set and Controllability. Suppose we have a linear system described by the state equation
Chapter 2 CONTROLLABILITY 2 Reachable Set and Controllability Suppose we have a linear system described by the state equation ẋ Ax + Bu (2) x() x Consider the following problem For a given vector x in
More informationComplex Eigenvalues. 1 Complex Eigenvalues
Complex Eigenvalues Today we consider how to deal with complex eigenvalues in a linear homogeneous system of first der equations We will also look back briefly at how what we have done with systems recapitulates
More informationSecond Order Linear Nonhomogeneous Differential Equations; Method of Undetermined Coefficients. y + p(t) y + q(t) y = g(t), g(t) 0.
Second Order Linear Nonhomogeneous Differential Equations; Method of Undetermined Coefficients We will now turn our attention to nonhomogeneous second order linear equations, equations with the standard
More informationChapter 17. Orthogonal Matrices and Symmetries of Space
Chapter 17. Orthogonal Matrices and Symmetries of Space Take a random matrix, say 1 3 A = 4 5 6, 7 8 9 and compare the lengths of e 1 and Ae 1. The vector e 1 has length 1, while Ae 1 = (1, 4, 7) has length
More informationIntroduction to Matrix Algebra
Psychology 7291: Multivariate Statistics (Carey) 8/27/98 Matrix Algebra - 1 Introduction to Matrix Algebra Definitions: A matrix is a collection of numbers ordered by rows and columns. It is customary
More informationExamination paper for TMA4115 Matematikk 3
Department of Mathematical Sciences Examination paper for TMA45 Matematikk 3 Academic contact during examination: Antoine Julien a, Alexander Schmeding b, Gereon Quick c Phone: a 73 59 77 82, b 40 53 99
More informationT ( a i x i ) = a i T (x i ).
Chapter 2 Defn 1. (p. 65) Let V and W be vector spaces (over F ). We call a function T : V W a linear transformation form V to W if, for all x, y V and c F, we have (a) T (x + y) = T (x) + T (y) and (b)
More informationA Brief Review of Elementary Ordinary Differential Equations
1 A Brief Review of Elementary Ordinary Differential Equations At various points in the material we will be covering, we will need to recall and use material normally covered in an elementary course on
More informationSolving Linear Systems, Continued and The Inverse of a Matrix
, Continued and The of a Matrix Calculus III Summer 2013, Session II Monday, July 15, 2013 Agenda 1. The rank of a matrix 2. The inverse of a square matrix Gaussian Gaussian solves a linear system by reducing
More informationLinear Maps. Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007)
MAT067 University of California, Davis Winter 2007 Linear Maps Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007) As we have discussed in the lecture on What is Linear Algebra? one of
More informationSECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS A second-order linear differential equation has the form 1 Px d y dx dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. Equations of this type arise
More informationMath 115A HW4 Solutions University of California, Los Angeles. 5 2i 6 + 4i. (5 2i)7i (6 + 4i)( 3 + i) = 35i + 14 ( 22 6i) = 36 + 41i.
Math 5A HW4 Solutions September 5, 202 University of California, Los Angeles Problem 4..3b Calculate the determinant, 5 2i 6 + 4i 3 + i 7i Solution: The textbook s instructions give us, (5 2i)7i (6 + 4i)(
More information9. Particular Solutions of Non-homogeneous second order equations Undetermined Coefficients
September 29, 201 9-1 9. Particular Solutions of Non-homogeneous second order equations Undetermined Coefficients We have seen that in order to find the general solution to the second order differential
More informationNOTES ON LINEAR TRANSFORMATIONS
NOTES ON LINEAR TRANSFORMATIONS Definition 1. Let V and W be vector spaces. A function T : V W is a linear transformation from V to W if the following two properties hold. i T v + v = T v + T v for all
More informationSubspaces of R n LECTURE 7. 1. Subspaces
LECTURE 7 Subspaces of R n Subspaces Definition 7 A subset W of R n is said to be closed under vector addition if for all u, v W, u + v is also in W If rv is in W for all vectors v W and all scalars r
More informationSection 6.1 - Inner Products and Norms
Section 6.1 - Inner Products and Norms Definition. Let V be a vector space over F {R, C}. An inner product on V is a function that assigns, to every ordered pair of vectors x and y in V, a scalar in F,
More informationORDINARY DIFFERENTIAL EQUATIONS
ORDINARY DIFFERENTIAL EQUATIONS GABRIEL NAGY Mathematics Department, Michigan State University, East Lansing, MI, 48824. SEPTEMBER 4, 25 Summary. This is an introduction to ordinary differential equations.
More informationminimal polyonomial Example
Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We
More informationCITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION
No: CITY UNIVERSITY LONDON BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION ENGINEERING MATHEMATICS 2 (resit) EX2005 Date: August
More informationDERIVATIVES AS MATRICES; CHAIN RULE
DERIVATIVES AS MATRICES; CHAIN RULE 1. Derivatives of Real-valued Functions Let s first consider functions f : R 2 R. Recall that if the partial derivatives of f exist at the point (x 0, y 0 ), then we
More informationInner Product Spaces
Math 571 Inner Product Spaces 1. Preliminaries An inner product space is a vector space V along with a function, called an inner product which associates each pair of vectors u, v with a scalar u, v, and
More informationMATH 304 Linear Algebra Lecture 18: Rank and nullity of a matrix.
MATH 304 Linear Algebra Lecture 18: Rank and nullity of a matrix. Nullspace Let A = (a ij ) be an m n matrix. Definition. The nullspace of the matrix A, denoted N(A), is the set of all n-dimensional column
More informationMATH10212 Linear Algebra. Systems of Linear Equations. Definition. An n-dimensional vector is a row or a column of n numbers (or letters): a 1.
MATH10212 Linear Algebra Textbook: D. Poole, Linear Algebra: A Modern Introduction. Thompson, 2006. ISBN 0-534-40596-7. Systems of Linear Equations Definition. An n-dimensional vector is a row or a column
More informationName: Section Registered In:
Name: Section Registered In: Math 125 Exam 3 Version 1 April 24, 2006 60 total points possible 1. (5pts) Use Cramer s Rule to solve 3x + 4y = 30 x 2y = 8. Be sure to show enough detail that shows you are
More informationLINEAR ALGEBRA W W L CHEN
LINEAR ALGEBRA W W L CHEN c W W L Chen, 1997, 2008 This chapter is available free to all individuals, on understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied,
More informationInner Product Spaces and Orthogonality
Inner Product Spaces and Orthogonality week 3-4 Fall 2006 Dot product of R n The inner product or dot product of R n is a function, defined by u, v a b + a 2 b 2 + + a n b n for u a, a 2,, a n T, v b,
More information1 Introduction to Matrices
1 Introduction to Matrices In this section, important definitions and results from matrix algebra that are useful in regression analysis are introduced. While all statements below regarding the columns
More informationVector and Matrix Norms
Chapter 1 Vector and Matrix Norms 11 Vector Spaces Let F be a field (such as the real numbers, R, or complex numbers, C) with elements called scalars A Vector Space, V, over the field F is a non-empty
More informationMathematics Course 111: Algebra I Part IV: Vector Spaces
Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are
More information160 CHAPTER 4. VECTOR SPACES
160 CHAPTER 4. VECTOR SPACES 4. Rank and Nullity In this section, we look at relationships between the row space, column space, null space of a matrix and its transpose. We will derive fundamental results
More informationMATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set.
MATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set. Vector space A vector space is a set V equipped with two operations, addition V V (x,y) x + y V and scalar
More informationMethods for Finding Bases
Methods for Finding Bases Bases for the subspaces of a matrix Row-reduction methods can be used to find bases. Let us now look at an example illustrating how to obtain bases for the row space, null space,
More informationEigenvalues and Eigenvectors
Chapter 6 Eigenvalues and Eigenvectors 6. Introduction to Eigenvalues Linear equations Ax D b come from steady state problems. Eigenvalues have their greatest importance in dynamic problems. The solution
More information1 Sets and Set Notation.
LINEAR ALGEBRA MATH 27.6 SPRING 23 (COHEN) LECTURE NOTES Sets and Set Notation. Definition (Naive Definition of a Set). A set is any collection of objects, called the elements of that set. We will most
More informationRecall the basic property of the transpose (for any A): v A t Aw = v w, v, w R n.
ORTHOGONAL MATRICES Informally, an orthogonal n n matrix is the n-dimensional analogue of the rotation matrices R θ in R 2. When does a linear transformation of R 3 (or R n ) deserve to be called a rotation?
More informationThe Determinant: a Means to Calculate Volume
The Determinant: a Means to Calculate Volume Bo Peng August 20, 2007 Abstract This paper gives a definition of the determinant and lists many of its well-known properties Volumes of parallelepipeds are
More informationSolving Systems of Linear Equations
LECTURE 5 Solving Systems of Linear Equations Recall that we introduced the notion of matrices as a way of standardizing the expression of systems of linear equations In today s lecture I shall show how
More information8 Square matrices continued: Determinants
8 Square matrices continued: Determinants 8. Introduction Determinants give us important information about square matrices, and, as we ll soon see, are essential for the computation of eigenvalues. You
More informationNonhomogeneous Linear Equations
Nonhomogeneous Linear Equations In this section we learn how to solve second-order nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form ay by cy G x where
More informationMATH2210 Notebook 1 Fall Semester 2016/2017. 1 MATH2210 Notebook 1 3. 1.1 Solving Systems of Linear Equations... 3
MATH0 Notebook Fall Semester 06/07 prepared by Professor Jenny Baglivo c Copyright 009 07 by Jenny A. Baglivo. All Rights Reserved. Contents MATH0 Notebook 3. Solving Systems of Linear Equations........................
More informationQuestion 2: How do you solve a matrix equation using the matrix inverse?
Question : How do you solve a matrix equation using the matrix inverse? In the previous question, we wrote systems of equations as a matrix equation AX B. In this format, the matrix A contains the coefficients
More informationIRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL. 1. Introduction
IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL R. DRNOVŠEK, T. KOŠIR Dedicated to Prof. Heydar Radjavi on the occasion of his seventieth birthday. Abstract. Let S be an irreducible
More information13 MATH FACTS 101. 2 a = 1. 7. The elements of a vector have a graphical interpretation, which is particularly easy to see in two or three dimensions.
3 MATH FACTS 0 3 MATH FACTS 3. Vectors 3.. Definition We use the overhead arrow to denote a column vector, i.e., a linear segment with a direction. For example, in three-space, we write a vector in terms
More informationMATH 423 Linear Algebra II Lecture 38: Generalized eigenvectors. Jordan canonical form (continued).
MATH 423 Linear Algebra II Lecture 38: Generalized eigenvectors Jordan canonical form (continued) Jordan canonical form A Jordan block is a square matrix of the form λ 1 0 0 0 0 λ 1 0 0 0 0 λ 0 0 J = 0
More informationThe Method of Partial Fractions Math 121 Calculus II Spring 2015
Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method
More informationMATH 551 - APPLIED MATRIX THEORY
MATH 55 - APPLIED MATRIX THEORY FINAL TEST: SAMPLE with SOLUTIONS (25 points NAME: PROBLEM (3 points A web of 5 pages is described by a directed graph whose matrix is given by A Do the following ( points
More informationVector Spaces. Chapter 2. 2.1 R 2 through R n
Chapter 2 Vector Spaces One of my favorite dictionaries (the one from Oxford) defines a vector as A quantity having direction as well as magnitude, denoted by a line drawn from its original to its final
More information1 0 5 3 3 A = 0 0 0 1 3 0 0 0 0 0 0 0 0 0 0
Solutions: Assignment 4.. Find the redundant column vectors of the given matrix A by inspection. Then find a basis of the image of A and a basis of the kernel of A. 5 A The second and third columns are
More informationDecember 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B. KITCHENS
December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B KITCHENS The equation 1 Lines in two-dimensional space (1) 2x y = 3 describes a line in two-dimensional space The coefficients of x and y in the equation
More informationTHREE DIMENSIONAL GEOMETRY
Chapter 8 THREE DIMENSIONAL GEOMETRY 8.1 Introduction In this chapter we present a vector algebra approach to three dimensional geometry. The aim is to present standard properties of lines and planes,
More informationCHAPTER 2. Eigenvalue Problems (EVP s) for ODE s
A SERIES OF CLASS NOTES FOR 005-006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 4 A COLLECTION OF HANDOUTS ON PARTIAL DIFFERENTIAL EQUATIONS
More informationChapter 6. Orthogonality
6.3 Orthogonal Matrices 1 Chapter 6. Orthogonality 6.3 Orthogonal Matrices Definition 6.4. An n n matrix A is orthogonal if A T A = I. Note. We will see that the columns of an orthogonal matrix must be
More informationOrthogonal Diagonalization of Symmetric Matrices
MATH10212 Linear Algebra Brief lecture notes 57 Gram Schmidt Process enables us to find an orthogonal basis of a subspace. Let u 1,..., u k be a basis of a subspace V of R n. We begin the process of finding
More informationLecture 1: Schur s Unitary Triangularization Theorem
Lecture 1: Schur s Unitary Triangularization Theorem This lecture introduces the notion of unitary equivalence and presents Schur s theorem and some of its consequences It roughly corresponds to Sections
More informationHOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!
Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following
More informationLinear Algebra Notes
Linear Algebra Notes Chapter 19 KERNEL AND IMAGE OF A MATRIX Take an n m matrix a 11 a 12 a 1m a 21 a 22 a 2m a n1 a n2 a nm and think of it as a function A : R m R n The kernel of A is defined as Note
More informationFactorization Theorems
Chapter 7 Factorization Theorems This chapter highlights a few of the many factorization theorems for matrices While some factorization results are relatively direct, others are iterative While some factorization
More information3.2 Sources, Sinks, Saddles, and Spirals
3.2. Sources, Sinks, Saddles, and Spirals 6 3.2 Sources, Sinks, Saddles, and Spirals The pictures in this section show solutions to Ay 00 C By 0 C Cy D 0. These are linear equations with constant coefficients
More informationDEFINITION 5.1.1 A complex number is a matrix of the form. x y. , y x
Chapter 5 COMPLEX NUMBERS 5.1 Constructing the complex numbers One way of introducing the field C of complex numbers is via the arithmetic of matrices. DEFINITION 5.1.1 A complex number is a matrix of
More informationLinear Algebra: Determinants, Inverses, Rank
D Linear Algebra: Determinants, Inverses, Rank D 1 Appendix D: LINEAR ALGEBRA: DETERMINANTS, INVERSES, RANK TABLE OF CONTENTS Page D.1. Introduction D 3 D.2. Determinants D 3 D.2.1. Some Properties of
More informationMAT188H1S Lec0101 Burbulla
Winter 206 Linear Transformations A linear transformation T : R m R n is a function that takes vectors in R m to vectors in R n such that and T (u + v) T (u) + T (v) T (k v) k T (v), for all vectors u
More informationLecture L3 - Vectors, Matrices and Coordinate Transformations
S. Widnall 16.07 Dynamics Fall 2009 Lecture notes based on J. Peraire Version 2.0 Lecture L3 - Vectors, Matrices and Coordinate Transformations By using vectors and defining appropriate operations between
More informationSecond-Order Linear Differential Equations
Second-Order Linear Differential Equations A second-order linear differential equation has the form 1 Px d 2 y dx 2 dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. We saw in Section 7.1
More information[1] Diagonal factorization
8.03 LA.6: Diagonalization and Orthogonal Matrices [ Diagonal factorization [2 Solving systems of first order differential equations [3 Symmetric and Orthonormal Matrices [ Diagonal factorization Recall:
More informationChapter 19. General Matrices. An n m matrix is an array. a 11 a 12 a 1m a 21 a 22 a 2m A = a n1 a n2 a nm. The matrix A has n row vectors
Chapter 9. General Matrices An n m matrix is an array a a a m a a a m... = [a ij]. a n a n a nm The matrix A has n row vectors and m column vectors row i (A) = [a i, a i,..., a im ] R m a j a j a nj col
More informationNOV - 30211/II. 1. Let f(z) = sin z, z C. Then f(z) : 3. Let the sequence {a n } be given. (A) is bounded in the complex plane
Mathematical Sciences Paper II Time Allowed : 75 Minutes] [Maximum Marks : 100 Note : This Paper contains Fifty (50) multiple choice questions. Each question carries Two () marks. Attempt All questions.
More informationa 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)
ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x
More informationMethods of Solution of Selected Differential Equations Carol A. Edwards Chandler-Gilbert Community College
Methods of Solution of Selected Differential Equations Carol A. Edwards Chandler-Gilbert Community College Equations of Order One: Mdx + Ndy = 0 1. Separate variables. 2. M, N homogeneous of same degree:
More informationSolution to Homework 2
Solution to Homework 2 Olena Bormashenko September 23, 2011 Section 1.4: 1(a)(b)(i)(k), 4, 5, 14; Section 1.5: 1(a)(b)(c)(d)(e)(n), 2(a)(c), 13, 16, 17, 18, 27 Section 1.4 1. Compute the following, if
More information26 Ideals and Quotient Rings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 26 Ideals and Quotient Rings In this section we develop some theory of rings that parallels the theory of groups discussed
More informationLinear Algebra I. Ronald van Luijk, 2012
Linear Algebra I Ronald van Luijk, 2012 With many parts from Linear Algebra I by Michael Stoll, 2007 Contents 1. Vector spaces 3 1.1. Examples 3 1.2. Fields 4 1.3. The field of complex numbers. 6 1.4.
More informationLectures notes on orthogonal matrices (with exercises) 92.222 - Linear Algebra II - Spring 2004 by D. Klain
Lectures notes on orthogonal matrices (with exercises) 92.222 - Linear Algebra II - Spring 2004 by D. Klain 1. Orthogonal matrices and orthonormal sets An n n real-valued matrix A is said to be an orthogonal
More informationProblem Set 5 Due: In class Thursday, Oct. 18 Late papers will be accepted until 1:00 PM Friday.
Math 312, Fall 2012 Jerry L. Kazdan Problem Set 5 Due: In class Thursday, Oct. 18 Late papers will be accepted until 1:00 PM Friday. In addition to the problems below, you should also know how to solve
More information6. Cholesky factorization
6. Cholesky factorization EE103 (Fall 2011-12) triangular matrices forward and backward substitution the Cholesky factorization solving Ax = b with A positive definite inverse of a positive definite matrix
More informationChapter 13: Basic ring theory
Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring
More informationLEARNING OBJECTIVES FOR THIS CHAPTER
CHAPTER 2 American mathematician Paul Halmos (1916 2006), who in 1942 published the first modern linear algebra book. The title of Halmos s book was the same as the title of this chapter. Finite-Dimensional
More informationMATH PROBLEMS, WITH SOLUTIONS
MATH PROBLEMS, WITH SOLUTIONS OVIDIU MUNTEANU These are free online notes that I wrote to assist students that wish to test their math skills with some problems that go beyond the usual curriculum. These
More information