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1 COORDINATES, EIGENVALUES, AND EIGENVECTORS Review of coordinates The standard basis The standard basis for R n is the basis E = {e, e, e n }, where e j is the vector with in the j-th position and zeros elsewhere Coordinates with respect to a basis Let B = {v, v, v n } be a basis for R n We know that for each vector x R n there are unique scalar coordinates c, c, c n such that () x = c v + c v + + c n v n We call the vector c R n having coordinates c, c c n the coordinate vector of x with respect to the basis B, and denote it by x B 3 First examples (a) x E = x for each x R n (b) For any basis B = {v, v, v n } of R n : v j B = e j ( j n) (c) 0 B = 0 for any basis B (the zero vector has the same representation with respect to any basis 4 Matrix calculation of coordinates Letting B = {v, v, v n } be a basis for R n we can rewrite equation () in matrix form as () x = V x B where V is the matrix having the vector v j for its j-th column, ie V = v, v, v n Date: March 6, 005

2 EIGENVALUES/EIGENVECTORS V is invertible because its columns are linearly independent (see, for example, Summary 39(v) of our textbook, pp 9 0), so we can compute the B coordinate representation of any vector x R n by solving () for x B : (3) x B = V x For example, if in R we take B = {v, v } to be our basis, where v = and 3 v =, then 5 5 V = and V 3 5 = 3 Suppose, for instance, that x = Then 5 x B = V x = 3 More generally, if x = x 7 = 4 is any vector in R x n then the same calculation shows that 5x + x x B = 3x x 5 The matrix of a linear transformation with respect to a basis Recall that each n n matrix induces a linear transformation T : R n R n by means of matrix multiplication: (4) T (x) = Ax (x R n ) We refer to A as the standard matrix for T The lesson of what s to follow is that it s not always the best matrix for T! For a basis B = {v, v, v n } for R n we define (5) T B = T (v ) B, T (v ) B T (v n ) B, ie T B is the matrix that has the B-coordinate representation of T (v j ) as its j-th column ( j n) We call T B the matrix of T with respect to the basis B 6 Example: B = E, the standard basis In this case T B = Ae, Ae, Ae n, the standard matrix for T

3 EIGENVALUES/EIGENVECTORS 3 7 Geometric examples in R In this next group of examples, L is a line through the origin in the direction of a non-zero vector v R, and w is a non-zero vector perpendicular to v B = {v, w} We take as our basis for R, not the standard one, but instead (a) Suppose T : R R is the transformation of orthogonal projection onto L Then T (v) = v and T (w) = 0, so T B = T (v) B, T (w) B = v B, 0 B = e, 0 = where we use the fact that v B = e and 0 B = 0 (see 3 (b) and (c)) Compare this simple and informative matrix for T with the one we got when we first introduced orthogonal projection Then we took v to be a unit vector with coordinates u and u, and did this computation: u T (x) = (x v)v = (x u + x u ) u = u x + u u x u u u x + u = u u x u u u, x, x so the standard matrix for the linear transformation of orthogonal projection onto L is the completely uninformative one u A = u u u u u (b) Suppose T is reflection in L Then T (v) = v as before, but now T (w) = w Thus T B = T (v) B, T (w) B = v B, w B = e, e = 0 0 Note that in this computation we used the fact that w B = w B ( linearity of coordinates Fact 34, page 39 of our textbook) and that v B = e and w B = e ( 3 (b)) (c) Suppose T is a shear in the direction of L, ie that T (v) = v and T (w) = kv + w for some fixed real number k Then using linearity of coordinates and 3(b) we see that k T B = 0 which is the same as the standard matrix for a horizontal shear

4 4 EIGENVALUES/EIGENVECTORS 8 Using T B to compute T (x) The utility of the matrix of a linear transformation T with respect to a basis B lies in the following formula: (6) T B x B = T (x) B (x R n ), which says that, if you re willing to give up the standard basis fro R n, and do all your calculations relative to the basis B, then you can use the matrix of T with respect to B in the same way you used to use the standard matrix Proof of formula (6): For x R n we have the unique representation x = c v + c v + + c n v n where c, c,, c n are the (standard) coordinates of x B By linearity of T : T (x) = c T (v ) + c T (v ) + + c n T (v n ), so by linearity of coordinates: T (x) B = c T (v ) B + c T (v ) B + + c n T (v n ) B, But the expression on the right is just the matrix product T B x B, so we are done Eigenvalues and Eigenvectors Definition Suppose A is an n n matrix and T the associated linear transformation of R n (so T (x) = Ax for every x R n ) We call a real number λ an eigenvalue of T (or of A) whenever there is a non-zero vector x R n such that T (x) = λx The vector x is called an eigenvector of T (or of A) corresponding to the eigenvalue λ Examples (a) Look back at the examples of 7, where L is a line through the origin in the direction of a non-zero vector v, and w is a non-zero vector perpendicular to v If T : R R is orthogonal projection onto L, then v and w are eigenvectors of T corresponding respectively to eigenvalues and 0 If T is reflection in L then v and w are still eigenvectors, but now the corresponding eigenvalues are and If T is a shear along L then v is an eigenvector with eigenvalue, but w is not an eigenvector

5 EIGENVALUES/EIGENVECTORS 5 (b) (see Problem 5, page 46) Suppose A = and v =, w = 4 3 Then one computes that Av = 5v and Aw = w, so v is an eigenvector for A corresponding to the eigenvalue 5 and w is an eigenvector corresponding to the eigenvalue Note that, with respect to the basis B = {v, w} of R, the matrix 5 0 for the linear transformation induced by A is just 0 (c) (see Problem 9 of 34) Here T is the linear transformation of R 3 induced by the matrix A = We are given a basis B = {v, v, v 3 } for R 3, where v =, v =, and v 3 = You can check easily that each of these basis vectors is an eigenvector: Av = 0 (eigenvalue zero), Av = v (eigenvalue ) and Av 3 = v 3 (eigenvalue ), and that therefore the matrix of the associated linear transformation T with respect to the basis B is T B = In each of the examples of the previous two sections, whenever we had a linear transformation T of R n that bequeathed to R n a basis B of eigenvectors, the matrix of T with respect to B turned out to be diagonal, ie all entries zero off the main diagonal 3 Notation The n n diagonal matrix with entry λ j in the j-th diagonal position will be denoted diag {λ, λ, λ n } Thus, for example, the 3 3 identity matrix is diag {,, }, while in (c), T B = diag {0,, } Diagonal matrices are the easiest ones for which to find eigenvalues and eigenvectors: 4 Proposition If A = diag {λ, λ, λ n } then the standard unit vector e j is an eigenvector for the eigenvalue λ j ( j n)

6 6 EIGENVALUES/EIGENVECTORS Proof Our hypothesis is that A = λ e, λ e,, λ n e n, ie that A is the matrix with λ j e j as its j-th column ( j n) But Ae j is the j-th column of A, so Ae j = λ j e j for each j between and n 5 Exercise Prove the converse: If an n n matrix A has the standard basis for R n as eigenvectors, then A is diagonal 6 The Diagonalization Theorem For a linear transformation T : R n R n, and a basis B for R n, the following two statements are equivalent (each implies the other): (a) B consists of eigenvectors of T (b) T B is a diagonal matrix Proof (a) (b): Suppose B = {v, v, v n } consists of eigenvectors of T, so that there are real numbers λ j ( j n) such that T (v j ) = λ j v j (j =,,, n) We must show that T B is diagonal Starting with the definition of T B, we compute: T B = T v B, T v B,, T v n B = λ v B, λ v B,, λ n v n B = λ v B, λ v B,, λ n v n B = λ e, λ e,, λ n e n = diag {λ, λ,, λ n } where, in this computation we used successively: the definition of T B (line ), the fact that T (v j ) = λ j v j (line ), linearity of coordinates (line 3), and finally the fact that v j B = e j (line 4) (b) (a): Now suppose B = {v, v, v n } is a basis for R n, and that T B = diag {λ, λ,, λ n } We must show that each v j is an eigenvector of T In fact, we ll see that T v j = λ j v j for j n Fix such a j, and note that the j-th column of T B is just λ j e j, ie, () T B e j = λ j e j

7 EIGENVALUES/EIGENVECTORS 7 Now the right-hand side of () is λ j v j B which, by linearity of coordinates, is λ j v j B Similarly, the left-hand side of () is T B v j B which, by (6), is T v j B Thus T v j B = λ j v j B, ie, the both T v j and λ j v j have the same B-coordinate vectors Thus T v j = λ j v j, as desired 7 Definition Call an n n matrix diagonalizable if R n has a basis consisting of eigenvectors of this matrix Warning: Not all square matrices are diagonalizable! For example it s geometrically clear that shear transfomations of R have (up to multiplication by a non-zero constant) just a single eigenvector, while rotations of R through any angle not an integer multiple of π have no eigenvectors at all! We ll see how to prove this kind of result analytically (ie without reference to pictures) in the next section 3 Efficient eigenvalue computation for matrices We ll base our computations for matrices on the notion of determinant a b 3 Definition The determinant of the matrix A = is: c d det(a) = ad bc The key to all the work of this section is: 3 Theorem For A a matrix, the system of linear equations Ax = 0 has nontrivial solution if and only if det(a) = 0 Proof In scalar form the system of equations is: ax + bx = 0 cx + dx = 0 Suppose first that neither a nor c is zero Then we can replace the second equation by c times the first minus a times the second, yielding the equivalent system: ax + bx = 0 (ad bc)x = 0

8 8 EIGENVALUES/EIGENVECTORS which has a nontrivial solution if and only if ad bc = 0 Suppose a = 0 Then det(a) = bc, which is zero precisely when either b or c is zero If b = 0 then we are back to one equation in two unknowns, so there are nontrivial (infinitely many) solutions If b 0 then any solution must have x = 0, whereupon the second equation implies cx = 0 If c = 0 (ie det(a) = 0) then there are infinitely many solutions: x = 0 and x arbitrary If, however c 0 (ie, det(a) 0) then x is also required to be zero, hence the only solution is the trivial one 33 Corollary The eigenvalues of a matrix A are the real numbers λ that satisfy its characteristic equation det(a λi) = 0 Proof This follows immediately from the above Theorem and the definition of eigenvalue More precisely, to say a real number λ is an eigenvalue of A means that there is a non-zero vector x R such that: Ax = λx ie Ax λx = 0 ie (A λi)x = 0 Thus λ is an eigenvector of A if and only if the homogeneous system (A λi)x = 0 has a nontrivial solution, which, by Theorem 3, happens if and only if det(a λi) = 0 34 Example Show that the matrix is diagonalizable A = 4 3 Proof Of course we know this already from (b) where the eigenvectors and eigenvalues of A were handed to us But suppose they weren t Then we write the characteristic equation: λ 0 = det(a λi) = det( 4 3 λ = ( λ)(3 λ) 8 = λ 4λ 5 = (λ 5)(λ + ) )

9 EIGENVALUES/EIGENVECTORS 9 so that λ = 5 and λ = are the solutions, and so these are the eigenvalues of A Now for λ = 5, 4 A λi = A 5I = 4 and the system (A 5I)x = 0 is easily seen to have solutions x = any scalar multiple of In particular, v = is the unique (up to multiplication by a non-zero scalar) eigenvector for the eigenvalue 5 I leave it to you to show by the same method that v = eigenvector for the eigenvalue is the unique It s easy to check that B = {v, v } is a basis for R, so A is diagonalizable, and the matrix of its linear transformation, with respect to the basis B is diag {5, } 0 35 Example: a non-diagonalizable matrix Let A = The characteristic 0 0 equation for this matrix is λ = 0, which has only λ = 0 as its solution The corresponding eigenvectors are the nontrivial solutions to the equation Ax = 0 (ie the non-zero vectors in the kernel of A), and these all the non-zero scalar multiples of the standard unit vector e Thus there aren t enough eigenvectors here to make a basis of R, so A is not diagonalizable 4 Matrix of T B vs standard matrix of T Suppose T is a linear transformation of R n whows standard matrix is A, ie T (x) = Ax (x R n ) Suppose B = {v, v, v n } is a basis for R n It s time to get serious about the relationship between A and T B Recall that in 8 we derived equation (3): x B = V x where V = v, v,, v n is the matrix having the vector v j as its j-th column Recall also that the invertibility of V derives from the fact that its columns are linearly independent

10 0 EIGENVALUES/EIGENVECTORS Thus we may compute: T B = T (v ) B, T (v ) B,, T (v n ) B = Av B, Av B,, Av n B = V Av, V Av,, V Av n = V A v, v,, v n = V AV where the first line is the definition of T B, the second uses the definition of T in terms of A, the third uses (3), and the fourth is the definition of matrix multiplication We have proved: 4 Theorem (Fact 344, Page 44 of our text) Suppose B = {v, v, v n } is a basis for R n and T a linear transformation of R n with standard matrix A Then T B = V AV, where V is the matrix whose j-th column is v j ( j n) 5 Application: Powers of matrices Theorem 4 is of great utility in computing large powers of matrices a calculation that would be difficult to do by hand in many circumstances The method depends on two crucial facts 5 Theorem If A, B and V are n n matrices, and V is invertible, then for each positive integer p: (V BV ) p = V B p V Proof We use the method of Mathematical Induction, which asserts that to prove the theorem we need only check that it s true for p =, and then show that: (*) Result true for an integer p Result true for p + The point is that once you ve established the case p = and proved (*), then (*) gives you for free the case p =, and then the case p = 3, and so on forever

11 EIGENVALUES/EIGENVECTORS For our result, the case p = is obviously true, so suppose we have proved the theorem for some p We must show that it s true for p + ; to this end write (V BV ) p+ = (V BV ) p (V BV ) = (V B p V )(V BV ) = (V B p )(V V )(BV ) = (V B p )I(BV ) = V B p+ V where the second line follows from our induction hypothesis that the theorem is true for p, the third line follows from the associate law of matrix multiplication, and the others are clear The next result asserts that for a diagonal matrix you can compute a power by taking the corresponding power of each element of the matrix 5 Theorem For any positive integer p, diag {λ, λ,, λ n } p = diag {λ p, λ p,, λ p n} Proof I leave this one mostly as an exercise First check that diag {λ, λ,, λ n }diag {µ, µ,, µ n } = diag {λ µ, λ µ,, λ n µ n } The result about powers then follows quickly from this by induction 5 53 Example Compute 4 3 Solution Recall that we ve already found that the vectors v = and v = are eigenvectors for A = (for eigenvalues 5 and - respectively) and that B = 4 3 {v, v } is a basis for R Thus, according to Theorem 4 above, 5 0 = T 0 B = V AV, where V = v, v = Thus A = V V,

12 EIGENVALUES/EIGENVECTORS so successive application of Theorem 5 and Theorem 5 yield, (5) A 5 = V V = V 0 ( ) 5 V A little calculation shows that V = /3 /3 /3 /3 more calculation (and the fact that ( ) 5 = ) yield A 5 = , after which (5) and a little 54 Example: A dynamical system Neighboring countries Elbonia and Pilikia are undergoing a relatively peaceful period in their histories Each is reducing its annual defense budget, but because of nervousness about the others intentions, each adds back in to its budget a percentage of its neighbors expenditures More precisely, Pilikia reduces its budget by 5%, but adds back in 5% of Elbonia s budget, while Elbonia reduces its budget by a whopping 75%, but adds back in 75% of Pilikia s defense budget Let P (t) denote Pilikia s defense budget for year t, and E(t) Elbonia s (in 005 Euros times one billion) Suppose initially P (0) = 3 and E(0) = Determine the behavior of P (t) and E(t) as t P (t) Solution The vector x(t) = describes the state of the system at year t We have E(t) (with amounts calculated in billions of Euros) ie, P (t + ) = 75P (t) + 5E(t) E(t + ) = 5P (t) + 75E(t) (5) x(t + ) = Ax(t) where A = Thus defense expenditures for both countries will be: x() = Ax(0) in year one, x() = Ax() = A x(0) in year two, and more generally in year n they will be x(n) = Ax(n ) = A x(n ) = A n x(0)

13 So it all comes down to computing A n EIGENVALUES/EIGENVECTORS 3 For this I leave it to you to check that the eigenvalues of A are and /, and the corresponding eigenvectors are v = for λ = and v = for λ = / Now B = {v, v } is a basis for R relative to which the linear transformation T whose standard matrix is A has a diagonal matrix: 0 T B = = V 0 / AV, where V = v, v = Thus for each positive integer n, n (/) n = 0 / hence solving for A n : Now one calculates easily that A n = V V = n = V A n V, n 0 0 (/) n V / / / / from which follows A n = + n n n + n Thus x(n) = A n x(0) = ( + n ) n 3 n + n = 5 + n 5 n 5/ Thus x(n) as n, ie both nations tend toward a yearly defense budget 5/ of 5 billion Euros I 7 # 4, 8, 9,, 36, 38 6 Exercises II 7 #8; this will be part of a future writing assignment III 73 #40, 4 These will be part of a future writing assignment 54 of these notes works out a special case of #40

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