Homework 7 M 373K by Mark Lindberg and Travis Schedler

Chapter 3, Exercise 5 Homework 7 M 373K by Mark Lindberg and Travis Schedler (a) Determine the basechange matrix in R, when the old basis is the standard basis E (e, e ) and the new basis is B (e + e, e e ) The basechange matrix will be B in matrix form, that is, P We can see that this works because EP e e e + e e e B (b) Determine the basechange matrix in R n when the old basis is the standard basis is E and the new basis is B (e n, e n,, e ) 0 The basechange matrix will be the reverse diagonal identity, that is, P 0 This is because 0 EP e e e n e n e n e n e e B 0 (c) Let B be the basis of R in which v e and v is a vector of unit length making an angle of 0 with v Determine the basechange matrix that relates E to B Let us assume that this is making an angle in the ( counterclockwise direction from v [ ] ) We see by some brief trigonometry that B, 0 3 Then P 0 3 This is because EP [ [ ] ] e e 0 3 0 3 B ( ) If instead the angle were clockwise, we would get P 0 Chapter 3, Exercise 55 How many subspaces of each dimension are there in (a) F 3 p? We have that that the number of subspaces of F 3 p of any given dimension is number of bases (linearly independent sets of n vectors) in a space of dim(3) number of bases (linearly independent sets of n vectors) in a given subspace of dim(n), because each basis of a space of dim(3) generates some subspace, and so we must divide by the number of bases that are contained within that subspace, so that we only count the bases which generate different subspaces Then we have: 3

n 0: There is only one basis (and indeed, only one set) of dimension 0: {} The subspace it generates is {0} (by definition), so its basis must be {}, and so we see that we have subspace of dimension 0 n : Since the matrix cannot be 0, the bases can be written in the form {v v 0} Let v f f f 3 Then there are p 3 possible vectors, minus the 0 vector, for a total of p 3 bases Each of these bases generates a subspace of dimension, which will have p elements (multiples of the basis), one of which will be the 0 vector Any of the others will be a basis, so we have p such bases Thus, there are a total of p3 p p + p + subspaces of dimension n : The bases are of the form {(v, v ) v, v 0, v, v linearly independent} Thus, we see that there are, as in the previous part, p 3 possible choices for the first vector There will be p multiples of this vector, including the 0 vector, so there are p 3 p linearly independent vectors to choose from for v Thus, there are a total of (p 3 )(p 3 p) bases Now consider the bases of this subspace There are p non-zero choices for the first vector, and p p (subtracting multiples of the first) choices for the second vector, giving a total of (p )(p p) bases for the subspace Thus, there are (p3 )(p 3 p) (p )(p p) p + p + subspaces of dimension n 3: By observation, we see that any basis of a subspace of dimension 3 must span F 3 p, and so there can only be subspace Fp 3 itself Note: the numbers are equal for k and 3 k because there is a bijection between subspaces W F 3 p of dimension k and subspaces W F 3 p of dimension n k given by taking perpendiculars with respect to dot product: W W {v F 3 p v w 0, w W } (Remark: although over R we would have W W {0}, this does not hold over other fields including F p, eg, Span(,, ) Span(,, ) in F 3 3, but it is still true that in Fn p, dim W n dim W and that (W ) W ) (b) F 4 p? We will use our earlier formula above, only slightly modified to see that there are in this case number of bases (linearly independent sets of n vectors) in a space of dim(4) number of bases (linearly independent sets of n vectors) in a given subspace of dim(n) n 0: As above, the only basis is {}, its corresponding subspace is {0}, and the only basis of said subspace is {}, so there is again subspace of dimension 0 n : Similarly to above, there are p 4 choices for the vector, and p bases for the generated subspace, giving p4 p p3 + p + p + subspaces of dimension n : There are p 4 choices for the first vector, and then p 4 p for the second, for a total of (p 4 )(p 4 p) bases In any given subspace, however, there are p choices for the first vector of the basis, and p p choices for the second Thus, there are a total of (p )(p p) bases contained in a given subspace Thus, there are (p4 )(p 4 p) (p )(p p) p4 + p 3 + p + p + subspaces of dimension n 3: By the same reasoning as above (abbreviated), we have (p 4 )(p 4 p)(p 4 p ) bases, and in a given subspace, (p 3 )(p 3 p)(p 3 p ) bases, for a total of (p4 )(p 4 p)(p 4 p ) (p 3 )(p 3 p)(p 3 p ) p3 + p + p + subspaces of dimension 3

n 4: By the same reasoning as above, we see that any basis of a subspace of dimension 4 must span F 4 p, and so there can only be subspace F 4 p itself Again, for the same reason as before, the numbers of spaces of dimension k and of dimension 4 k coincide for all k, by the bijection W W 3 Chapter 3, Exercise 6 Prove that the space R n n of all n n real matrices is the direct sum of the space of symmetric matrices (A t A) and the space of skew-symmetric matrices (A t A) It isn t totally clear from the wording here whether we are supposed to also verify that the symmetric matrices actually form a vector subspace as well as the skew-symmetric matrices This is easy to verify: observe a symmetric matrix times any real number is still symmetric and the sum of symmetric matrices is symmetric, and same for skew-symmetric matrices So they do indeed form subspaces Consider the intersection of the spaces This means that A A t A, and this is only true when A 0, and thus, the intersection is {0} Now we must show that it spans the set We can see that any matrix A R n n can be written as (A+At A t ) (A+At )+ (A At ) Then note that ( (A + At )) t (A + At ) t (At + A) (A + At ), and so the first matrix is symmetric Similarly, ( (A At )) t (A At ) t (At A) (A + At ), and so the second matrix is skew-symmetric Thus, we have decomposed A into the sum of a symmetric and a skew-symmetric matrices, and so this sum spans R n n Then by part (b) of proposition 366 in Artin, we have that R n n is a direct sum of (A t A) (A t A) 4 Chapter 4, Exercise 5 (a) Let U and W be vector spaces over a field F Show that the operations (u, w) + (u, w ) (u + u, w + w ) and c(u, w) (cu, cw) on pairs of vectors make the product set U W into a vector space It is called the product space I will show that these operations fulfill the requirements on page 84 of the textbook Let u, u, u 3 U and w, w, w 3 W, a, b F : (i) Associativity: ((u, w ) + (u, w ))+(u 3, w 3 ) (u +u, w +w )+(u 3, w 3 ) (u + u + u 3, w + w + w 3 ) (u, w ) + (u + u 3, w + w 3 ) (u, v ) + ((u, w ) + (u 3, w 3 )), by the associativity of U and V by the definition of vector spaces Identity: (u, v ) + (0, 0) (u + 0, v + 0) (u, v ) (0 + u, 0 + v ) (0, 0)+(u, v ), by the identity of U and V by the definition of vector spaces 3 Inverse: (u, v )+( u, v ) (u u, v v ) (0, 0) ( u +u, v + v ) ( u, v ) + (u, v ), by the inverses of U and V by the definition of vector spaces 4 Commutativity: (u, v ) + (u, v ) (u + u, v + v ) (u + u, v + v ) (u, v ) + (u, v ), by the commutativity of U and V by the definition of vector spaces (ii) (u, w ) (u, w ) (u, w ), by the multiplicative identity of U and V by the definition of vector spaces (iii) (ab)(u, w ) ((ab)u, (ab)w ) (a(bu ), a(bw )) a(bu, bw ) a(b(u, w )), by the associativity of U and V under scalar multiplication by the definition of vector spaces 3

(iv) Distributivity: (a+b)(u, w ) ((a+b)u, (a+b)w ) (au +bu, aw +bw ) (au, aw )+ (bu, bw ) a(u, w ) + b(u, w ), by the distributivity of U and V under scalar multiplication by the definition of vector spaces a((u, w ) + (u, w )) a(u + u, w + w ) (a(u + u ), a(w + w )) (au + au, aw + aw ) (au, aw ) + (au, aw ) a(u, w ) + a(u, w ), by the distributivity of U and V under scalar multiplication by the definition of vector spaces (b) Let U and W be subspaces of a vector space V Show that the map T : U W V defined by T (u, w) u + w is a linear transformation Let u, u U, w, w W, c F (i) T ((u, w )+(u, w )) T ((u +u, w +w )) (u +u )+(w +w ) u +w + u + w T ((u, w )) + T ((u, w )), by the associativity and commutativity of addition in the vector space V (ii) T (c(u, w )) T ((cu, cw )) cu + cw c(u + w ) ct ((u, w )), by the distributivity of scalar multiplication over addition in the vector space V (c) Express the dimension formula for T (same as in the previous part) in terms of the dimensions of subspaces of V By the dimension formula, dim(u W ) dim(ker(t ))+dim(im(t )) We can see that the ker(t ) is going to be all u U and w W such that u + w id v 0 But then u w, which implies that w U, and then since U is a group under addition, w U Similarly, w u, which implies that u W, and u W Thus, we see that we will get 0 if and only if each vector is also an element of the other vector space, that is, u, w U W Then ker(t ) (x, x), where x U W, and so there are dim(u W ) such elements x and thus dim(ker(t )) dim(u W ) Then we see that im(t ) is going to be all elements that can be expressed as a linear combination of the basis vectors of u and w, since for any u, w, each of u and w can be written as a linear combination of their respective bases, and therefore u + w can be written as the sum of these two combinations Thus, dim(im(t )) dim(u + W ), by definition Then we examine the elements of U W Let B U {u,, u r } and B W {w,, w m } be bases for U and W respectively Then we see that dim(u) r and dim(w ) m by definition Let u U and w W, and by definition of a basis, there is a unique set a,, a r, b,, b m such that u a u + + a r u r, and w b w + + a m w m Then (u, w) (a u + + a r u r, b m + + a m w m ) a (u, 0) + + a r (u r, 0) + b (0, w ) + + b m (0, w m ) We see that since this is the form of all vectors in U W, the set ((u, 0),, (u r, 0), (0, w ),, (0, w m )) spans U W, by definition We also see that, if (u, w) (0, 0), then 0 a u + + a r u r and 0 b w + + a m w m, which means that a a r 0 and b b m 0, by the fact that the bases of U and W are linearly independent, and so, since every coefficient must be 0, by the same property, ((u, 0),, (u r, 0), (0, w ),, (0, w m )) must be linearly independent Since it is a linearly independent spanning set, by definition, it is a basis Thus, there are r + m vectors in this basis, so dim(u W ) r + m dim(u) + dim(w ) In conclusion, dim(u) + dim(w ) dim(u + W ) + dim(u W ) 5 Chapter 4, Exercise Let A and B be matrices Determine the matrix of the 4

operator T : M AMB on the space F of matrices, with respect to the basis (e, e, e, e ) of F ( ) Let the basis be written as C,,, so c e, c e, a a c 3 e, and c 4 e, A b b and B Then we see that a a b b e : a a T (e ) 0 b b a a b b a 0 b b a 0 b b a b a b a b a b 0 a b + a b + a b + a b a b Thus, T (e ) a b c +a b c +a b c 3 +a b c 4, and we write it as T a b a b a b e : a a T (e ) b b a a b b 0 a b b 0 a b b a b a b a b a b ] a b [ 0 + a b [ ] + a b + a b a b Thus, T (e ) a b c +a b c +a b c 3 +a b c 4, and we write it as T a b a b a b e 3 : a a T (e 3 ) b b a a 0 b b a 0 b b a 0 b b a b a b a b a b 0 a b + a b + a b + a 0 b 5

a b Thus, T (e 3 ) a b c +a b c +a b c 3 +a b c 4, and we write it as T 3 a b a b a b e 4 : a a T (e 4 ) b b a a b b 0 a b b 0 a b b a b a b a b a b 0 a b + a b + a b + a 0 b a b Thus, T (e 4 ) a b c +a b c +a b c 3 +a b c 4, and we write it as T 4 a b a b a b Then by proposition 45 from the textbook, we see that the matrix of the operator is a b a b a b a b T T T 3 T 4 a b a b a b a b a b a b a b a b a b a b a b a b 6 Chapter 4, Exercise 69 Find all real matrices such that A I, and describe geometrically the way they operate by left multiplication on R a b a b a b These matrices must be of the form A such that AA c d c d c d [ a ] + bc ab + bd 0 ca + dc cb + d I Thus, a + bc d + bc and b(a + d) c(a + d) 0 Now we consider cases: a + d 0: Then by the cancellation property of roots, b c 0 Then a d, and so either a d or a d (If a d [ or ] [ a d ], we violate 0 0 the assumption) Thus, the two matrices we get are, 0 a + d 0: Then d a, and we can conclude that bc a, but nothing else a b Thus, the matrices must be of the form, with the given restriction on bc c a a b Thus, A I, I, or such that bc a c a Geometrically, I is the identity and does no transformation I is the 80 degree rotation about the origin (alternatively, sends vectors to their negative 6

Now we will consider the third, crazy case Then det(λi A) λ a b c λ + a (λ a)(λ + a) bc λ a bc λ (a + bc) λ (λ + )(λ ) When (λ + )(λ ) 0, then λ ± Let t corresponding eigenvectors be g, g They make a basis for R because they are linearly independent Any vector v R can thus be written as d(g ) + f(g ), where d, f R are constants Then Av Ad(g ) + Af(g ) d(ag )+f(ag ) dg fg, by the definition of an eigenvector Thus, this is a reflection of g across g, and since the eigenvectors do not need to be perpendicular, this is an oblique reflection Alternative solution: The condition A I implies that the only possible complex eigenvalues are ± The Jordan form of A can only have one block, because a Jordan block of size ( ) ( ) λ 0 λ 0 cannot square to the identity (this is easy to verify; in this case λ λ λ which is not the identity, even if λ ±) So the Jordan form of A is either ±I (in which case P AP ±I for some P and hence A ±I itself), ( so ) that A is the 0 or 80 degree 0 rotation about the origin, or else it is the matrix In the latter case, in some 0 basis g, g, we have Ag g and Ag g, so the transformation is an oblique reflection ag + bg ag bg for all a, b R 7 Chapter 4, Exercise 7 Prove that A is an idempotent matrix, ie, that A A, and find its Jordan form It is straightforward to see that and so A is idempotent A AA + + + + + + + + + A Now consider the possible Jordan forms N P AP of A such that A A In this case we also have N P A P P AP N, so we can see which Jordan forms satisfy N N We see that the eigenvalues (the elements on the diagonal of N) can only be zero or one, and squaring any Jordan block of size will not get back itself (for blocks of size two this follows from the computation in the previous problem; and we can restrict to eigenvalues of 0 and ) So the possible Jordan forms are just the diagonal matrices with s and 0 s on the diagonal In the case at hand, A is a rank one matrix, since all 7

nonzero rows are multiples of each other (equivalently this holds for the columns) and A is nonzero So the Jordan form, up to rearranging the blocks, must be 0 0 8 Let T : V V be a linear transformation from a finite-dimensional vector space V to itself For any basis B V, we can form the matrix A of T in the basis (so that if A (a ij ) are the entries, then T (v j ) i a ijv i, or in other words, the i-th column of A expresses T (v j ) in terms of the basis B) Consider the characteristic polynomial χ A χ A (x) : det(xi A) where x is viewed as a variable Show that χ A does not depend on the choice of B, so gives a well-defined characteristic polynomial for T, call it χ T Show that the roots of χ T are the eigenvalues of T (recall an eigenvalue λ is an element of F such that T v λv for some nonzero v V ) Hint: For any different choice of basis, A is replaced by P AP for some invertible matrix P, and hence xi A is replaced by xi P AP P (xi A)P Recall also that det(λi A) 0 if and only if (λi A)X 0 has a nonzero solution If we chose a different B BP, then the transformation of A to A is given by A P AP, as per proposition 435 from Artin Then we see that χ A det(xi A ) det(xi P AP ) det(xip P P AP ) det(p xip P AP ) det(p (xi A)P ) det(p ) det(xi A) det(p ) det(xi A) det(p ) det(p ) det(xi A) χ A and so the value of χ does not depend on the basis, and is therefore well defined Thus, we call the characteristic polynomial χ T Now note that by theorem from the textbook xi T is invertible if and only if the system (xi T )X 0has only the trivial solution X 0 Since we also know that xi T is invertible if and only if det(xi T ) 0, then we can see that if det(xi T ) 0, then the only solution to (xi T )v 0 is v 0, and so by definition this is not an eigenvalue But if det(xi T ) 0, then there is some non-zero v V such that (xi T )v 0, and so xiv T v 0, or xiv T v, or xv T v, and so by definition, x is an eigenvalue of T 9 Again let T : V V be a linear transformation with V finite-dimensional Define the characteristic polynomial of T, χ T (x), as in the previous problem (ie, for any basis B of V, if A is the matrix of T in the basis B, then χ T (x) det(xi A)) Prove the following: (a) The polynomial χ T (x) is monic (ie, has leading coefficient ) and is of degree n By definition, for any matrix A of T, we have χ T (x) det(xi A) Let B : xi A and let a ij and b ij xδ ij a ij be the entries of A and B, respectively, in the i-th 8

row and j-th column Now expanding this with the determinant formula, χ T (x) det(b) σ S n b σ() b nσ(n) Since each b ij is a polynomial of degree in x, this gives a polynomial of degree n The only term in the sum of degree n is the one b b b nn (x a ) (x a nn ) This has x n as its leading term (b) We have χ T (x) x n if T is nilpotent, ie, T n 0 for some n Bonus: Prove the converse over F C, using the Jordan Normal Form theorem (Theorem 470) As per Schedler s office hours, we can use the result of problem 68 from chapter 4 in Artin, which tells us that since T is nilpotent, there is a basis of V such that the 0 matrix of T is upper triangular, with diagonal entries 0 That is, A, 0 x 0 0 x 0 and by problem 8, χ T det(xi A) det x 0 x 0 x det xn, by a property of the determinants of upper triangular x matrices (c) We have χ T (x) (x ) n if T is unipotent, ie, T n I for some n Hint: This easily reduces to (b) Bonus: Again prove the converse By various definitions, the only definitions of unipotent matrices that we could find outside of this homework, T is unipotent if T I is nilpotent Then by the result of the previous part of this problem, y n χ T I (y), and we see that χ T (y) det(iy (T I)) det(i(y+)+t ) χ T (y+) Let x y+ Then χ T (x) (y) n (x ) n Alternate solution: By definition, T I is nilpotent Then by the same problem cited in the previous part of this problem, there is a basis V such that the matrix 0 of T I is of the form A But then the matrix of T is A A +I 0 x 0 Then χ 0 x 0 T (x) det(xi A) det x x 0 x det (x )n x (d) If T is a rotation of the plane R, show that χ T (x) x ( cos θ)x+, and show that the roots of this are e ±iθ (which by definition is cos θ ± i sin θ) Bonus: Conclude 9

( ) cos θ sin θ that these are the complex eigenvalues of the rotation matrix A, sin θ cos θ and compute eigenvectors with these eigenvalues cos(θ) sin(θ) The rotation matrix on R, A, rotates through an angle of θ sin(θ) cos(θ) degrees We see that the characteristic polynomial χ A (x) det(xi A) x cos(θ) sin(θ) sin(θ) x cos(θ) x x cos(θ) + cos (θ) + sin (θ) x x cos(θ) + This is 0 when, by the quadratic formula, x cos(θ)± 4 cos (θ) 4 cos(θ) ± sin (θ) cos(θ) ± i sin(θ) e ±iθ, by Euler s formula and some trig identities 0 Again let T be as in the previous problem Show that, if λ is an eigenvalue of T, and dim ker(t λi) m, ie, the eigenspace E λ : ker(t λi) {v V T v λv} has dimension m, then (x λ) m χ T (x) (that is, (x λ) m is a factor of the polynomial χ T (x)) Do not use the Jordan normal form theorem: instead, take a basis of E λ and extend it to a basis of V, and then write the matrix of T in this basis and take its determinant You may use the fact that ( we observed ) earlier in the semester: the determinant of a block A B upper-triangular matrix is det(a) det(d) (see Chapter, Miscellaneous problem 0 D for a generalization of this identity to some non-upper triangular matrices!) Let λ be some eigenvalue of T The eigenspace has dimension m, as given, and so a basis of this eigenspace can be written as B λ {v, v,, v m }, where v i V But dim(v ) n m, so this may not span V, and we must extend it, adding w V This can be written as B V {v, v,, v m, w m+,, w n }, using w to differentiate the newly added vectors Now we will express T as a matrix with respect to B v By definition, T (v k ) λv k 0v + + λv k + + 0w n, where only the v k entry has coefficient λ, and all others have a coefficient 0 Then with respect to B v, the vector will appear as follows: λe λe m C, where C is some n (n m) matrix Since there are only m of the first type, we can separate this into a matrix λi m above 0, which is a block matrix of size (n m) m We can similarly break C into B above [ D, where ] B λim B is m (n m) and D is (n m) (n m) Thus, we can write it as, and 0 D ( ) λim B χ T (x) det xi 0 D (x λ)i m B 0 xi n m D m) det(xi n m D) (x λ) n det(xi n m D) by the referenced problem from an earlier homework Thus, by the definition of divides, (x λ) n χ T (x) 0