MATH 6 Homework solutions Problems from Dr Kazdan s collection 7 (a) Compute the dimension of the intersection of the following two planes in R 3 : x y z, 3x 3y z (b) A map L: R 3 R is defined by the matrix L : (kernel) of L 3 3 Find the nullspace (a) Row reduction: 3 3 9 4 4 9 so the solution of these equations is x y z c 9 4 9 So the dimension of the intersection of the planes is (b) The kernel of L is the one-dimensional set of vectors given above Let U and V both be two-dimensional subspaces of R 5, and let W U X V Find all possible values for the dimension of W Since the intersection of the two subspaces is contained in U, it can have dimension at most And the dimension can be, or To see this we can give examples: trx, x,,, su X tr,,, x 4, x 5 su has dimension trx, x,,, su X trx,,,, x 5 su has dimension trx, x,,, su X trx, x,,, su has dimensino Let A be an n n matrix of real or complex numbers Which of the following statements are equivalent to: the matrix A is invertible? (a) The columns of A are linearly independent (b) The columns of A span R n (or C n ) (c) The rows of A are linearly independent (d) The kernel of A is tu
(e) The only solution of the homogeneous equations Ax is x (f) The linear transformation T A : R n R n (or T A : C n C n ) defined by A is one-to-one (g) The linear transformation T A : R n R n (or T A : C n C n ) defined by A is onto (h) The rank of A is n (i) The adjoint (ie, transpose or, in the case of complex A the conjugate transpose), A, is invertible (j) det A Answer: All of them Since A is n-by-n, it determines a mapping from R n to R n The image of A is the span of the columns of A, so if the n columns of A are linearly independent, then they span R n and so A is surjective, hence invertible (and vice versa) A is also injective, since there s no nonzero linear combination of the columns of A that yields zero The determinant of A is nonzero since the determinant of A is the reciprocal of the determinant of A 8 Let A and B be n n matrices with AB Give a proof or counterexample for each of the following (a) Either A or B (or both) (b) BA (c) If det A 3, then B (d) If B is invertible then A (e) There is a vector v such that BAv (a) False: Let A and B (b) The example in part (a) also shows this is false (c) True: If det A 3 then A is invertible and then A AB A which gives B (d) True: If B is invertible then ABB B so A (e) This is true, since detpbaq pdet Bqpdet Aq pdet Aqpdet Bq detpabq det, so BA is not invertible so its kernel contains more than the zero vector 5 Proof or counterexample In these L is a linear map from R to R, so its representation will be as a matrix (a) If L is invertible, then L is also invertible (b) If Lv 5v for all vectors v, then L w p{5qw for all vectors w
3 (c) If L is a rotation of the plane by 45 degrees counterclockwise, then L is a rotation by 45 degrees clockwise (d) If L is a rotation of the plane by 45 degrees counterclockwise, then L is a rotation by 35 degrees counterclockwise (e) The zero map (v for all vectors v) is invertible (f) The identity map (Iv v for all vectors v) is invertible (g) If L is invertible, then L (h) If Lv for some non-zero vector v, then L is not invertible (i) The identity map (say from the plane to the plane) is the only linear map that is its own inverse: L L These are all one-liners (a) The inverse of L is L, so L is invertible, too (b) True since 5 p5vq v, multiplication by 5 is the inverse of multiplication by 5 (c) True obvious (d) True since 45 35 36 (e) False the map is clearly not one-to-one (f) True since I I (g) True since Lpq, we ll have L (h) True since then L is not one-to-one (i) False I is also its own inverse 36 Let V be an n-dimensional vector space and T : V V a linear transformation such that the image and kernel of T are identical (a) Prove that n is even (b) Give an example of such a linear transformation T (a) Since dim V dim ker T dim V dim ker T is even dim im T and dim ker T dim im T (since im T ker T ) we have (b) Suppose dim V and let te, e u be a basis for V Define T via T pe q and T pe q e
4 Then both the image of T and the kernel of T are spanned by e The matrix of T is 37 Let V R be a linear subspace of dimension 4 and consider the family A of all linear maps L: R R 9 each of whose nullspace contains V Show that A is a linear space and compute its dimension We can choose the basis of R so that the first four vectors in it are a basis for V So any linear map in A must map the vectors r,,, s, r,,, s, r,,, s and r,,, s to zero That means the first four columns of the matrix of the linear map contain only zeros The rest of this 9-by- matrix is arbitrary So the dimension of the space A is 9 p 4q 9 7 63 4 Say A P Mpn, Fq has rank k Define L : tb P Mpn, Fq BA u and R : tc P Mpn, Fq AC u Show that L and R are linear spaces and compute their dimensions Since the n-by-n matrix A has rank k, then dim im A k and dim ker A n k Any matrix B in L has the property that the kernel of B contains the image of A so if we choose the right basis, the first k columns of B are zero and the rest of B is arbitrary (as in problem 37) So the dimension of L is npn kq n nk Similarly, any matrix C in R has the property that the image of C is contained in the kernel of A If we choose the right basis (so the first n k basis vectors are a basis for the kernel of A, we have that the first n k columns of A are zero and the last k columns of A are independent so the last k rows of C must be zero so that AC The dimension the space R of such C s is thus npn kq n nk 54 Every real upper triangular n n matrix pa ij q with a ii, i,,, n is invertible Proof or counterexample This is true, since the determinant of the matrix will be 55 Let L: V V be a linear map on a vector space V (a) Show that ker L ker L and, more generally, ker L k ker L k for all k (b) If ker L j ker L j for some integer j, show that ker L k ker L k for all k j Does your proof require that V is finite dimensional? (c) Let A be an n n matrix If A j for some integer j (perhaps j n), show that A n
5 (a) If v P ker L, then L v LpLvq Lpq so v P ker L This shows ker L ker L Likewise if v P ker L k then L k v LpL k vq Lpq so v P ker L k, which shows ker L k ker L k (b) Suppose v R ker L j but v P ker L j Then the vector Lpvq P ker L j but Lpvq R ker L j So we have found a vector in ker L j that is not in ker L j and so we cannot have ker L j ker L j We have thus shown that ker L j ker L j implies that ker L j ker L j This is the contrapositive of the statement we were trying to prove, so that statement is true as well, ie, ker L j ker L j implies that ker L j ker L j This proof does not require that the dimension of V is finite (c) If A j for some j, then dim ker A j n By part (b), we must have dim ker A dim ker A for all powers,, until the dimension reaches n Since the dimensions are whole numbers, the maximum n must be reached in n or fewer steps, so the dimension of the kernel will be n for A n (or possibly already for a lower power) A B Notation: Let M be an pn kq pn kq block matrix partitioned into the n n C D matrix A, the n k matrix B, the k n matrix C and the k k matrix D W X Let N be another matrix with the same shape as M Y Z 76 Show that the naive matrix multiplication AW BY AX BZ MN CW DY CX DZ is correct If we think about the elements of M, then Likewise If P MN, then p ij m ij a ij if i n and j n m ij b i,jn if i n and n j n k m ij c in,j if n i n k and j n m ij d in,jn if n i n k and n j n k n ij w ij if i n and j n n ij x i,jn if i n and n j n k n ij y in,j if n i n k and j n n ij z in,jn if n i n k and n j n k k, and we ll have to break the sum up into n p ij n k ln
6 so that we can take into account the partitioning of M and N For instance, if i n and j n then p ij Next, if i n and n p ij n n k ln j n k then k ln a il w lj a il x l,jn ķ ķ b il y lj paw q ij pby q ij b il z l,jn paxq i,jn pbzq ijn For the lower left corner of P MN, or n i n k and j n we have p ij n k ln c in,l w lj ķ d in,l y lj pcw q in,j pdy q in,j Finally for the lower right corner of P, where n i n k and n j n k we have p ij n k ln Summarizing, we have c in,l x l,jn ķ d in,l z l,jn pcxq in,jn pdzq in,jn p ij paw q ij pby q ij if i n and j n p ij paxq i,jn pbzq ijn if i n and n j n k p ij pcw q in,j pdy q in,j if n i n k and j n p ij pcxq in,jn pdzq in,jn if n i n k and n j n k This is the block form given in the problem P MN AW BY AX BZ CW DY CX DZ 77 [Inverses] (a) Show that matrices of the above form but with C are a sub-ring (b) If C, show that M is invertible if and only if both A and D are invertible and find a formula for M involving A, etc (c) More generally, if A is invertible, show that M is invertible if and only if the matrix H : D CA B is invertible in which case A A BH CA A BH M H CA H (d) Similarly, if D is invertible, show that M is invertible if and only if the matrix K : A BD C is invertible in which case K K BD M D CK D D CK BD
7 (e) For which values of a, b, and c is the following matrix invertible? What is the inverse? a b b b b c a c a S : c a c a A B (f) Let the square matrix M have the block form M C so D If B and C are square, show that M is invertible if and only if both B andc are invertible, and find an explicit C formula for M [Answer: M : B B AC ] (a) Subring means that this set of matrices is closed under addition and matrix multiplication For closure under addition, we have A B W X A W B X, D Z D Z which is another matrix of this form Similarly, A B W X AW AX BZ D Z DZ, so the set of pn kq-by-pn kq matrices with a k-by-n block of zeros in the lower left corner is a subring of the set of all pn kq-by-pn kq matrices (b) Since the identity element has the form In n k k n I k we need (using the multiplication in part (a)) AW I n and DZ I k So A and D have to be invertible and we must have W A and Z D Then we must choose X to satisfy AX BZ AX BD, and so X A BD We conclude A B A A BD D A B if A and D are invertible, and D D is not invertible otherwise
8 (c) Now we re trying to invert A B M C D under the assumption that A is invertible We start from A B W X AW BY AX BZ C D Y Z CW DY CX DZ, and see that we have to solve AW BY I n AX BZ n k CW CX DY k n DZ I k Let s look at the first and third of these equations first, since they involve the matrix A which is known to be invertible and only the two unknown matrices W and Y We re going to do a kind of row reduction on this system of two equations If we multiply the first equation by A on the left, and then subtract C times this new first equation from the second we get: AW BY I n W A BY A W A BY A CW DY k n CW DY k n DY CA BY CA The second of these equations can be rewritten pd CA BqY CA so if the k-by-k matrix H D CA B is invertible we can solve for Y as Y H CA If H is not invertible then there may be no solution for Y or infinitely many solutions either way, the big matrix M will not be invertible Now we have to complete our solution for the inverse of M under the assumption that H is invertible Since we know that Y H CA, we can substitute this back into the first equation and solve for W : W A BpH CA q A ùñ W A A BH CA We treat the other two equations for X and Z similarly: AX BZ n k X A BZ n k X A BZ n k CX DZ I k CX DZ I k DZ CA BZ I k Once again, the second of these can be written pd CA BqZ I k and there s H again Therefore Z H We put this back into the first equation and get X A BH n k ùñ X A BH
9 We conclude that, if A and H D CA B are invertible, then A B A A BH CA A BH C D H CA H (d) This is just like part (c), except this time we assume that D is invertible We proceed as before and first consider the equations for W and Y (working on the second equation first): AW BY I n CW DY k n AW BY I n D CW Y k n AW BD CW I n D CW Y k n The first of these equations is pa BD CqW I n, or KW I n where K A BD C We conclude that W K, and use this to solve the second equation: D CK Y k n ùñ Y D CK We treat the other two equations for X and Z similarly: AX BZ n k AX BZ n k CX DZ I k D CX Z D AX BD CX BD D CX Z D Once again, the first of these can be written pa BD CqX BD and there s K again Therefore X K BD We put this back into the second equation and get D CK BD Z D ùñ Z D D CK BD We conclude that, if D and K A BD C are invertible, then A B K K BD C D D CK D D CK BD Something that is worthwhile to do is to check that the answers to (b) and (c) agree if A and D are both invertible (e) We can partition S as follows: S a bv T cv ai n where v is an pnq-by- column vector containing all s It s clear that the matrix is not invertible if a, since it will have n identical rows So we assume a and then we can use parts (c) and (d) For instance, the matrix H from part (c) will be the pn q-by-pn q matrix H ai n a vvt Now H : R n R n, and if x is a vector perpendicular to v then Hx ai n x a vvt x ax
since v T x v x And Hv ai n v a vvt v av pn q a v a pn q v a So v will be in the kernel of H (and H will not be invertible) if a pn q Likewise the matrix K from part (d) will be the -by- matrix K a a vt v a pn q a So K will be invertible (ie, nonzero) if a pn q a, ie, if a pn q We conclude that S is invertible if and only if a and a pn q Now we set about finding S Since both (c) and (d) apply, we ll use the easier one It will be easier to use the formula from part (d), since K is a -by- matrix which is easy to invert Let λ a pn q, and assume that a and λ For the purposes of applying the formula from part (d), we note that A a, B bv T, C cv and D ai n Also K λ{a so that K a{λ (the matrices A, K and K are -by- matrices so we treat them as numbers) We also have D p{aqi n So the formula from part (d) gives S a λ c λ v a I n b λ vt aλ vvt a a pn q c a pn q v a I n b a pn q vt apa pn qq vvt Note that vv T is an pn q-by-pn q matrix all of whose entries are So S a a pn q c a pn q c a pn q c a pn q c a pn q b a pn q a pn q apa pn qq apa pn qq apa pn qq apa pn qq b a pn q apa pn qq a pn q apa pn qq apa pn qq apa pn qq b b a pn q a pn q apa pn qq apa pn qq apa pn qq apa pn qq a pn q apa pn qq apa pn qq a pn q apa pn qq apa pn qq (f) In this part we are assuming that k n and A We ll use the technique from part (b) since B W X BY BZ, C D Y Z CW DY CX DZ
we see that we need BY I BZ CW DY CX DZ I We see immediately that B must be invertible and so Y B and Z Then the last equation becomes CX I so C must also be invertible and X C Finally the third equation gives B CW DB or W C DB We conclude that C D is invertible if and only if both B and C are invertible and B C DB C C D B 8 (a) Find a cubic polynomial ppxq with the properties that ppq, ppq, pp3q, and pp4q 5 Is there more than one such polynomial? (b) Given any points px, y q, px, y q, px 3, y 3 q, px 4, y 4 q with the x i s distinct, show there is a unique cubic polynomial ppxq with the properties that ppx i q y i (a) If we write ppxq a a x a x Gauss elimination: 3 9 7 4 6 64 5 4 3 6 8 3 9 7 4 6 64 a 3 x 3 then we ll have to solve the linear system a a a a 3 5 3 9 7 4 6 64 4 4 3 6 4 4 6 8 4 3 We conclude that a 3, a 3, a 3 so a 5 3 and a So the unique cubic polynomial with the given properties turns out to be quadratic and it is ppxq 5 3 x 3 x (b) It would help if we could choose a basis for the space of cubic polynomials which is more adapted to our problem For example, if we could choose p pxq so that p px q, p px q,
p px 3 q, p px 4 q and then choose p pxq so that p px q, p px q, p px 3 q, p px 4 q etc, we could then set ppxq y p pxq y p pxq y 3 p 3 pxq y 4 p 4 pxq p q There is such a basis, since we know that the polynomial px x qpx x 3 qpx x 4 q is zero at x, x 3 and x 4 and has the value px x qpx x 3 qpx x 4 q for x x, we could choose Likewise we can choose p pxq p pxq px x qpx x 3 qpx x 4 q px x qpx x 3 qpx x 4 q px x qpx x 3 qpx x 4 q px x qpx x 3 qpx x 4 q p 3 pxq px x qpx x qpx x 4 q px 3 x qpx 3 x qpx 3 x 4 q and then formula p q above gives the solution to the problem p 4 pxq px x qpx x qpx x 3 q px 4 x qpx 4 x qpx 4 x 4 q January notes 3 Let S and T be transformations in LpV, V q (sometimes called endomorphisms of V ) If ST T S we say that S and T commute (duh) (a) If S and T commute, show that pst q n S n T n for n (b) If S and T are any invertible operators in LpV, V q, then pst q T S (c) If S and T are commutative, invertible operators, then S and T commute (a) We do this by induction It s certainly true for n that pst q S T ST So assume that pst q n S n T n and we ll prove this implies pst q n S n T n Well, pst q n pst q n pst q S n T n ST, so what we really have to prove is that T n S ST n for all n We do this by induction as well It s certainly true for n (since T I and I commutes with everything) and for n (since that is the statement that S and T commute, which is given) So now assume that T n S ST n Multiply on the left by T and get T n S T ST n ST T n ST n, and we re done (b) To show this, we ll verify that pst qpt S q I But this is clear since pst qpt S q SpT T qs SIS SS I (c) From (b), we have pst q T S and pt Sq S T But since ST T S it s true that pst q pt Sq So T S S T 4 Find the matrix P of the orthogonal projection of R onto the line ax by and show that P P Orthogonal projection onto the line ax vector rb, as And proj rb,as v by is the same as orthogonal projection onto the v rb, as rb, as rb, as rb, as
3 We can express this as So and P pa b q a b rb v abv, abv a v s b ab ab a P b b ab ab a b a b a b a ab ab a b a b ab a b a b ab a a b a b v v b pa b q abpa b q pa b q abpa b q a pa b q P 34 Solve the following systems of linear equations you ll have to use all three row operations sometimes and there will be interpretation issues! pbq 4x 3 3x 4 b x x 4x 3 x 4 b x x 8x 3 4x 4 b 3 Under what conditions on b, b, b 3 is there a solution? Use row reduction FIrst, swap the first two rows: 4 3 b 4 b 4 b 4 3 b 8 4 b 3 8 4 b 3 Next, subtract the first row from the third row: 4 b 4 3 b 4 b 4 3 b 8 4 b 3 4 3 b 3 b Next, multiply the second row by 4 : 4 b 4 3 b 4 3 b 3 b 4 b 3 4 4 b 4 3 b 3 b Next, subtract 4 times the second row from the third: 4 b 3 4 4 b 4 b 3 4 4 b 4 3 b 3 b b 3 b b The matrix is now in row-echelon form if b 3 b b, and there is a contradiction (the last equation is essentially ) if b 3 b b
4 So we conclude that this system of equations has solutions if and only if b 3 b b Since in this case the non-zero rows start in columns and 3, we find the particular solution by setting x and x 4 equal to zero, and conclude that a particular solution is x x x 3 x 4 b b 4 b To get solutions of the associated homogeneous system, which row-reduces to: 4 3 4, we first set x and x 4, and then set x and x 4 and get that the general solution of the homogeneous system is c c 3 4 We conclude that the original system has solutions if and only if b 3 b b and in this case the general solution is x b b x x 3 4 b c c 3 4 x 4 January notes 7 Let A be an n-by-n matrix, Prove that the following are equivalent: (a) A is invertible (b) The only solution of Ax is x (c) For any vector b, the equation Ax b has at most one solution (d) For any vector b, the equation Ax b has at least one solution (e) The columns of A are linearly independent (f) The columns of A span R n (g) The rows of A are linearly independent (h) The rows of A span R n (i) det A
5 We should be getting good at these: (a) implies (b) since if A is invertible, then A Ax A so x (b) implies (c) since the difference between two solutions of Ax b would be nonzero and satisfy Ax (c) implies (d) since the dimension of ker A being zero implies the dimension of im A is n, so all vectors in R n are in the image of A (d) implies (e) since the span of the columns is the image of A so the n columns must be linearly independent (e) implies (f) since the span of n linearly independent vectors is n-dimensional (f) implies (g) since the columns being independent implies A is invertible, and since A is square, the same n-by-n matrix A works on the left and on the right A A I gives linear combinations of the rows of A to produce a spanning set of row vectors so (g) and hence (h) are true If the determinant were zero, then the rows would be dependent, so (h) implies (i) Similarly, since the determinant is nonzero, A is invertible and (i) implies (a) 9 (a) Show that the equation of the line through the two points pa, bq and pc, dq is det x a y b c d (b) Show that the equation of the parabola through the three points pa, bq, pc, dq and pe, fq (with a, c and e all distinct) is x x y det a a b c c d e e f (a) The simplest way to do this is to write out the determinant: pad q pb dqx pc aqy This is clearly the equation of a line and if we put x a and y b we get pad q pba adq p abq so the point pa, bq is on the line, and likewise for the point pc, dq (b) We don t really have to write out the determinant You can see that if you expand the determinant along the first row, you ll get an equation that looks like c c x c x c 3 y, which is the equation of a parabola And the points pa, bq, pc, dq and pe, fq are all on the parabola since setting px, yq equal to any of them gives a determinant with two equal rows, so the equation will be satisfied