EE263 homework 8 solutions
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1 EE263 Prof S Boyd EE263 homework 8 solutions 37 FIR filter with smll feedbck Consider cscde of 00 one-smple delys: u z z y () Express this s liner dynmicl system x(t + ) = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t) (b) Wht re the eigenvlues of A? (c) Now we dd simple feedbck, with gin α = 0 5, to the system: u z z y Express this s liner dynmicl system α x(t + ) = A f x(t) + B f u(t), y(t) = C f x(t) + D f u(t) (d) Wht re the eigenvlues of A f? (e) How different is the impulse response of the system with feedbck (α = 0 5 ) nd without feedbck (α = 0)? Solution For the system without feedbck, we cn tke A = , B = 0 0, C = [ 0 0 ], D = 0 The eigenvlues of A re ll zero In fct, A hs only one Jordn block The impulse response of this system is very simple It s nothing more thn single unit impulse t t = 00 The system simply delys the input signl 00 smples
2 Now we dd feedbck For this system we cn tke A = 0 0, B = C = [ 0 0 ], D = 0 The only difference is the entry 0 5 in the upper righthnd corner Let s find the eigenvlues of A We cn use some determinnt formuls, or just look t the poles of the denomintor of the trnsfer function The trnsfer function is z z 00, so its poles re the zeros of z This gives poles z = 0 5/00 e 2πk/00, k = 0,,99 These re points uniformly spced, 36 prt, on circle of rdius 0 5/00 = 0893 In prticulr, the eigenvlues re distinct, so A is digonlizble Note tht the eigenvlues of this A, compred to the cse without feedbck, re drmticlly different Note lso tht smll chnge to the mtrix (chging one entry from 0 to 0 5 ) hs huge chnge on the eigenvlues It s very esy to find the impulse response of the system with feedbck If we put in unit pulse t t = 0, unit pulse comes out t t = 00 Then t t = 200, nother pulse comes out with mplitude 0 5 At t = 300, we get nother one, with mplitude 0 0 And so on In other words, we get very fint echo, with period of 00 The first echo is brely detectble, nd the second nd subsequent ones re not present, for ny prcticl purpose Thus, the feedbck hs essentilly no effect on the system This impulse response is not wht you might hve predicted from the eigenvlues While the mgnitude of the eigenvlues mkes sense, most of them re complex, so we d expect to see some strong oscilltions But the impulse response is much simpler thn you might hve guessed We ve seen the feedbck hs essentilly no effect on the impulse response of the system, but it cuses n extreme chnge in eigenvlues Why? There re lots of nswers, but the most importnt messge is tht the poles cn give you good qulittive feel for the impulse response, when there s just hndful of them But when there re mny of them, it need not The extreme sensitivity of the eigenvlues to the chnge in the one entry is due to the Jordn form of A When mtrix is digonlizble, the eigenvlues re differentible functions of the entries But when the mtrix hs Jordn blocks, the eigenvlues cn chnge drmticlly with smll chnges in the mtrix 2
3 42 Norm expressions for qudrtic forms Let f(x) = x T Ax (with A = A T R n n ) be qudrtic form () Show tht f is positive semidefinite (ie, A 0) if nd only if it cn be expressed s f(x) = Fx 2 for some mtrix F R k n Explin how to find such n F (when A 0) Wht is the size of the smllest such F (ie, how smll cn k be)? (b) Show tht f cn be expressed s difference of squred norms, in the form f(x) = Fx 2 Gx 2, for some pproprite mtrices F nd G How smll cn the sizes of F nd G be? Solution: () We know tht the norm expression f(x) = Fx 2 is positive semidefinite qudrtic form simply becuse f(x) 0 for ll x nd f(x) = x T Ax with A = F T F 0 In this problem we will show the converse, ie, ny positive semidefinite qudrtic form f(x) = x T Ax cn be written s norm expression f(x) = Fx 2 Suppose the eigenvlue decomposition of A 0 is QΛQ T, with Q T Q = I nd Λ = dig(λ,,λ n ) where λ i re the eigenvlues of A Since λ i 0 (becuse A 0) then Λ /2 = dig( λ,, λ n ) is rel mtrix Let F = Λ /2 Q T R n n Then we hve Fx 2 = x T F T Fx = QΛ /2 Λ /2 Q T = x T Ax = f(x) To get smllest F suppose tht Rnk(A) = r Therefore, A R n n hs exctly r nonzero eigenvlues λ,,λ r Suppose Λ + = dig(λ,,λ r ) Hence, the eigenvlue decomposition of A cn be written s A = [ Q Q 2 ] [ Λ + 0 r (n r) 0 (n r) r 0 (n r) (n r) ] [ Q T nd s result A = Q Λ + Q T where Q R n r Now we cn tke F = Λ /2 + Q T R r n Therefore, k cn be s smll s r, ie, k = Rnk(r) Note tht k cnnot be ny smller thn Rnk(A) becuse A = F T F implies tht Rnk(A) k (b) In generl, qudrtic form need not to be positive semidefinite In this problem we show tht ny qudrtic form cn be decomposed into its positive nd negtive prts In other words, we cn write f(x) s the difference of two norm expressions, ie, f(x) = Fx 2 Gx 2 Suppose A hs n positive eigenvlues λ,,λ n, n 2 negtive eigenvlues λ n +,, λ n +n 2, nd therefore n n n 2 zero eigenvlues Let Λ + = dig(λ,,λ n ), Λ = dig( λ n +,, λ n +n 2 ) The eigenvlue decomposition of A cn be written s A = [ ] Q Q 2 Q 3 Q T 2 Λ + 0 n n 2 0 n (n n n 2 ) 0 n2 n Λ 0 n2 (n n n 2 ) 0 (n n n 2 ) n 0 (n n n 2 ) n 2 0 (n n n 2 ) (n n n 2 ) 3 ] Q T Q T 2 Q T 3
4 so A = Q Λ + Q T Q T 2 Λ Q 2 Now simply tke F = Λ /2 + Q T R n n nd G = Λ /2 Q T 2 R n2 n It is esy to verify tht A = F T F G T G nd therefore x T Ax = Fx 2 Gx 2 In fct, this method gives the smllest sizes for F nd G 43 Congruences nd qudrtic forms Suppose A = A T R n n () Let Z R n p be ny mtrix Show tht Z T AZ 0 if A 0 (b) Suppose tht T R n n is invertible Show tht T T AT 0 if nd only if A 0 When T is invertible, TAT T is clled congruence of A nd TAT T nd A re sid to be congruent This problem shows tht congruences preserve positive semidefiniteness Solution: () By definition, ll we hve to show is tht x T (Z T AZ)x 0 for ll x R p But x T (Z T AZ)x = (Zx) T A(Zx) nd by considering Zx s n element in R n, since A 0 we hve (Zx) T A(Zx) 0 nd we re done (b) The if prt ws shown in problem (0) (simply tke Z = T) For the only if prt we hve to show tht T T AT 0 implies A 0 By definition, it suffices to prove tht x T Ax 0 for ll x R n Suppose tht x is n rbitrry element in R n Since T is invertible, T exists nd let y = T x R n Using the fct tht T T AT 0 we hve y T (T T AT)y 0 But y T (T T AT)y = (T x) T (T T AT)(T x) = x T T T (T T AT)T x = x T Ax, nd therefore y T (T T AT)y 0 implies x T Ax 0 nd we re done 44 Positive semidefinite (PSD) mtrices () Show tht if A nd B re PSD nd α R, α 0, then so re αa nd A + B (b) Show tht ny (symmetric) submtrix of PSD mtrix is PSD (To form symmetric submtrix, choose ny subset of {,,n} nd then throw wy ll other columns nd rows) (c) Show tht if A 0, A ii 0 (d) Show tht if A 0, A ij A ii A jj In prticulr, if A ii = 0, then the entire ith row nd column of A re zero Solution: () To show tht αa 0 we verify tht x T (αa)x 0 for ll x But x T (αa)x = α(x T Ax) nd since x T Ax 0 (A 0) nd α 0, we immeditely get x T (αa)x 0 Agin, to show tht A + B 0 we show tht x T (A + B)x 0 for ll x This is esy becuse x T (A + B)x = x T Ax + x T Bx nd A,B 0 imply tht x T Ax,x T Bx 0 nd therefore x T (A + B)x 0 4
5 (b) Suppose tht A = A T 0 Any symmetric submtrix of A cn be written s Z T AZ for some suitble mtrix Z For exmple, if A R 3 3 nd we wnt to pick the submtrix formed by the first nd third columns nd rows we simply tke Z = so tht Z T AZ = [ ] A A 2 A 3 A 2 A 22 A 23 A 3 A 23 A = [ A A 3 A 3 A 33 ] The ide here is to pick the columns of Z s the unit vectors corresponding to the column/row numbers we wnt to keep In this exmple, we wnted to keep the first nd third columns/rows so we took Z = [e e 3 ] In generl, consider the m m symmetric submtrix of A which consists of elements of A tht re only on the columns nd rows i,,i m of A Then it is esy to verify tht (submtrix formed from columns/rows i,,i m ) = Z T AZ, Z = [e i e im ], where e ij is the i j th unit vector in R n Using the result of problem (0), A 0 implies tht Z T AZ 0 nd therefore ny symmetric submtrix of A is lso positive semidefinite (c) This is esy We cn simply use the result of the previous prt (A ii R is symmetric submtrix of A), or more directly, use the fct tht A 0 implies e T i Ae i 0 nd note tht e T i Ae i is nothing but A ii (d) Choose ny 2 2 symmetric submtrix of A, sy [ Aii A Ã = ij A ij According to problem (0b) this (symmetric) submtrix is positive semidefinite nd therefore its eigenvlues re nonnegtive Hence, the determinnt of the submtrix (which is equl to the product of the eigenvlues) is lso nonnegtive In other words [ ] detã = Aii A ij = A ii A jj A 2 ij 0 A ij A jj nd immeditely we get A 2 ij A ii A jj or A ij A ii A jj In prticulr, if A ii = 0 then A ij 0 or A ij = 0 (for ny j) nd the entire ith row (nd hence ith column since A is symmetric) should be zero A jj ] 5
6 46 Grm mtrices Given functions f i : [,b] R, i =,,n, the Grm mtrix G R n n ssocited with them is defined by () Show tht G = G T 0 G ij = f i (t)f j (t) dt (b) Show tht G is singulr if nd only if the functions f,,f n re linerly dependent Solution: () First of ll it is obvious tht G = G T becuse G ij = f i (t)f j (t) dt = f j (t)f i (t) dt = G ji Define the vector function f : [,b] R n s f(t) = [f (t) f 2 (t) f n (t) ] T Hence since f(t)f(t) T = = G = f (t) f 2 (t) f n (t) f(t)f(t) T dt [ f (t) f 2 (t) f n (t) ] f (t)f (t) f (t)f 2 (t) f 2 (t)f (t) f 2 (t)f 2 (t) f n (t)f (t) f n (t)f 2 (t) f (t)f n (t) f 2 (t)f n (t) f n (t)f n (t) Now to show tht G 0, we consider ny x = [x x n ] T R n nd verify tht x T Gx 0 We hve x T Gx = x T ( f(t)f(t) T dt)x = x T f(t)f(t) T x dt = (x T f(t)) 2 dt, nd therefore x T Gx 0 since it is the re under the nonnegtive function (x T f(t)) 2 from to b (b) Note tht if G 0 is nonsingulr it does not hve ny zero eigenvlues (in fct ll eigenvlues should be strictly positive) nd therefore x T Gx > 0 for ll x or G > 0 We show tht if f,,f n re linerly independent then G > 0 nd vice vers In the previous prt we showed tht x T Gx = (x T f(t)) 2 dt = 6 (x f (t) + + x n f n (t)) 2 dt
7 Therefore, x T Gx > 0 for ll x s long s no liner combintion of the functions f,,f n is zero, ie, x f (t) + + x n f n (t) 0 In other words, G > 0 if nd only if the functions f,,f n re linerly independent nd we re done 48 Express n (x i+ x i ) 2 in the form x T Px with P = P T Is P 0? P > 0? Solution: Let x = [x x n ] T R n We hve n (x i+ x i ) 2 = = = = n n n n (e T i+x e T i x) 2 ( (e T i+ e T i )x ) 2 ( (e T i+ e T i )x ) T ( (e T i+ e T i )x ) x T (e i+ e i )(e T i+ e T i )x ( n ) = x T (e i+ e i )(e T i+ e T i ) x ( n ) = x T (e i+ e T i+ e i+ e T i e i e T i+ + e i e T i ) x = x T Px Let E(i,j) := e i e T j R n n (ie, mtrix with ll zero elements except for t the (i,j)th entry) The mtrix P R n n cn be written in terms of E(i,j) s follows: P = = = n n n (e i+ e T i+ e i+ e T i e i e T i+ + e i e T i ) (E(i +,i + ) E(i +,i) E(i,i + ) + E(i,i)) E(i +,i + ) n E(i +,i) n E(i,i + ) + n E(i,i) But n E(i +,i + ) = 0, n E(i,i) = 0 7
8 nd n E(i +,i) = , n E(i,i + ) = Therefore P = is tridigonl mtrix Clerly, since the qudrtic form is nothing but the sum of squres ( n (x i+ x i ) 2 ) it is nonnegtive for ny x nd therefore P 0 However, P is not strictly greter thn zero becuse for exmple x = [ ] T 0 gives x T Px = n ( ) 2 = 0 49 Suppose A nd B re symmetric mtrices tht yield the sme qudrtic form, ie, x T Ax = x T Bx for ll x Show tht A = B Hint: first try x = e i (the ith unit vector) to conclude tht the entries of A nd B on the digonl re the sme; then try x = e i + e j Solution: With x = e i, x T Ax = x T Bx gives A ii = B ii, i =, 2,,n With x = e i + e j, x T Ax = x T Bx gives nd therefore e T i Ae i + e T i Ae j + e T j Ae i + e T j Ae j = e T i Be i + e T i Be j + e T j Be i + e T j Be j A ii + A ij + A ji + A jj = B ii + B ij + B ji + B jj, i,j =, 2,,n, so tht A ij +A ji = B ij +B ji This, long with A ii = B ii mens tht A+A T = B +B T Finlly, using the fct tht A = A T nd B = B T, we conclude tht A = B 4 Suppose tht A R n n Show tht A < implies I + A is invertible Solution: We show tht A < implies I +A invertible by proving the contrpositive sttement: I + A singulr implies A The proof is very simple this wy For I + A singulr, there exists nonzero vector v such tht (I + A)v = 0, or Av = v For this v, we hve Av = Now, Av Ax mx v v x 0 = A Thus, A, which proves the x contrpositive 8
9 43 Eigenvlues nd singulr vlues of symmetric mtrix Let λ,,λ n be the eigenvlues, nd let σ,,σ n be the singulr vlues of mtrix A R n n, which stisfies A = A T (The singulr vlues re bsed on the full SVD: If Rnk(A) < n, then some of the singulr vlues re zero) You cn ssume the eigenvlues (nd of course singulr vlues) re sorted, ie, λ λ n nd σ σ n How re the eigenvlues nd singulr vlues relted? Solution: Since A is symmetric it cn be digonlized using n orthogonl mtrix Q s A = Q λ λ n Q T where λ,,λ n re the eigenvlues of A Suppose tht the columns of Q re ordered such tht λ λ 2 λ n Thus sgnλ A = Q sgnλ n λ Q T λ n Now we define sgnλ U = Q sgnλ n λ, Σ =, V = Q λ n Clerly, U is n orthogonl mtrix becuse UU T = QQ T = I Now A = UΣV T is SVD of A, nd we conclude tht σ i = λ i 42 Two representtions of n ellipsoid In the lectures, we sw two different wys of representing n ellipsoid, centered t 0, with non-zero volume The first uses qudrtic form: E = { x x T Sx }, with S T = S > 0 The second is s the imge of unit bll under liner mpping: with det(a) 0 E 2 = { y y = Ax, x }, () Given S, explin how to find n A so tht E = E 2 (b) Given A, explin how to find n S so tht E = E 2 (c) Wht bout uniqueness? Given S, explin how to find ll A tht yield E = E 2 Given A, explin how to find ll S tht yield E = E 2 9
10 Solution: First we will show tht E 2 = { y y T (AA T ) y } () E 2 = {y y = Ax, x } = { y A y = x, x } since A is invertible squre mtrix = { y A y } = { y y T A T A y } = { y y T (AA T ) y } Since S is symmetric positive definite, the eigenvlues of S re ll positive nd we cn choose n orthonorml eigenvectors So S = QΛQ T where Λ = dig(λ,,λ n ) > 0 nd Q is orthogonl Let Λ 2 = dig( λ,, λ n ) If we let A = Q(Λ 2) = QΛ 2, (AA T ) = (QΛ 2 Λ 2 Q T ) = (QΛ Q T ) = QΛQ T = S Therefore, by () A = QΛ 2 yields E = E 2 By (), S = (AA T ) yields E = E 2 Uniqueness: We show tht E S = E T S = T (2) where E S = {x x T Sx }, E T = {x x T Tx }, S T = S > 0 nd T T = T > 0 It is cler tht if S = T, then E S = E T Now we show tht E S = E T x T Sx = x T Tx, x R n Without loss of generlity let s ssume x 0 R n such tht x T 0 Sx 0 > x T 0 Tx 0 = α 0 If we let x = x 0 / α, then x T Tx =, but x T Sx >, thus x E T but x / E S, nd therefore E S E T Finlly, E S = E T x T Sx = x T Tx, x R n S = T by the uniqueness of the symmetric prt in qudrtic form Hence S is unique Given S, find ll A tht yield E = E 2 The nswer is A = QΛ 2 V T where V R n n is ny orthogonl mtrix nd S T = S = QΛQ T > 0 where Q is orthogonl nd Λ = dig(λ,,λ n ) > 0 Let the singulr vlue decomposition of A be A = UΣV T where U,V R n n re orthogonl nd Σ = dig(σ,,σ n ) > 0 (since det(a) 0) By (), E 2 = { y y T (AA T ) y } = { y y T (UΣ 2 U T ) y } = { y y T UΣ 2 U T y } 0
11 Thus, if E = E 2, then S = UΣ 2 U T by (2) Therefore U = Q nd Σ = Λ 2, nd V cn be ny orthogonl mtrix You cn lso see why A s re different only by right-side multipliction by n orthogonl mtrix by the following rgument By (2) nd (), AA T = S Let A = [ ã ã 2 ã n ] T Then we hve, ã i 2 = (S ) ii, ã T i ã j = (S ) ij, nd cos θ ij = (S ) ij (S ) ii (S ) jj This mens tht the row vectors of ny A stisfying AA T = S hve the sme length nd the sme ngle between ny two of them So the rows of A cn vry only by the ppliction of n identicl rottion or reflection to ll of them These re the trnsformtions preserving length nd ngle, nd correspond to orthogonl mtrices Since we re considering row vectors, the orthogonl mtrix should be multiplied on the right 433 Uncovering hidden liner explntion Consider set of vectors y,,y N R n, which might represent collection of mesurements or other dt Suppose we hve y i Ax i + b, i =,,N, where A R n m, x i R m, nd b R n, with m < n (Our min interest is in the cse when N is much lrger thn n, nd m is smller thn n) Then we sy tht y = Ax+b is liner explntion of the dt y We refer to x s the vector of fctors or underlying cuses of the dt y For exmple, suppose N = 500, n = 30, nd m = 5 In this cse we hve 500 vectors; ech vector y i consists of 30 sclr mesurements or dt points But these 30-dimensionl dt points cn be explined by much smller set of 5 fctors (the components of x i ) This problem is bout uncovering, or discovering, liner explntion of set of dt, given only the dt In other words, we re given y,,y N, nd the gol is to find m, A, b, nd x,,x N so tht y i Ax i + b To judge the ccurcy of proposed explntion, we ll use the RMS explntion error, ie, ( ) /2 J = y i Ax i b 2 N One rther simple liner explntion of the dt is obtined with x i = y i, A = I, nd b = 0 In other words, the dt explins itself! In this cse, of course, we hve y i = Ax i + b, so the RMS explntion error is zero But this is not wht we re fter To be useful explntion, we wnt to hve m substntilly smller thn n, ie,
12 substntilly fewer fctors thn the dimension of the originl dt (nd for this smller dimension, we ll ccept nonzero, but hopefully smll, vlue of J) Generlly, we wnt m, the number of fctors in the explntion, to be s smll s possible, subject to the constrint tht J is not too lrge Even if we fix the number of fctors s m, liner explntion of set of dt is not unique Suppose A, b, nd x,,x N is liner explntion of our dt, with x i R m Then we cn multiply the mtrix A by two (sy), nd divide ech vector x i by two, nd we hve nother liner explntion of the originl dt More generlly, let F R m m be invertible, nd g R m Then we hve Thus, y i Ax i + b = (AF )(Fx i + g) + (b AF g) Ã = AF, b = b AF g, x = Fx + g,, x N = Fx N + g is nother eqully good liner explntion of the dt In other words, we cn pply ny ffine (ie, liner plus constnt) mpping to the underlying fctors x i, nd generte nother eqully good explntion of the originl dt by ppropritely djusting A nd b To stndrdize or normlize the liner explntion, it is usully ssumed tht N x i = 0, N x i x T i = I In other words, the underlying fctors hve n verge vlue zero, nd unit smple covrince (You don t need to know wht covrince is it s just definition here) Finlly, the problem () Explin clerly how you would find hidden liner explntion for set of dt y,,y N Be sure to sy how you find m, the dimension of the underlying cuses, the mtrix A, the vector b, nd the vectors x,,x N Explin clerly why the vectors x,,x N hve the required properties (b) Crry out your method on the dt in the file linexp_dtm vilble on the course web site The file gives the mtrix Y = [y y N ] Give your finl A, b, nd x,,x N, nd verify tht y i Ax i b by clculting the norm of the error vector, y i Ax i b, for i =,,N Sort these norms in descending order nd plot them (This gives good picture of the distribution of explntion errors) By explicit computtion verify tht the vectors x,,x N obtined, hve the required properties Solution () We hve to find b nd x,,x N tht minimize J = NJ 2 = (y i Ax i b) T (y i Ax i b) 2
13 Tking the grdient with respect to b nd setting it to zero gives, b ( J) = 2(y i Ax i b)( ) = 0 Since we wnt N x i = 0, we get b = N y i Let X be the mtrix [x x N ] R m N Let z i = y i b, i =,,N, nd Z = [ z z N ] R n N Note tht Z is known from the dt fter b is clculted s shown bove The mtrix (AX) R n N nd hs rnk t most rnk m Then n J = z i Ax i 2 = (Z kj (AX) kj ) 2 = Z AX 2 F k= j= Thus minimizing J is minimizing the Frobenius norm of the mtrix (Z AX) where AX is t most rnk m Let the SVD of Z be Z = UΣV T, where U R n r, Σ R r r, V R N r nd r is the rnk of Z The choice of m depends on the singulr vlues σ,,σ r obtined for the prticulr dt A good choice of m would be when there is significnt jump in the singulr vlues, ie, σ m σ m+ ; or when the singulr vlue becomes smll enough (σ m+ is negligible) Thus we pick vlue for m Then the m rnk pproximtion to Z is m (AX) = σ i u i vi T = U m Σ m Vm, T where U m R n m, Σ m R m m, V m R N m We pick A = N U m Σ m, nd x i s N times the ith column of V T m Then N x i x T i = N ( NV m ) T ( NV m ) = I In order to show tht N N x i = 0, consider Z = z i = (y i b) = 0, where is the vector of ones of size N The vector is in the nullspce of Z which we cn write s UΣV T = 0 The mtrix UΣ is full rnk, therefore V T = 0 Hence V T m = 0 s V m re the first m columns of the mtrix V This mens N V T m = N [x x N ] = N 3 x i = 0
14 Thus we hve found b = N y i, A = N U m Σ m, [x x N ] = NV T m, with the required properties (b) The following Mtlb code implements solution method described in the prt () We observe tht there re 3 significnt singulr vlues, nd therefore we tke m = 3 linexp_dt; [n,n] = size(y); b = sum(y ) /N; for i = :N Z(:, i) = Y(:, i) - b; end [U S V] = svd(z); A = /sqrt(n)*u(:, :3)*S(:3, :3); X = sqrt(n)*v(:, :3) ; error = Z - A*X; for :N errnorm(i) = norm(error(:, i)); end plot(sort(errnorm,2, descend )); xlbel( index ); ylbel( errors ); Here is explicit check shows tht X stisfies given properties >> sum(x,2)/n = 0e-5 * >> X*X /N = Plot of the sorted error norm is shown below: 4
15 y î Ax î b sorted index î Solution to dditionl exercise Some simple mtrix inequlity counter-exmples () Find (squre) mtrix A, which hs ll eigenvlues rel nd positive, but there is vector x for which x T Ax < 0 (Give A nd x, nd the eigenvlues of A) Morl: You cnnot use positivity of the eigenvlues of A s test for whether x T Ax 0 holds for ll x Wht is the correct wy to check whether x T Ax 0 holds for ll x? (You re llowed to find eigenvlues in this process) (b) Find symmetric mtrices A nd B for which neither A B nor B A holds Of course, we d like the simplest exmples in ech cse Solution () A = [ 3 0 ] (b) A = dig(, 2), B = dig(2, ) 5
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