# Linear Algebra 1A - solutions of ex.4

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1 Liner Algebr A - solutions of ex.4 For ech of the following, nd the inverse mtrix (mtritz hofkhit if it exists - ( 6 6 A, B (, C 3, D, 4 4 ( E i, F (inverse over C for F. i Also, pick n invertible mtrix of these (not B. nd express it nd its inverse s product of elementry mtrices. -solution- Let us write the extended mtrix (A I nd perform elementry row opertions on it till we get mtrix of the form (L B, where L is the (cnonicl reduced echelon mtrix (mtritzt mdregot cnonit. Then, if L I, we conclude tht A is invertible with inverse equls to B, nd otherwise, A is not invertible. ( 6 4 ( R 6R ( ( 4 R R 6 R R / 3 ( ( R R /( 3 R R R ( Hence A is invertible, nd A * Remrk: it's n x mtrix, so we could rst prove the clim in q. here, nd use the formul to sy tht the mtrix is invertible nd to nd it's inverse. (Check tht you get the sme result! For B (, it's invertible, with inverse B (. For C we write similrly 3 4 R R R ; R 3 R R R 3 + R R ;

2 R R /( 3 7 R 3 R 3 7R ; R R 3R R 3 R 3 ( R R R 3; R 3 7 R R 3 9 Thus, C is invertible, with inverse C ; (One might check tht it's indeed the inverse of C by clculting CC. Let us now write down the elementry mtrices tht correspond to ech elementry row opertion here. Denote them by E,..., E 9 ; Note tht when we perform two elementry opertions t time, we should write mtrix for ech seprtely. So, E is switching the rst nd second rows, hence (we perform this opertion on the identity mtrix E R 3 R 3 + R, tht is E Further, E 4 E E 8 7. E would be R R R, nd E 3 would be, E 3 ;, nd (in the order the opertions re written, E 6 3, E 7, E 9 ; Now, the proccess is s follows: (A I (E A turns out to be I, E {}}{ E I (E E A E E {}}{ ( E 9 E 8 E E A E 9 E 8 E E (I E 9 E 8 E E. Hence, we hve A E 9 E 8 E E, nd lso A (E 9 E 8 E E E E E8 E 9 ;

3 (Here we use the property tht if A nd B re invertible, then their product is lso invertible, nd (AB B A ; (severl times; or prove it by induction for product of n N mtrices. Also, the fct tht ny elementry mtrix is invertible. How to nd E etc.? There re some possible wys. (For instnce, using the usul proccess of nding the inverse, s it ws done for the mtrix C. But, knowing wht ws the elemntry opertion performed, we cn write down the mtrix of the opposite opertion, getting, for instnce, E gin switching them. E (the opposite opertion of switching rows nd is (the opposite opertion of R R R is R R + R. And so on. (The opposite of multiplying row by c is multiplying it by c. For D - write R R R ; R 3 R 3 + R R R ; R (new R (new + R 3 R R We're lmost there (nd cn continue (two more opertions, but we cn see tht in ny cse, the mtrix L (on the LHS would not be row-equivlent to the identity mtrix I, since there would be one row of zeros. Hence, D is not invertible. For E, write R 4 R R R 4 ; R (new ; R R R ; R 3 R 3 R ; R 4 R (new /( ; R 3 R 3 /( ; R 4(new R 4(new /( 3

4 R 3 R 3 R 4 Hence, E is invertible, nd E Finlly, for F ( i i R R R R R R 3 ; ;, let us use q.. (Which works for ny eld, so lso for C. Note tht : d bc i ( i. Hence, F is invertible, nd ( ( i i F ( F. i i (Well, so F F. Hence F n F n N (you cn prove it by induction. Remrk: you cn use the usul method insted of q.. ( b Let A. Prove tht A is invertible (hkh if nd only if c d d bc, ( d b nd tht if d bc, then A d bc. c -proofs- (Wht follows holds for ny eld F. But we should lwys be creful not to devide by zero (of some eld somewhere. Here it does not hppen. Proof : Let us nd the conditions under which A is invertible, using the usul method. Write (A I ( b c d (. Then we hve the big mtrix row to get. Two possibilities from here: ( b c d. Switch the two 4

5 ( c d b. Two cses here: R R ( d (. c (so lso d bc. In this cse, we hve. b (One cn continue bit more, but it's possible to conclude mmiditely tht - In this cse, we would not be ble to get in the entry,. So the reduced nchelon form of A is not I, hence, A is not invertible. (. c. Hence we cn continue by performing R R /c nd getting to ( d c c. b (.. If b (nd hence, lso d bc, then I is not the reduced enchelon form of A, hence A is not invertible. ( d (.. If b, devide the second row by b getting c c R b ( R d c R d bc c ; b So in this cse, A is invertible, nd ( ( A d ( bc c d b {}}{ d b b bc c d bc c (. Then perform R R / getting ( ( b c d b R R cr d bc c. Agin, two options from here: (. d bc. Tht is, multiplying by, d bc. In this cse, the second row of the mtrix on the left hnd side is zero; Hence, we conclude tht A is row-equivlent to not invertible mtrix (wht we get is A's (cnonicl reduced enchelon form, which is not the identity mtrix, I. Thus, A is not invertible. (. d bc d bc. Tht is d bc. So perform R R /( d bc to get ( b R R b R c ( d bc d bc + bc (d bc d bc+bc (d bc d d bc b (d bc c d bc d bc ( d d bc b d bc c d bc d bc ( d b So in this cse, A is invertible, nd indeed A d bc. c To sum up, looking t ll the cses, we see tht A is invertible if nd only if d bc, nd( in ll such cses the inverse is indeed d b A d bc. c

6 Remrk: one cn perform the similr clcultions ( bit more complicted for 3x3 mtrix. Anyhow, you will soon study the generl cse of nxn mtrix. Proof : : If d bc nd we tke the mtrix written bove, one cn check tht indeed multiplying it by A gives the identity mtrix. (Multiplying from both sides, before you solve q.3.; Hence A would be invertible, with this (unique inverse mtrix. : To the other direction; ssume tht A is invertible. Then it hs n inverse mtrix (which is unique. Then the rows of A cnnot be dependent (i.e., it cnnot be tht exists constnt λ F such tht λr R (or λr R. This is becuse otherwise we could perform n elementry row opertion to get mtrix with zero row; this mtrix is not invertible, but is row equivlent to A. This is contrdiction! Suppose we hd d bc ; tht is, d bc. ( If, then bc, so b or c is zero. (We re in eld :. This cn be mentioned lso in proof in vriouse stges.. If ( b, then the rst row is zero, which is impossible. If c, then b A. So gin (check it: if one of b or d is zero, we done, nd d otherwise, we cn cler on row using the other we would hve t lest one free vrible; So A would not be invertible. Hence, when (nd d bc, A is not invertible. Which is contrdiction. ( So we should hve. In fct, sme rgument proves tht none of, b, c, d cn be zero (if d bc In this cse, d bc. And, since b is lso non-zero, we hve d b c : λ (dene λ to be this constnt. Hence we hve R λ R. Which cnnot hold if A is invertible! So nlly, if A is invertible, d bc. Now, the ( inverse is unique; So it is enough to check tht the given mtrix d b d bc (which is dened(! gives the identity mtrix when c multiplying it by A from both sides. (Note tht one should prove tht d bc in the second direction; Becuse multiplying mtrices give you the desired nswer only if the mtrices re dened nd if you're sure you don't get the zero mtrix there insted of the identity (in some step there, depends on how you ttempt to prove the clim. 3 Let A be mxn mtrix, nd let B be nxm mtrix. Suppose tht m > n. Prove tht AB is not invertible. (Hint: consider the system Bx. -proof- Consider the system Bx : homogeneous system with n equtions in m vribles. It hs solution (the trivil, x, nd it hs independent vribles (becuse the number of dependent vribles cnnot exceed the number of rows, n, 6

7 so the number of independent vribles is t lest m n >. Hence, the system hs non-zero solution: v in R m, such tht Bv. But then we hve AB(v A(Bv A (for this v in R m (or F m. Thus the homogeneous system tht corresponds to the mxm mtrix AB hs non-zero solution. Hence, by theorem proved in clss, AB in not invertible. (The theorem sttes: A squre mtrix mxm, A, is invertible if nd only if the system Ax b hs unique solution for ll b R m. (And one cn prove tht this lst hppends if nd only if the homogeneous system Ax hs unique solution. 4 Let A nd B be two squre mtrices (mtritzot ribuiyot of the sme size. ( Prove tht if AB is invertible nd A is invertible, then B is invertible. (b Prove tht if AB is invertible, then both A nd B re invertible. (c Prove tht if AB I, then BA I. -proofs- There could be severl proofs. For instnce, one cn rst prove (b (without ( by proving tht B is invertible, nd then A is invertible. Anyhow, let us prove in the order suggested here. ( Suppose A is invertible. Then A is lso invertible (with (A A, hence the product of the two invertible mtrices A nd AB give gin n invertible mtrix: so B is invertible. (b Well, let us prove tht A is invertible, nd then use (. Suppose tht A is not invertible. A is invertible if nd only if A t is invertible, hence, A t is sumed to be not invertible. Then there exists vector v such tht A t v. (# We know tht AB is invertible. Then (AB t B t A t is lso invertible. Now, if using (# we get (B t A t v B t (A t v B t, where v. But then B t A t is not invertble, so AB is not invertible. Contrdiction! So A is invertible. And using (, B is lso invertible. (c Now, suppose tht AB I. Then AB is invertible (it is I so, it's inverse is lso I. But then by prt (b, A nd B re invertible; So exist A nd B. Tking AB I nd multiplying it by B form the right, A B. So indeed, BA BB I. Remrk: Agin, there could be severl proofs, so you re encorged to try to nd better one (for ech prt; Py ttention you do not use wht is not proven yet or wht sould be proven. Remrk: (Everything here is bout squre mtrices. Prt (c here is pretty importnt: it shows tht (for instnce for mtrix A, if AB I (tht is, A is invertible from the right (or the left, depends on how you decide to cll it, then we hve BA I (so A is invertible lso from the left. 7

8 Recll tht by denition, in order for A to be invertible, we require the existnce of mtrix B such tht AB BA I (tht is, A should be invertible from both sides with n inverse B (from both sides. But now, 4(c tells you tht invertbility from one side is enough. (*It is importnt here tht the dimention of things here is nite. 6 Let A, B nd P be squre mtrices of the sme size, nd suppose they stisfy the reltion B P AP (we suppose tht P is invertible. ( Prove tht for ny nturl k we hve B k P A k P. (b Let f(x be polynomil in x. (Then we cn consider f(a etc. Prove tht if f(a (the zero mtrix of the sme size, then lso f(b. Let A denote squre mtrix which stises the identity A 4 + 3A 3 + A 4I. Prove tht A is invertible nd express the mtrix A in terms of A. -proof- We hve A 4 + 3A 3 + A 4I, so A(A 3 + 3A + I 4I, A (A3 +3A +I 4 I. Then A is invertible, nd A (A3 +3A +I 4. -proofs- ( Let us prove the identity B k P A k P by induction on k. For k it holds (it is given. {}}{ Suppose it holds for k. Then B k+ B B k P A P P A k P P A A k P P A k+ P (the desired result for k+; This completes the proof. (b Suppose f is of some degree n nd f(x n x n + n x n + + x+. Then f(a n A n + n A n + + A + I. (Where,,, n re in some eld F. In R, if you like. Suppose now tht f(a. Using prt ( we get P IP {}}{ f(b n B n + n B n + + B+ I n P A n P + n P A n P + + P AP + P IP. Now, using the two properties of mtrix ddition nd multipliction: (pilug (C + DE CE + DF, E(C + D EC + ED, nd lso properties of multipliction by sclrs we get f(b P ( n A n P + n A n P + + AP + IP P ( n A n + n A n + + A + IP P f(ap P P. 7(hlf *-ed Let J be n invertible mtrix, nd let A be mtrix which stises A t JA J. Prove tht A is invertible, nd tht A stises (A t JA J. -proof- I 8

9 * Note tht J is given to be invertible, tht is invertible from both sides, hence it's squre mtrix. A is lso ment to be squre (of the sme order, lthough it ws not stted... First, multiplying both sides of A t JA J by J (which exists from the left, we get J A t JA I. By question 4(c (or mybe by wht ws proven in clss, if A is invertible from the one side (from the left here, it's invertible lso from the right, hence, by denition, it is invertible, nd A J A t J. Now for the second prt. Let us tke the trnspose of both sides of A t JA J, to get A t J t A J t. (# Now, multiplying both sides of this lst equlity by (J t from the left, we get (J t A t J t A I So lso A (J t A t J t (J t A t J t. Hence we hve I I {}}{{}}{ (A t JA ((J t A t J t t J A JA J J A J AA J. As desired. (Remrk: we use three results here: I. If C nd D re two squre mtrices of the sme order, then (CD t D t C t ; II. If squre mtrix C is invertible, then C t is lso invertible, nd (C t (C t ; III. For two squre invertible mtrices of the sme order, C nd D, their product CD is lso invertible, nd (CD D C ; (Remrk: one cn lso void using step (#, nd then would probbly hve to tke trnspose of some mtrix t some point. 8 Solve the eqution ( 3 ( 3 X ( 3 3 where the indeterminte X is x mtrix. (Tht is, nd X. (There re t lest two dierent wys to do it. Hint: using the theory would be shorter. -solution- Solution ( - The eqution ( is of the form ( A X B C, where A, B, C ; Let us note tht (by question no. here, for instnce the mtrices A nd B re invetible. So multiplying both sides of the eqution by A from the left, we get A A X B A C, tht is XB A C; And then multiplying both sides of this lst eqution by B from the right, we get X A CB, thus solving for X. Now, we only should mke the computtions. By question,, 9

10 ( A 3 3 B 3 ( ( 3 So nlly ( X A CB 3 (. ( ; ( 3 3 ; ( 3 3 ( 3 ( 3 ( b Solution - Tke X, multiply everything (in the eqution, c d to get some ( system of ( 4 liner equtions ( on 4( vribles (, b, c nd d; solve it. 3 b b 3 + b Here: ( c d ( c + d 3c + d 3 + 3b + c + d 9 + 6b + 3c + d b + c + d 6 + 4b + 3c + d (Compre this to the rst solution. 9 Suppose tht A is nilpotent mtrix (exists N N such tht A N. Prove I A nd I + A re both invertible nd nd their inverse mtrices. (Hint: consider this problem for N rst. (Hint: think of the innite geometric series (sidr hndsit nd its sum. -Proofs- Let us rst prove the clim bout I A. Mybe we should not hve given hint, becuse tht mde one think mostly in direction of one proof, while there could be other proofs. So here re two proofs. Proof - We hve (for rel/complex numbers (or ny other eld or even not eld the formul x n ( x( + x x n. (Proof - by induction. Or by opening brckets (which ctully uses induction. Or by the formul of sum of geometric series (sidr hndsit (which itself is proven usung induction. Now, generlly we cnnot pply numericl formuls for mtrices (becuse there is no commuttivity!; But in this cse, it holds for squre mtrix in plce of x, becuse everything here commutes! (CI IC I for ny mtrix C, nd C p C q C q C p for ny nturl p nd q. So we hve (gin, the proof is by induction, replcing x by A. Or just using the formul for A stting tht everything commutes - I A n (I A(I + A A n. Let N be the order of nilpotency of A (tht is, A N, A N. Then pplying the formul for N, I {}}{ A N (I A(I + A A N ; So I (I A(I + A A N.

11 Hence I A is invertible, nd it's inverse is (I + A A N. Remrk: We hve shown tht I A is invertible from one side, but ctully I (I + A A N (I A lso holds, using the sme formul. Or otherwise, From question 4(c here, we cn conclude emmiditely tht it's invertible from the other side s well. Remrk: We could use ny M > N insted of N (s A M would lso be zero nd would rrive t the sme result. Proof - We wnt to prove tht I A is invertible. Suppose it is not. Then there exists non-zero vector v (of the pproporite size, depending on the mrix size such tht (I Av. (Tht is, non-zero solution of the homogeneous system corresponding to the mtrix (I A. So we hve Iv Av, tht is Av Iv, so Av v. (s Iv v, check why! [So if A is nxn, then it represents function A : R n R n given by A(v Av for ny v R n ; And then the condition Av v mens tht A does not move this specil vector v.] But now one cn note tht lso A v v [tht is, v styes in plce fter pplying A.] This is becuse A v A(Av A(v v. Gret! Wht bout A 3, A 4 nd so on? [Well, if we pply A nd it does not move v, then pplying A once more (tht is, pplying A would not move v s well. And so on.] Let us prove by induction, tht A n v v for ll n N : For n, indeed Av v. Assume the clim is true for n. A n vv by ssumption Then A n+ v A(A n {}}{ v Av v; Which completes the proof. But recll tht A is nilpotent mtrix. Let N be the nilpotency order of A (tht is A N nd A N. Then by wht we proved, we hve A N v v. But A N, so v A N v. Contrdiction! So I A is invertible. (One cn lso formulte this proof not d bsurdum (be shlil, but s follows: ssume tht (I Av (tht is, v is solution of the homogeneous system corresponding to I A, prove tht v, nd conclude tht, becuse the homogeneous system hs unique solution, the mtrix I A is invertible. Now, bout I + A. Agin, more thn one proof cn be given. Proof - Similr to the rst proof bove, but using slightly dierent formul (with lternting signs - x n ( + x( x + x.. + ( k x k ( n x n ; Proof - Similr to the second proof bove, but with Av v (so things would be very similr, but with lternting signs.

12 Proof 3 - And the shortest: Use wht hs lredy been proven: If A is nilpotent (or order N, then A is nilpotent of the sme order, becuse generlly ( A n ( n A n. (So use it for N, getting ( A N ; nd lso for N (which is not needed here. Then using the bove clim for ( A, we know tht the mtrix I ( A I + A is invertible. A squre mtrix is clled upper tringulr (meshulshit el'yon if ll its entries below the min digonl (lkhson rshi re zero. Tht is, A nxn ( ij is upper tringulr if ij for ll j < i ( i, j n. For exmple, A 8 3, B ( 9 re upper tringulr, while C 8 is not (becuse of the in entry (3,. 3 An upper tringulr mtrix is clled strictly upper tringulr (we my cll it el'yon mmsh, if it is upper tringulr with zeros lso on the min digonl. ( Prove tht the product of two upper tringulr mtrices is n upper tringulr mtrix. (b Prove tht the product of two strictly upper tringulr mtrices is strictly upper tringulr mtrix. (c* Prove tht n upper tringulr mtrix is invertible if nd only if ll digonl entries re not zero. (d Clculte theinverse of the following upper tringulr mtrix: b A c 3 (you my ssume tht A is upper tringulr s well. (* Why? -proofs- ( Let A nxn ( ij nd B nxn (b ij be two upper tringulr mtrices of the sme order n. Tht is, ij b ij for ll j < i (where i, j n. Let us compute [AB] pq for q < p. (between nd n. [AB] qp n pk b kq k p k for k<p {}}{ pk b kq + n kp pk, s q<p k {}}{ b kq (b Similrly to (, but with ij b ij for ll j i. (Check tht it works! (c Let A ( ij be n upper tringulr mtrix, tht is

13 .... A looks like nn (Strs re for mtrix entries we do not cre bout so much. First, suppose there exists n entry kk. Then the homogeneous eqution Ax hs non-trivil (non-zero solution: tke ll vrible except the k-th to be zero (x... x k x k+... x n nd x k. (Indeed, it is solution of this system, checking every eqution. But then A cnnot be invertible. (Remrk: s usul, there cn be other solutions suggested. For instnce, switch some rows nd mke some other elementry opertions to get mtrix tht is not equivlent to the identity mtrix (for exmple, mtrix tht hs some row with zeros in ll entries. Now, the other direction: suppose jj for ll j n. We wnt to prove tht in such cse, A is invertible. Proof (using derug, tht is bringing A into the cnonicl form, nd observing tht it is row equivlent to the identity mtrix. We proceed s it is done usully in when bringing A to it's cnonicl form. Let us (for (* of (d look lso t the big system (A I. Wht hppends there whould be stted in []-s. First, look t the rst row. ; Devide the rst row by to get new mtrix ( E A. This chnges the rst row only (remining the mtrix upper-tringulr. [The identity mtrix becomes stys lmost the sme, but with t entry,. It is lso upper-tringulr, s is the identity.] The stte so fr is - n.... E A..... ; nn Now, the mtrix is upper tringulr, so ll the rst colomn under entry, contins zeros. So no need to perform nything with it. Next, look t the second row. The entry, is zero. The entry, is. Devide the second row by, nd cler out the entry, in the second colomn. Tht is, perform R R R, where the by entry-tg we shll from now on denote the current stte of entry, (it is, but it is not so interesting for our purpose. 3

14 [In the right hnd side, there pers something in the rst row (insted of zeros. Still n upper-tringulr mtrix.] So fr, we hve n ; nn Let proceed by induction: we clim tht we cn perform elementry row opertions nd get the identity mtrix. We wnt to show tht we cn ech time get mtrix of the form ( Ik,k ; (# A kk;nn.. kk kn nn nd proceed to the next stge. (Here, * is some mtrix of order (k-x(n-k+; is ll zeros nd is of order (n-k+x(k-; nd by A kk;nn we men the rows nd colomns k to n of the initil mtrix A (without chnges. [The right hnd side on this stged is climed to be still upper-tringulr. See next []-s.] Well, for k it is known (if we just consider the initil mtrix. (or strt from k if you wnt. Now, suppose we hve performed the required opertions nd got to the form in (#. Let us look t the k-th row. kk (nd it ws not chnged yet. Devide the k-th row by kk. Then, for ech l < k perform R l R l lk R k(new; You would rrive t mtrix of the form ; (s required!.. kn nn 4

15 [By the ssumption, there's still n upper-tringulr mtrix on the RHS. Performing the opertions bove on this mtrix shll chnge entries below the min digonl.] So by induction, we get the identity mtrix on the LHS nd some upper tringulr mtrix on the RHS: (A I (I B; [with B being upper-tringulr.] Then n upper-tringulr mtrix with no zeros on the min digonl, A, is invertible, nd it's inverse is lso upper-tringulr. Proof (proving tht there is unique solution to the homogeneous system tht corresponds to A, nd concluding tht A is invertible. Let us look t the homogeneous system of equtions Ax. Strting t the lst row, x n. Substitute it in row (n -th row to get {}}{ n,n x n + something x n, so x n. So, by induction on k, let us prove tht x n k. (For k > n there is just no mening to this sttement.. For k, it's lredy known. (for l, we my sy. Suppose it's known for ll l k (tht is, x n l for ll l k. Let us prove tht then the clim is true for l k + (tht is, for ll l k +. Let us look t the n (k + -th row. The corresponding eqution is n (k+,n (k+ x n (k+ + n (k+,n (k++ x n (k++ + n (k+,n (k++ x n (k n (k+,n x n + n (k+,n x n Which, by the ssumption reduces to n (k+,n (k+ x n (k+. We ssumed tht the digonl entries re non-zero, hence x n (k+. This completes the inductive proof. So now, we know tht the homogeneous system Ax hs unique solution (the trivil solution. Then it follows (by theorem studied in clss tht A is invertible. (d (* Using the rst proof of (c, we cn see how the RHS chnges (cn crefully monitor it, by induction, nd conclude tht it is upper-tringulr. So we my tke A d e f g h (knowing by (c tht it exists, nd by k (d tht it is lso upper tringulr. Now, AA b c 3 d e f g h k. So we must solve the system d e + g f + h + bk g h + ck 3k

16 d g 3k for d, e,..., k. Well, d, g e + g, k 3, hence the f + h + bk h + ck e + system reduces to f + h + 3 b. h + 3 c We cn write it in mtrix nd solve, but it's possible solve immeditely: e, h 6 c nd then f 3 b + 6 c. To sum up, A 3 b + 6 c 6 c. 3 6

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