ax + by = f cx + dy = g ( $ 0 1 (ag! cf ) "'

Transcription

1 C H A P T E R 4 Determiats Suppose we wat to solve the system of equatios ax + by = f cx + dy = g where a, b, c, d, f, g! F. It is easily verified that if we reduce the augmeted matrix to reduced row-echelo form we obtai 1 0 ( fd! gb) "!% ( 0 1 (ag! cf ) " where Î = ad - cb. We must therefore have Î " 0 if a solutio is to exist for every choice of f ad g. If A! Mì(F) is the matrix of coefficiets of our system, we call the umber Î the determiat of A, ad write this as det A. While it is possible to proceed from this poit ad defie the determiat of larger matrices by iductio, we prefer to take aother more useful approach i developig the geeral theory. We will fid that determiats arise i may differet ad importat applicatios. Recall that uless otherwise oted, we always assume that F is ot of characteristic 2 (see Exercise ). 170

2 4.1 DEFINITIONS AND ELEMENTARY PROPERTIES DEFINITIONS AND ELEMENTARY PROPERTIES Recallig our discussio of permutatios i Sectio 1.2 we make the followig defiitio. If A = (aáé) is a x matrix over a field F, we defie the determiat of A to be the scalar det A =! " S (sg!" )a 1"1 a 2" 2!!!a " where ßi is the image of i = 1,..., uder the permutatio ß. We frequetly write the determiat as det A = a 11! a 1 " " a 1! a!!. Note that our defiitio cotais! terms i the sum, where each term is a product of factors aij, ad where each of these terms cosists of precisely oe factor from each row ad each colum of A. The determiat of a x matrix A is said to be of order. We will sometimes deote the determiat of A by \A\. Note that the determiat is oly defied for a square matrix. Example 4.1 We leave it to the reader to show that i the case of a 2 x 2 matrix, our defiitio agrees with the elemetary formula a c b d = ad! cb!!. I the case of a 3 x 3 matrix, we have det A =! " (sg!" )a 1"1 a 2" 2 a 3" 3 = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32!!!!!!!!!!!!!!!!!a 13 a 22 a 31 a 11 a 23 a 32 a 12 a 21 a 33!!. The reader may recogize this from a more elemetary course whe writte i the memoic form

3 172 DETERMINANTS a 11 a 12 a 13 a 21 a 22 a 23 _ a 31 a 32 a 33 Here, we are to add together all products of terms coected by a (+) lie, ad subtract all of the products coected by a (-) lie. We will see i a later sectio that this 3 x 3 determiat may be expaded as a sum of three 2 x 2 determiats. Recall that a diagoal matrix A = (aáé) is defied by the property that aáé = 0 for i " j. We therefore see that if A is ay diagoal matrix, the det A = a 11!!!a = sice oly the idetity permutatio results i solely ozero factors (see also Theorem 4.5 below). I particular, we have the simple but importat result det I = 1. We ow proceed to prove several useful properties of determiats.! Theorem 4.1 For ay A! Mñ(F) we have det AT = det A. Proof Cosider oe of the terms (sg ß)a1ß1 ~ ~ ~ aß i det A. The aáßi is i the ith row ad ßith colum of A, ad as i varies from 1 to, so does ßi. If we write ay factor aáßi as aáé, the j = ßi so that i = ßîj. By Theorem 1.5 we kow that sg ßî = sg ß, ad hece we ca write our term as (sg ß) aßî1 1 ~ ~ ~ aßî = (sg œ) aœ1 1 ~ ~ ~ aœ where œ = ßî. Therefore, sice Sñ = {œ = ßî: ß! Sñ}, we have i=1 a ii det A =! " S (sg!" )a 1!"1!!!a!" =! S (sg!)a 1!1!!!a!!!.

4 4.1 DEFINITIONS AND ELEMENTARY PROPERTIES 173 But atáé = aéá so that det A T =! " S (sg!")a T 1!"1!!!a T!" =! " S (sg!")a "1!1!!!a "! ad hece det A = det AT. It will be very coveiet for us to view the determiat of A! Mñ(F) as a fuctio of the row vectors Aá. Whe we wish to do this, we will write det A = det(aè,..., Añ). Now cosider a matrix A! Mñ(F) ad assume that Aè = rbè + scè where Bè = (b11,..., b1) ad Cè = (c11,..., c1) are ay two arbitrary (row) vectors i F, ad r, s! F. We the have det A = det(a 1,!!,!A ) = det(rb 1 + sc 1,!A 2,!!,!A ) =! " S (sg!" )(rb 1!"1 + sc 1!"1 )a 2!" 2!a!" = r! " S (sg!" )b 1!"1 a 2!" 2!a!"!!!!!!!!!!!!!!!!!!!!!!!!!!+s! " S (sg!" )c 1!"1 a 2!" 2!a!"!!. If we ow let B be the matrix such that Bá = Aá for i = 2,..., ad C be the matrix such that Cá = Aá for i = 2,..., we see that det A = r det B + s det C. Geeralizig slightly, we summarize this result as a theorem for easy referece. Theorem 4.2 Let A! Mñ(F) have row vectors Aè,..., Añ ad assume that for some i = 1,..., we have Aá = rbá + scá where Bá, Cá! F ad r, s! F. Let B! Mñ(F) have rows Aè,..., Ai-1, Bá, Ai+1,..., Añ ad C! Mñ(F) have rows Aè,..., Ai-1, Cá, Ai+1,..., Añ. The det A = r det B + s det C.

5 174 DETERMINANTS Corollary 1 Let A! Mñ(F) have rows Aè,..., Añ ad suppose that for some i = 1,..., we have A i = k! j=1 r j B j where Bé! F for j = 1,..., k ad each ré! F. The det A = det(a 1,!!,!A i!1,!" k j=1 r j B j,!a i+1,!!,!a ) k = r j det( A 1,!!,!A i!1,!b j,!a i+1,!!,!a )!!. j=1 Proof This follows at oce by iductio from the theorem. Corollary 2 If ay row of A! Mñ(F) is zero, the det A = 0. Proof If ay row of A is zero, the clearly oe factor i each term of the sum det A = Íß Sñ(sg ß)a1ß1 ~ ~ ~ aß will be zero. (This also follows from Theorem 4.2 by lettig r = s = 0.) Corollary 3 If A! Mñ(F) ad r! F, the det(ra) = r det A. Proof Sice ra = (raáé) we have det(ra) =! " S (sg!" )(ra 1!"1 )!!!(ra!" ) = r! " S (sg!" )a 1!"1!!!a!" = r det A!!. For ay A! Mñ(F) ad ß! Sñ, we let ßA deote the matrix with rows Aß1,..., Aß. For example, if A ad ß are give by the! A =! 4 5 6!!!!!!!!!!ad!!!!!!!!!! =! " 3 1 2% " 7 8 9% " 7 8 9%! A =! 1 2 3!!

6 4.1 DEFINITIONS AND ELEMENTARY PROPERTIES 175 Theorem 4.3 For ay A! Mñ(F) ad ß! Sñ we have Proof First we ote that by defiitio det(ßa) = (sg ß)det A. det(ßa) = det(aß1,..., Aß) = ̓ Sñ (sg ƒ) aß1 ƒ1 ~ ~ ~ aß ƒ. Now ote that for each i = 1,..., there exists a j! {1,..., } such that ßj = i. The j = ßîi ad ƒj = ƒßîi = œi (where we have defied ƒßî = œ) so that aßj ƒj = ai œi. Sice ß is fixed we see that Sñ = {œ = ƒßî: ƒ! Sñ}, ad sice by Theorem 1.4 we have sg ƒ = sg(œß) = (sg œ)(sg ß), it follows that det(ßa) = Íœ Sñ (sg œ)(sg ß) a1 œ1 ~ ~ ~ a œ = (sg ß)det A. Note that all we really did i this last proof was rearrage the terms of det(ßa) to where each product cotaied terms with the rows i atural (i.e., icreasig) order. Sice the sum is over all ƒ! Sñ, this is the the same as summig over all œ! Sñ where œ = ƒßî. Corollary 1 If B! Mñ(F) is obtaied from A! Mñ(F) by iterchagig two rows of A, the det B = -det A. Proof If ß is a traspositio, the sg ß = -1 so that det B = det(ßa) = -det A. Corollary 2 If A! Mñ(F) has two idetical rows, the det A = 0. Proof If we let B be the matrix obtaied by iterchagig the two idetical rows of A, the det A = det B = -det A implies that det A = 0. Let us make two remarks. First, the reader should realize that because of Theorem 4.1, Theorems 4.2 ad 4.3 alog with their corollaries apply to colums as well as to rows. Our secod rather techical remark is based o the material i Sectio 1.5. Note that our treatmet of determiats has made o referece to the field of scalars with which we have bee workig. I particular, i provig Corollary 2 of Theorem 4.3, what we actually showed was that det A = -det A, ad hece 2det A = 0. But if F happes to be of characteristic 2, the we ca ot coclude from this that det A = 0. However,

7 176 DETERMINANTS i this case it is possible to prove the corollary directly through use of the expasio by miors to be discussed i Sectio 4.3 (see Exercise ). This is why we remarked earlier that we assume our fields are ot of characteristic 2. I fact, for most applicatios, the reader could just as well assume that we are workig over either the real or complex umber fields exclusively. 4.2 ADDITIONAL PROPERTIES OF DETERMINANTS I this sectio we preset a umber of basic properties of determiats that will be used frequetly i much of our later work. I additio, we will prove that three fudametal properties possessed by ay determiat are i fact sufficiet to uiquely defie ay fuctio that happes to have these same three properties (see Theorem 4.9 below). Theorem 4.4 Suppose A! Mñ(F) ad let B! Mñ(F) be row equivalet to A. (a) If B results from the iterchage of two rows of A, the det B = -det A. (b) If B results from multiplyig ay row (or colum) of A by a scalar k, the det B = k det A. (c) If B results from addig a multiple of oe row of A to aother row, the det B = det A. Proof By Corollary 1 of Theorem 4.3, a type å elemetary row trasformatio merely chages the sig of det A. Next, Theorem 4.2 shows that a type trasformatio multiplies det A by a ozero scalar (choose r = costat ad s = 0 i the statemet of the theorem). Now cosider a type trasformatio that adds k times row j to row i. The Bá = Aá + kaé so that applyig Theorem 4.2 ad Corollary 2 of Theorem 4.3 we have det B = det(b 1,!!,!B i,!!,!b j,!!,!b ) = det(a 1,!!,!A i + ka j,!!,!a j,!!,!a ) = det(a 1,!!,!A i,!!,!a j,!!,!a )!!!!!!!!!!!!!!!!!!!!!!+k det(a 1,!!,!A j,!!,!a j,!!,!a ) = det A + k! 0 = det A!!. Corollary If R is the reduced row-echelo form of a matrix A, the det R = 0 if ad oly if det A = 0. Proof This follows from Theorem 4.4 sice A ad R are row-equivalet.

8 4.2 ADDITIONAL PROPERTIES OF DETERMINANTS 177 A matrix A! Mñ(F) is said to be upper-triagular if aáé = 0 for i > j. Similarly, A is said to be lower-triagular if aáé = 0 for i < j. I other words, a upper-triagular matrix has all zeros below the mai diagoal, ad a lowertriagular matrix has all zeros above the mai diagoal. Theorem 4.5 If A! Mñ(F) is a triagular matrix, the det A = %i ˆ=1 aáá. Proof If A is lower-triagular, the A is of the form! a ! 0 a 21 a ! 0 a 31 a 32 a 33 0! 0!!. " " " " " " a 1 a 2 a 3 a 4! a % Sice det A = Íß Sñ (sg ß)a1 ß1 ~ ~ ~ a ß, we claim that the oly ozero term i the sum occurs whe ß is equal to the idetity permutatio. To see this, cosider the ozero term t = a1 ß1 ~ ~ ~ a ß for some ß! Sñ. Sice aáé = 0 for i < j, we must have ß1 = 1 or else a1 ß1 = 0 = t. Now cosider a2 ß2. Sice ß1 = 1 we must have ß2 " 1, ad hece the fact that a2 ß2 = 0 for 2 < ß2 meas that oly the ß2 = 2 term will be ozero. Next, sice ß1 = 1 ad ß2 = 2, we must have ß3 " 1 or 2 so that a3 ß3 = 0 for 3 < ß3 meas that oly a3 3 ca cotribute. Cotiuig i this maer, we see that oly the term t = aèè ~ ~ ~ aññ is ozero, ad hece det A = a 11!!!a =! a ii!!. If A is a upper-triagular matrix, the the theorem follows from Theorem 4.1. Corollary If A = (aij) is diagoal, the det A = %iaii. It is a obvious corollary of this theorem that det I = 1 as we metioed before. Aother extremely importat result is the followig. Theorem 4.6 A matrix A! Mñ(F) is sigular if ad oly if det A = 0. Proof Let R be the reduced row-echelo form of A. If A is sigular, the r(a) < so that by Theorem 3.9 there must be at least oe zero row i the matrix R. Hece det R = 0 by Theorem 4.2, Corollary 2, ad therefore det A = 0 by the corollary to Theorem 4.4. i=1

9 178 DETERMINANTS Coversely, assume that r(a) =. The, by Theorem 3.10, we must have R = Iñ so that det R = 1. Hece det A " 0 by the corollary to Theorem 4.4. I other words, if det A = 0 it follows that r(a) <. We ow prove that the determiat of a product of matrices is equal to the product of the determiats. Because of the importace of this result, we will preset two differet proofs. The first is based o the ext theorem. Theorem 4.7 If E! Mñ(F) is a elemetary matrix ad A! Mñ(F), the det(ea) = (det E)(det A). Proof Recall from Theorem 3.22 that e(a) = e(i)a = EA. First ote that if e is of type å, the det E = -det I = -1 (Theorem 4.3, Corollary 1), ad similarly det e(a) = -det A. Hece i this case we have det(ea) = det e(a) = (-1)det A = (det E)(det A). If e is of type, the usig Theorem 4.2 we have det E = det e(i) = k det I = k so that det(ea) = det e(a) = k det A = (det E)(det A). Fially, if e is of type, the Theorem 4.5 shows us that det E = det e(i) = 1 ad hece det(ea) = det e(a) = det A!!(see the proof of Theorem 4.4) =(det E)(det A)!!. This proves the theorem for each of the three types of elemetary row operatios. Theorem 4.8 Suppose A, B! Mñ(F). The det(ab) = (det A)(det B). Proof 1 If either A or B is sigular, the so is AB (Corollary to Theorem 3.20). Hece (by Theorem 4.6) it follows that either det A = 0 or det B = 0, ad also det(ab) = 0. Therefore the theorem is true i this case. Now assume that A ad B are both osigular. From Theorem 3.24 we may write A = Eè ~ ~ ~ Er so that repeated applicatio of Theorem 4.7 yields

10 4.2 ADDITIONAL PROPERTIES OF DETERMINANTS 179 det AB = det(e 1!E r B) = det E 1 det(e 2!E r B) = det E 1 det E 2 det(e 3!E r B) =!!!= det E 1!det E r det B = det E 1!det E r!2 det(e r!1 det E r )det B =!!!= det(e 1!E r )det B = (det A)(det B)!!. Proof 2 If C = AB, the Cá = (AB)á = Íé aáébé for each i = 1,..., (see Sectio 3.6). From Corollary 1 of Theorem 4.2 we the have detc = det(c 1,!!,!C ) = det(! j1 a 1 j1 B j1,!!,!! j a j B j ) =! j1!!!! j a 1 j1!a j det(b j1,!!,!b j )!!. Now, accordig to Corollary 2 of Theorem 4.3 we must have jé " jm (for k " m) so that we eed cosider oly those terms i this expressio for det C i which (jè,..., jñ) is a permutatio of (1,..., ). Therefore detc =! " S a 1!"1!a!" det(b "1,!!,!B " ) ad hece by Theorem 4.3 we have detc =! " S a 1!"1!a!" (sg!" )det(b 1,!!,!B ) = (det B)! " S (sg!" )a 1!"1!a!" = (det B)(det A)!!.!! Corollary If A! M(F) is osigular, the det Aî = (det A)î. Proof If A is osigular, the Aî exists by Theorem 3.21 so that AAî = I. Hece applyig Theorem 4.8 shows that 1 = det I = det(aaî) = (det A)(det Aî). This implies det Aî = (det A)î. We ow show that three of the properties possessed by ay determiat are i fact sufficiet to uiquely defie the determiat as a fuctio D:

11 180 DETERMINANTS M(F) F. By way of termiology, a fuctio D: M(F) F is said to be multiliear if it is liear i each of its compoets. I other words, if D(A) = D(Aè,..., Añ) ad Aá = Bá + Cá for ay i = 1,..., the D(A) = D(Aè,..., Ai-1, Bá + Cá, Ai+1,..., Añ) = D(Aè,..., Ai-1, Bá, Ai+1,..., Añ) + D(Aè,..., Ai-1, Cá, Ai+1,..., Añ) ad if Aá = kbá for ay k! F, the D(A) = D(Aè,..., kbá,..., Añ) = k D(Aè,..., Bá,..., Añ). Note Theorem 4.2 shows that our fuctio det A is multiliear. Next, we say that D: M(F) F is alteratig if D(A) = 0 wheever A has two idetical rows. From Corollary 2 of Theorem 4.3 we see that det A is alteratig. To see the reaso for the word alteratig, suppose that Aá = Aé = Bá + Cá ad D is both multiliear ad alteratig. The 0 = D(A) = D(Aè,..., Aá,..., Aé,..., Añ) = D(Aè,..., Bá + Cá,..., Bá + Cá,..., Añ) = D(Aè,..., Bá,..., Bá + Cá,..., Añ) + D(Aè,..., Cá,..., Bá + Cá,..., Añ) = D(Aè,..., Bá,..., Bá,...., Añ) + D(Aè,..., Bá,..., Cá,..., Añ) + D(Aè,..., Cá,..., Bá,..., Añ) + D(Aè,..., Cá,..., Cá,..., Añ) = 0 + D(Aè,..., Bá,..., Cá,..., Añ) + D(Aè,..., Cá,..., Bá,..., Añ) + 0 so that D(Aè,..., B,..., C,..., Añ) = -D(Aè,..., C,..., B,..., Añ). Thus, to say that D is alteratig meas that D(A) chages sig if two rows of A are iterchaged. Fially, let {Eá} be the row vectors of Iñ (ote that Eè,..., Eñ form the stadard basis for F). The, as we saw i Theorem 4.5, det(eè,..., Eñ) = det I = 1. If we cosider a permutatio ß! Sñ, the from Theorem 4.3 we see that

12 4.2 ADDITIONAL PROPERTIES OF DETERMINANTS 181 det(eß1,..., Eß) = (sg ß)det(Eè,..., Eñ) = sg ß. We are ow i a positio to prove the uiqueess of the determiat fuctio. Theorem 4.9 Let D: M(F) F be a multiliear ad alteratig fuctio with the additioal property that D(I) = 1. If Dÿ: M(F) F is ay other fuctio with these properties, the Dÿ = D. I particular, the determiat is the oly such fuctio. Proof It follows from the above discussio that our fuctio det has all three of the properties give i the theorem, ad hece we must show that it is the oly such fuctio. Let Aè,..., Añ be ay set of vectors i F, ad defie the fuctio Î: F ª ~ ~ ~ ª F F by Î(Aè,..., Añ) = D(Aè,..., Añ) - Dÿ(Aè,..., Añ). We must show that Î(Aè,..., Añ) = 0. It should be clear that Î is multiliear ad alteratig, but that Î(I) = Î(Eè,..., Eñ) = D(I) - Dÿ(I) = 0. Sice {Eá} is the stadard basis F, it follows that for ay Aá! F we have Aá = Íé cáé Eé for some set of scalars cáé. Usig this ad the properties of Î, we the have!(a 1,!!,!A ) =!(" j1 c 1 j1 E j1,!!,!" j c j E j ) = " j1!" j c 1 j1!c j!(e j1,!!,!e j )!!. At this poit, each jé is summed from 1 to. However, Î is alteratig so that Î(Ejè,..., Ejñ) = 0 if jé = jm for ay k, m = 1,...,. Therefore the ozero terms occur oly whe (jè,..., jñ) is some permutatio of (1,..., ) ad hece Î(Aè,..., Añ) = Íß Sñ c1 ß1 ~ ~ ~ c ß Î(Eß1,..., Eß). Sice D ad Dÿ are alteratig, we have D(Eß1,..., Eß) = (sg ß) D(Eè,..., Eñ) = (sg ß) D(I) = sg ß

13 182 DETERMINANTS ad similarly for Dÿ (ote that this follows from Theorem 1.2). Therefore we fid that Î(Eß1,..., Eß) = D(Eß1,..., Eß) - Dÿ(Eß1,..., Eß) = sg ß - sg ß = 0 ad hece Î(Aè,..., Añ) = 0. Suppose that D is a multiliear ad alteratig fuctio o the set of all - square matrices over F. (Here we do ot require that D(I) = 1.) If we write the rows of a matrix A! Mñ(F) as Aè,..., Añ the A i =! j=1 a ij E j where {Eé} are the rows of the idetity matrix Iñ. Exactly as we did i the precedig proof for the fuctio Î, we may write D(A) = D(Aè,..., Añ) = D(Íjè aèjèejè,..., Íjñ ajñejñ) = Íjè ~ ~ ~ jñ a1jè ~ ~ ~ ajñ D(Ejè,..., Ejñ) = Íß Sñ a1 ß1 ~ ~ ~ a ß D(Eß1,..., Eß) = Íß Sñ (sg ß) a1 ß1 ~ ~ ~ a ß D(Eè,..., Eñ) = (det A) D(I). Note that this is actually a quite geeral formula, ad says that ay multiliear ad alteratig fuctio D defied o A! M(F) is just the determiat of A times D(I). We will use this formula later i the chapter to give a simple proof of the formula for the determiat of a block triagular matrix. Exercises 1. Compute the determiats of the followig matrices directly from the defiitio: " 1!!2!!3% "!!2!!0!!1% (a)!!! 4!2!!3!!!!!!!!!!!!!!(b)!!!!!3!!2!3 2!!5!1!1!3!!5

14 4.2 ADDITIONAL PROPERTIES OF DETERMINANTS Cosider the followig real matrix: " 2!1!!9!1% 4!3!1!2 A =!!. 1!4!!3!2 3!2!!1!4 Evaluate det A by reducig A to upper-triagular form ad usig Theorem Usig the defiitio, show that a b 1 b 2 b 3 b 4 c 1 c 2 c 3 c 4 d 1 d 2 d 3 d 4 = a 1 b 2 b 3 b 4 c 2 c 3 c 4 d 2 d 3 d 4!!. 4. Evaluate the determiat of the followig matrix:! 0 0! ! 1 0!!. " " " " " 1 0! 0 0% 5. If A is osigular ad Aî = AT, show that det A = ±1 (such a matrix A is said to be orthogoal). 6. Cosider a complex matrix U! Mñ(ç). (a) If U* = (uáé*), show that det U* = (det U)*. (b) Let U = U*T (this is called the adjoit or cojugate traspose of U, ad is ot to be cofused with the classical adjoit itroduced i the ext sectio). Suppose U is such that U U = UU = I (such a U is said to be uitary). Show that we may write detu = e i! for some real ƒ. 7. If A is a x matrix ad k is a scalar, show that: (a) det(ka) = k det A usig Theorem 4.4(b). (b) det A = (det A).

15 184 DETERMINANTS 8. (a) If A is a real x matrix ad k is a positive odd iteger, show that Ak = Iñ implies that det A = 1. (b) If A = 0 for some positive iteger, show that det A = 0. (A matrix for which A = 0 is said to be ilpotet.) 9. If the aticommutator [A, B]+ = AB + BA = 0, show that A ad/or B i M(F) must be sigular if is odd. What ca you say if is eve? 10. Suppose C is a 3 x 3 matrix that ca be expressed as the product of a 3 x 2 matrix A ad a 2 x 3 matrix B. Show that det C = 0. Geeralize this result to x matrices. 11. Recall that A is symmetric if AT = A. If A is symmetric, show that det(a + B) = det(a + BT). 12. Recall that a matrix A = (aáé) is said to be atisymmetric if AT = -A, i.e., atáé = -aéá. If A is a atisymmetric square matrix of odd size, prove that det A = (a) Recall (see Exercise 3.6.7) that if A! M(F), the Tr A = Íá aáá. If A is a 2 x 2 matrix, prove that det(i + A) = 1 + det A if ad oly if Tr A = 0. Is this true for ay size square matrix? (b) If \aij\ «1, show det(i + A) _ 1 + Tr A. 14. Two matrices A ad Aæ are said to be similar if there exists a osigular matrix P such that Aæ = PAPî. The operatio of trasformig A ito Aæ i this maer is called a similarity trasformatio. (a) Show that this defies a equivalece relatio o the set of all matrices. (b) Show that the determiat is ivariat uder a similarity trasformatio. (c) Show that the trace (Exercise 3.6.7) is also ivariat. 15. Cosider the matrices " 2!!0!1% "!3!2!4% A =! 3!!0!!2!!!!!!!!!!!!!!!!B =!!1!0!2 4!3!!7!2!3!!3 (a) Evaluate det A ad det B.

16 4.2 ADDITIONAL PROPERTIES OF DETERMINANTS 185 (b) Fid AB ad BA. (c) Evaluate det AB ad det BA. 16. Show that a 1 b 1 + xa 1 c 1 + yb 1 + za 1 a 1 b 1 c 1 a 2 b 2 + xa 2 c 2 + yb 2 + za 2 = a 2 b 2 c 2!!. a 3 b 3 + xa 3 c 3 + yb 3 + za 3 a 3 b 3 c Fid all values of x for which each of the followig determiats is zero: (a)!! x! x! x! 2 1 x x!!!!!!!!!!!!(b)!! x 1 x x x 1 1 x x 2 (c)!! Show that (a) det(aè + Aì, Aì + A3, A3 + Aè) = 2 det(aè, Aì, A3). (b) det(aè + Aì, Aì + A3, A3 + A4, A4 + Aè) = Give a matrix A, the matrix that remais after ay rows ad/or colums of A have bee deleted is called a submatrix of A, ad the determiat of a square submatrix is called a subdetermiat. Show that the rak of a matrix A is the size of the largest ovaishig subdetermiat of A. [Hit: Thik about Theorem 3.9, Corollary 2 of Theorem 4.2, ad Theorem 4.4.] 20. Show that the followig determiat is zero: a 2 (a +1) 2 (a + 2) 2 (a + 3) 2 b 2 (b +1) 2 (b + 2) 2 (b + 3) 2 c 2 (c +1) 2 (c + 2) 2 (c + 3) 2 d 2 (d +1) 2 (d + 2) 2 (d + 3) 2 [Hit: You eed ot actually evaluate it.]

17 186 DETERMINANTS 21. Show that = 0!!. 22. (a) If E is a elemetary matrix, show (without usig Theorem 4.1) that det ET = det E. (b) Use Theorem 3.24 to show that det AT = det A for ay A! M(F). 23. Use the material of this sectio to give a proof (idepedet of Chapter 3) that the product of osigular matrices is osigular. 4.3 EXPANSION BY MINORS We ow tur our attetio to methods of evaluatig determiats. Sice Sñ cotais! elemets, it is obvious that usig our defiitio of det A becomes quite impractical for ay much larger tha four. Before proceedig with the geeral theory of miors, let us first preset a method of evaluatig determiats that is based o Theorem 4.5. All we have to do is reduce the matrix to triagular form, beig careful to keep track of each elemetary row operatio alog the way, ad use Theorem 4.4. Oe example should suffice to illustrate the procedure. Example 4.2 Cosider the matrix A give by The we have " 2!1 3 % A =! 1 2!1!!.!3 0 2!!2!1!!3 det A =!!1!!2!1!3!!0!!2!!1! " (1 2)A 1 = 2!!1!2!1!3!0!!2

18 4.3 EXPANSION BY MINORS 187 1!1 2!!!3 2 = 2 0!!5 2!!5 2 "!A 1 + A 2 0!!3 2!!13 2 " 3A 1 + A 3 1!1 2!!!3 2 = 2 0!!5 2!!5 2 0!!0!!!5 " (3 5)A 2 + A 3 = (2)(1)(5/2)(5) = 25. The reader should verify this result by the direct calculatio of det A. We ow begi our discussio of the expasio by miors. Suppose A = (aáé)! M(F), ad ote that for ay term ar ßr i the defiitio of det A ad for ay s = 1,..., we may factor out all terms with ßr = s ad the sum over all s. This allows us to write det A =! "S (sg! )a 1!!1!!!a r!! r!!!a!! = a rs (sg! )a 1!!1!!!a r1!! (r1) a r+1!! (r+1)!!!a!! s=1 = a rs % s=1! "S,!! r=s a rs where Íß S, ßr = s meas to sum over all ß! Sñ subject to the coditio that ßr = s, ad a rs! = % (sg" )a 1!"1!!!a r1!" (r1) a r+1!" (r+1)!!!a!"!!. " S,!" r=s The term aærs is called the cofactor of ars. Sice the sum is over all ß! Sñ subject to the coditio ßr = s, we see that aærs cotais ( - 1)! terms. Ideed, it should be apparet that aærs looks very much like the determiat of some ( - 1) x ( - 1) matrix. To see that this is i fact the case, we defie the matrix Ars! M-1(F) to be the matrix obtaied from A! M(F) by deletig the rth row ad sth colum of A. The matrix Ars is called the rsth mior matrix of A, ad det Ars is

19 188 DETERMINANTS called the rsth mior of A. We ow prove the method of calculatig det A kow as the expasio by miors. Theorem 4.10 have where Suppose A = (aáé)! M(F). The for ay r = 1,..., we " a rs det A = a rs! s=1 a rs! = ("1) r+s det A rs!!. Proof We saw above that det A = Ís ˆ= 1arsaærs where each aærs depeds oly o those elemets of A that are ot i the rth row or the sth colum. I particular, cosider the expasio det A = aèèaæèè + ~ ~ ~ + aèñaæèñ ad look at the coefficiet aæèè of aèè. By defiitio, we have a 11! = " S,!"1=1 (sg!" )a 2!" 2!!!a!" where each term i the sum is a product of elemets, oe from each row ad oe from each colum of A except for the first row ad colum, ad the sum is over all possible permutatios of the remaiig - 1 colums. But this is precisely the defiitio of det Aèè, ad hece aæèè = det Aèè. (Remember that the rth row of Aèè is (Aèè)r = (ar+1 2,..., ar+1 ).) We ow eed to fid the coefficiet aærs for ay ars. To do this, we start with the matrix A, ad the move the rth row up to be the ew first row, ad move the sth colum left to be the ew first colum. This defies a ew matrix B such that bèè = ars ad Bèè = Ars (ote this implies that det Bèè = det Ars). Movig the rth row ivolves r - 1 iterchages, ad movig the sth colum ivolves s - 1 iterchages. Hece applyig Corollary 1 of Theorem 4.3 to both rows ad colums, we see that det B = (-1)r+s-2 det A = (-1)r+s det A. If we expad det B by the first row ad expad det A by the rth row, we fid b b 1p! = det B = ("1) r+s det A = ("1) r+s a 1p rp a rp!!!. p=1 Now remember that the set {bèè,..., bèñ} is just the set {ar1,..., ar} take i a differet order where, i particular, bèè = ars. Sice the rth row of A is arbitrary, we may assume that arj = 0 for all j " s. I this case, we have p=1

20 4.3 EXPANSION BY MINORS 189 det B = b11bæ11 = (-1)r+s det A = (-1)r+s ars aærs where b11 = ars, ad therefore or bæ11 = (-1)r+s aærs aærs = (-1)r+s bæèè. At the begiig of this proof we showed that aæ11 = det Aèè, ad hece a idetical argumet shows that bæèè = det Bèè. Puttig all of this together the results i aærs = (-1)r+s bæèè = (-1)r+s det Bèè = (-1)r+s det Ars. The reader may fid it helpful to repeat this proof by movig the rth row dow ad the sth colum to the right so that bññ = ars. I this case, istead of aèè we cosider a! = % " S,!" = (sg!" )a 1!"1!!!a 1!" (1) which is just det Aññ sice Aññ! M-1(F) ad the sum over all ß! Sñ subject to ß = is just the sum over all ß! S-1. It the follows agai that bæññ = det Bññ = det Ars ad bæññ = (-1)r+s aærs. Corollary 1 Usig the same otatio as i Theorem 4.10, for ay s = 1,..., we have det A = " a rs a rs!!!. r=1 (Note that here det A is expaded by the sth colum, whereas i Theorem 4.10 det A was expaded by the rth row.) Proof This follows by applyig Theorem 4.10 to AT ad the usig Theorem 4.1. Theorem 4.10 is called expasio by miors of the rth row, ad Corollary 1 is called expasio by miors of the sth colum. (See also Exercise ) Corollary 2 Usig the same otatio as i Theorem 4.10, we have

21 190 DETERMINANTS " a a ks! = 0 if k r ks s=1 " r=1 a rk a rk! = 0 if k s!!. Proof Give A = (aáé)! M(F), we defie B! M(F) by Bá = Aá for i " r ad Br = AÉ where r " k. I other words, we replace the rth row of A by the kth row of A to obtai B. Sice B has two idetical rows, it follows that det B = 0. Notig that Brs = Ars (sice both mior matrices delete the rth row), we see that bærs = aærs for each s = 1,...,. We therefore have 0 = det B = " b rs b rs! = " b rs a rs! = " a ks a rs!!!. s=1 Similarly, the other result follows by replacig the sth colum of A by the kth colum so that brs = ark, ad the usig Corollary 1. s=1 Example 4.3 Cosider the matrix A give by " 2!1!!5% A =! 0!!3!!4!!. 1!!2!3 To illustrate the termiology, ote that the (2, 3) mior matrix is give by " A 23 = 2!1 % 1!!2 ad hece the (2, 3) mior of A is det A23 = 4 - (-1) = 5, while the (2, 3) cofactor of A is (-1)2+3 det A23 = -5. We ow wish to evaluate det A. Expadig by the secod row we have s=1 det A = a 21 a 21! + a 22 a 22! + a 23 a 23! = 0 + ("1) 4 (3) 2!!5 1 "3 + ("1)5 (4) 2 "1 1!!2 = 3("6 " 5) " 4(4 +1) = "53!!. The reader should repeat this usig other rows ad colums to see that they all yield the same result.

22 4.3 EXPANSION BY MINORS 191 Example 4.4 Let us evaluate det A where A is give by "!!5!!4!2!!1%!!2!!3!!1!2 A =!!.!5!7!3!!9!1!2!1!!4 I view of Theorem 4.4, we first perform the followig elemetary row operatios o A: (i) Aè Aè - 2Aì, (ii) A3 A3 + 3Aì, (iii) A4 A4 + Aì. This results i the followig matrix B: " 1!2!0!!5% 2!3!1!2 B =!!. 1!2!0!3 3!1!0!2 Sice these were all type operatios it follows that det B = det A, ad hece expadig by miors of the third colum yields oly the sigle term 1!2!5 det A = (!1) 2+3 1!!2!3!!. 3!!1!2 This is easily evaluated either directly or by reducig to a sum of three 2 x 2 determiats. I ay case, the result is det A = 38. We are ow i a positio to prove a geeral formula for the iverse of a matrix. Combiig Theorem 4.10 ad its corollaries, we obtai (for k, r, s = 1,..., ) a a ks! = " det A (1a) rs kr s=1 a a rk! = " det A (1b) rs ks r=1 Sice each aæáé! F, we may use the them to form a ew matrix (aæáé)! M(F). The traspose of this ew matrix is called the adjoit of A (or sometimes the classical adjoit to distiguish it from aother type of adjoit to be discussed later) ad is deoted by adj A. I other words, adj A = (aæáé)t.

23 192 DETERMINANTS Notig that Iáé = áé, it is ow easy to prove the followig. Theorem 4.11 For ay A! M(F) we have A(adj A) = (det A)I = (adj A)A. I particular, if A is osigular, the A!1 = adj A det A!!. Proof Usig (adj A)sr = aærs, we may write equatio (1a) i matrix form as A(adj A) = (det A)I ad equatio (1b) as (adj A)A = (det A)I. Therefore, if A is osigular the det A " 0 (Theorem 4.6), ad hece A(adj A)/det A = I = (adj A)A/det A. Thus the uiqueess of the iverse (Theorem 3.21, Corollary 1) implies that Aî = (adj A)/det A. It is importat to realize that the equatios ad A(adj A) = (det A)I (adj A)A = (det A)I are valid eve if A is sigular. We will use this fact i Chapter 8 whe we preset a very simple proof of the Cayley-Hamilto Theorem. Example 4.5 Let us use this method to fid the iverse of the matrix "!1!!2!1 % A =!!!0!!3!2!!2!1!!0 used i Example Leavig the details to the reader, we evaluate the cofactors usig the formula aærs = (-1)r+s det Ars to obtai aæèè = -2, aæèì = -4,

24 4.3 EXPANSION BY MINORS 193 aæ13 = -6, aæ21 = -1, aæ22 = -2, aæ23 = 3, aæ31 = -7, aæ32 = -2, ad aæ33 = -3. Hece we fid "!2!1!7% adj A =!!4!2!2!!.!6!3!3 To evaluate det A, we may either calculate directly or by miors to obtai det A = -12. Alteratively, from equatio (1a) we have "!1!!2!!1% "!2!1!7% (det A)I = A(adj A) =!!!0!!3!2!!4!2!2!!2!1!!0!6!3!3 "!12!!0!!0 % " 1 0 0% =!!!0!12!!0 =!12! 0 1 0!!0!!0! so that we agai fid that det A = -12. I ay case, we see that A!1 = adj A " %!12 =! ! which agrees with Example 3.11 as it should. If the reader thiks about Theorem 3.9, Corollary 2 of Theorem 4.2, ad Theorem 4.4 (or has already worked Exercise ), our ext theorem should come as o real surprise. By way of more termiology, give a matrix A, the matrix that remais after ay rows ad/or colums have bee deleted is called a submatrix of A. (A more precise defiitio is give i Sectio 4.6.) Theorem 4.12 Let A be a matrix i Mmx(F), ad let k be the largest iteger such that some submatrix B! Mk(F) of A has a ozero determiat. The r(a) = k. Proof Sice B is a k x k submatrix of A with det B " 0, it follows from Theorem 4.6 that B is osigular ad hece r(b) = k. This meas that the k rows of B are liearly idepedet, ad hece the k rows of A that cotai the rows of B must also be liearly idepedet. Therefore r(a) = rr(a) k. By defiitio of k, there ca be o r x r submatrix of A with ozero determiat

25 194 DETERMINANTS if r > k. We will ow show that if r(a) = r, the there ecessarily exists a r x r submatrix with ozero determiat. This will prove that r(a) = k. If r(a) = r, let Aæ be the matrix with r liearly idepedet rows Aiè,..., Ai. Clearly r(aæ) = r also. But by defiitio of rak, we ca also choose r liearly idepedet colums of Aæ. This results i a osigular matrix AÆ of size r, ad hece det AÆ " 0 by Theorem 4.6. Exercises 1. Verify the result of Example 4.2 by direct calculatio. 2. Verify the result of Example Verify the terms aæáé i Example Evaluate the followig determiats by expadig by miors of either rows or colums: 2!1!!5 (a)!! 0!!3!!4 1!!!2!3!2!5!5!3!7!8!2!3 (b)!!!1!1!4!2!3!9!1!3!!3!!2!!2!3!!1!4!!2!1 (c)!!!!4!!5!1!0!1!4!!2!7 3 1!!0!4!!2!!1 2 0!!1!0!!5!!1 0 4!1!1!1!!2 (d)!! 0 0!!0!2!!0!!1 0 0!!0!0!!1!1 0 0!!0!1!!0!!1 5. Let A! M(F) be a matrix with 0 s dow the mai diagoal ad 1 s elsewhere. Show that det A = - 1 if is odd, ad det A = 1 - if is eve. 6. (a) Show that the determiat of the matrix is (c - a)(c - b)(b - a).! 1 a a 2! 1 b b 2 " 1 c c 2 %

26 4.3 EXPANSION BY MINORS 195 (b) Cosider the matrix Vñ! M(F) defied by Prove that " 1 x 1 x 2 1! x!1 1 % 1 x 2 x 2 2! x!1 2!!. " " " " 1 x x 2! x!1 " detv = (x j! x i ) i< j where the product is over all pairs i ad j satisfyig 1 i, j. This matrix is called the Vadermode matrix of order. [Hit: This should be doe by iductio o. The idea is to show that det Vñ = (x2 - x1)(x3 - x1) ~ ~ ~ (x-1 - x1)(x - x1) det V-1. Perform elemetary colum operatios o Vñ to obtai a ew matrix Væñ with a 1 i the (1, 1) positio ad 0 s i every other positio of the first row. Now factor out the appropriate term from each of the other rows.] 7. The obvious method for decidig if two quadratic polyomials have a commo root ivolves the quadratic formula, ad hece takig square roots. This exercise ivestigates a alterative root free approach. (While we will defie roots of polyomials i a later chapter, we assume that the reader kows that xà is a root of the polyomial p(x) if ad oly if p(xà) = 0.) (a) Show that det A = 1!(x 1 + x 2 ) x 1 x !(x 1 + x 2 ) x 1 x 2 1!(y 1 + y 2 ) y 1 y !(y 1 + y 2 ) y 1 y 2 = (x 1! y 1 )(x 1! y 2 )(x 2! y 1 )(x 2! y 2 )!!. (b) Usig this result, show that the polyomials a 0 x 2 + a 1 x + a 2!!!!!!(a 0! 0) b 0 x 2 + b 1 x + b 2!!!!!!(b 0! 0)

27 196 DETERMINANTS have a commo root if ad oly if a 0 a 1 a a 0 a 1 a 2 b 0 b 1 b 2 0 = 0!!. 0 b 0 b 1 b 2 [Hit: Note that if xè ad xì are the roots of the first polyomial, the (x - xè)(x - xì) = x2 + (aè/aà)x + aì/aà ad similarly for the secod polyomial.] 8. Show that!!x!!0!0!0!!!0 a 0!1!!x!0!0!!!0 a 1!!0!1!x!0!!!0 a 2!!"!!"!"!"!!" "!!0!!0!0!0!!1 a +1 + x = x + a!1 x!1 +!!!+a 0!!. Explai why this shows that give ay polyomial p(x) of degree, there exists a matrix A! M(F) such that det(xi - A) = p(x). (We will discuss the matrix A i detail i Chapter 8.) 9. Cosider the followig real matrix: " a!!b!!c!!d% b!a!!d!c A =!!. c!d!a!!b d!!c!b!a Show that det A = 0 implies that a = b = c = d = 0. [Hit: Fid AAT ad use Theorems 4.1 ad 4.8.] 10. Cosider the usual xy-plae 2. The the two vectors x = (xè, xì) ad y = (yè, yì) defie a quadrilateral with vertices at the poits (0, 0), (xè, xì), (yè, yì) ad (xè + yè, xì + yì). Show that (up to a sig) the area of this quadrilateral is give by

28 4.3 EXPANSION BY MINORS 197 x 1 x 2 y 1 y 2!!. [Hit: If the vectors x ad y are both rotated by a agle œ to obtai the ew vectors xæ = (x1æ, x2æ) ad yæ = (y1æ, y2æ), the clearly the area of the quadrilateral will remai uchaged. Show (usig Example 1.2) it is also true that x 1! x 2! y 1! y 2! = x 1 x 2 y 1 y 2 ad hece you may choose whatever œ you wish to simplify your calculatio.] What do you thik the differet sigs mea geometrically? We shall have quite a bit more to say o this subject i Chapter Let u, v ad w be three vectors i 3 with the stadard ier product, ad cosider the determiat G(u, v, w) (the Gramia of {u, v, w}) defied by G(u,!v,!w) = u,!u u,!v u,!w v,!u v,!v v,!w w,!u w,!v w,!w!!. Show that G(u, v, w) = 0 if ad oly if {u, v, w} are liearly depedet. As we shall see i Chapter 11, G(u, v, w) represets the volume of the parallelepiped i 3 defied by {u, v, w}.) 12. Fid the iverse (if it exists) of the followig matrices: " 1!1!2% (a)!! 1!!2!0 4!!1!3 "!2!2!3% (c)!!!!4!3!6!1!1!2! (b)!!! " 3 2 1% "!!8!!2!!5% (d)!!!!7!!3!4!!9!6!!4

29 198 DETERMINANTS! (e)!! " %! ( f )!! " % " 2!3!!2 4 % 4!6!!5 5 (g)!! 3!5!!2 14 2!2! Fid the iverse of cos! % "si! "si! (!!. "cos! 14. Show that the iverse of the matrix is give by " a 1!a 2!a 3!a 4 % a Q = 2!!a 1!a 4!!a 3 a 3!!a 4!!a 1!a 2 a 4!a 3!!a 2!!a 1 Q T Q!1 = a 12 + a 22 + a 32 + a Suppose that a -square matrix A is ilpotet (i.e., Ak = 0 for some iteger k > 0). Prove that Iñ + A is osigular, ad fid its iverse. [Hit: Note that (I + A)(I - A) = I - A2 etc.] 16. Let P! M(F) be such that P2 = P. If " 1, prove that Iñ - P is ivertible, ad that (I! "P)!1 = I + " 1! " P!!. 17. If A = (aáé) is a symmetric matrix, show that (aæáé) = (adj A)T is also symmetric. 18. If a, b, c!, fid the iverse of

30 4.3 EXPANSION BY MINORS 199 "!!1!!a!b%!!a!!1!c!!.!b!c!1 19. Prove that Corollary 2 of Theorem 4.3 is valid over a field of characteristic 2. [Hit: Use expasio by miors.] 20. (a) Usig Aî = (adj A)/det A, show that the iverse of a upper (lower) triagular matrix is upper (lower) triagular. (b) If a " 0, fid the iverse of! a b c d 0 a b c!!. 0 0 a b " a% 21. Let A! M( ) have all iteger etries. Show that the followig are equivalet: (a) det A = ±1. (b) All etries of Aî are itegers. 22. For each of the followig matrices A, fid the value(s) of x for which the characteristic matrix xi - A is ivertible.! (a)!! 2 0! (b)!! 1 1 " 0 3% " 1 1%! (c)!!! " 0 1 0% " 0!1!!2% (d)!!! 0!1!!3 0!0!1 23. Let A! M(F) have exactly oe ozero etry i each row ad colum. Show that A is ivertible, ad that its iverse is of the same form. 24. If A! M(F), show that det(adj A) = (det A) Show that A is osigular if ad oly if adj A is osigular.

31 200 DETERMINANTS 26. Usig determiats, fid the rak of each of the followig matrices: " % (a)!!! "! % (b)!!!!! ! DETERMINANTS AND LINEAR EQUATIONS Suppose that we have a system of equatios i ukows which we write i the usual way as! a ij x j = b i!,!!!!!!!!!!i = 1,!!,!!!. j=1 We assume that A = (aáé)! M(F) is osigular. I matrix form, this system may be writte as AX = B as we saw earlier. Sice A is osigular, Aî exists (Theorem 3.21) ad det A " 0 (Theorem 4.6). Therefore the solutio to AX = B is give by (adj A)B X = A!1 B = det A!!. But adj A = (aæáé)t so that (adj A) x j = ji b i a"! = ij b! i!!. det A det A i=1 From Corollary 1 of Theorem 4.10, we see that Íábáaæáé is just the expasio by miors of the jth colum of the matrix C whose colums are give by Ci = Ai for i " j ad Cj = B. We are thus led to the followig result, called Cramer s rule. Theorem 4.13 If A = (aáé)! M(F) is osigular, the the system of liear equatios has the uique solutio i=1! a ij x j = b i!,!!!!!!!!!!i = 1,!!,! j=1 x j = 1 det A det(a1,!!,!a j!1,!b,!a j+1,!!,!a )!!.

32 4.4 DETERMINANTS AND LINEAR EQUATIONS 201 Proof This theorem was actually proved i the precedig discussio, where uiqueess follows from Theorem However, it is istructive to give a more direct proof as follows. We write our system as ÍAi xá = B ad simply compute usig Corollary 1 of Theorem 4.2 ad Corollary 2 of Theorem 4.3: det(a1,..., Aj-1, B, Aj+1,..., A) = det(a1,..., Aj-1, ÍAi xá, Aj+1,..., A) = Íxá det(a1,..., Aj-1, Ai, Aj+1,..., A) = xé det(a1,..., Aj-1, Aj, Aj+1,..., A) = xé det A. Corollary A homogeeous system of equatios! j=1 a ij x j = 0!,!!!!!!!!!!i = 1,!!,! has a otrivial solutio if ad oly if det A = 0. Proof We see from Cramer s rule that if det A " 0, the the solutio of the homogeeous system is just the zero vector (by Corollary 2 of Theorem 4.2 as applied to colums istead of rows). This shows that the if the system has a otrivial solutio, the det A = 0. O the other had, if det A = 0 the the colums of A must be liearly depedet (Theorem 4.6). But the system Íéaáéxé = 0 may be writte as ÍéAj xj = 0 where Aj is the jth colum of A. Hece the liear depedece of the Aj shows that the xj may be chose such that they are ot all zero, ad therefore a otrivial solutio exists. (We remark that this corollary also follows directly from Theorems 3.12 ad 4.6.) Example 4.6 Let us solve the system 5x + 2y +!!!z = 3 2x!!!!y + 2z = 7 x + 5y!!!!z = 6 We see that A = (aáé) is osigular sice 5!!2!!1 2!1!!2 =!26 " 0!!. 1!!5!1

33 202 DETERMINANTS We the have x =!1 26 3!!2!!1 7!1!!2 6!!5!1 = (!1 26)(52) =!2 y =!1 26 5!3!!1 2!7!!2 1!6!1 = (!1 26)(!78) = 3 z =!1 26 5!!2!3 2!1!7 1!!5!6 = (!1 26)(!182) = 7!!. Exercises 1. Usig Cramer s rule, fid a solutio (if it exists) of the followig systems of equatios: (a)!!3x +!!y!!!z = 0 x!!!y + 3z =!1 2x + 2y +!!z = 7 (c)!!2x! 3y +!!!z = 10!x + 3y + 2z =!2 4x + 4y + 5z =!!4 (b)!!2x +!!y + 2z =!!!0 3x! 2y +!!z =!!!1!x + 2y + 2z =!7 (d)!!x + 2y! 3z +!!t =!9 2x +!!!y + 2z!!!t =!!3!x +!!!y + 2z!!!t =!!0 3x + 4y +!!z + 4t =!!3 2. By calculatig the iverse of the matrix of coefficiets, solve the followig systems: (a)!!2x! 3y +!!z = a x + 2y + 3z = b 3x!!!y + 2z = c (b)!!x + 2y + 4z = a!x + 3y! 2z = b 2x!!!y +!!!z = c

34 4.4 DETERMINANTS AND LINEAR EQUATIONS 203 (c)!!2x +!!y + 2z! 3t = a 3x + 2y + 3z! 5t = b 2x + 2y +!!z!!!!t = c 5x + 5y + 2z! 2t = d (d)!!6x + y + 4z! 3t = a 2x! y!!!!!!!!!!!!!!!= b x + y +!!!z!!!!!!!= c!3x! y! 2z +!!t = d 3. If det A " 0 ad AB = AC, show that B = C. 4. Fid, if possible, a 2 x 2 matrix X that satisfies each of the give equatios:! (a)!! 2 3! X 3 4! = 1 2 " 1 2% " 2 3% " 2 1%! (b)!! 0 1! X 1 1! = 2 1 " 1 0% " 0 1% " 3 2% 5. Cosider the system ax + by =! + "t cx + dy = +t where t is a parameter, " 0 ad a b c d! 0!!. Show that the set of solutios as t varies is a straight lie i the directio of the vector! a " c b d% 1!( ") %!!. 6. Let A, B, C ad D be 2 x 2 matrices, ad let R ad S be vectors (i.e., 2 x 1 matrices). Show that the system AX + BY = R CX + DY = S ca always be solved for vectors X ad Y if

35 204 DETERMINANTS a 11 a 12 b 11 b 12 a 21 a 22 b 21 b 22 c 11 c 12 d 11 d 12! 0!!. c 21 c 22 d 21 d BLOCK MATRICES There is aother defiitio that will greatly facilitate our discussio of matrix represetatios of operators to be preseted i Chapter 7. I particular, suppose that we are give a matrix A = (aáé)! Mmx(F). The, by partitioig the rows ad colums of A i some maer, we obtai what is called a block matrix. To illustrate, suppose A! M3x5( ) is give by " 7!5!!5!4!1% A = 2!1!3!0!!5!!. 0!8!!2!1!9 The we may partitio A ito blocks to obtai (for example) the matrix where! A = A 11 A 12 " A 21 A 22 % A 11 = ( 7 5 5)!!!!!!!!!!!!!!!!!!!A 12 = ( 4!1) " 2!1!3% " A 21 =!!!!!!!!!!!!!!!A 22 = 0!!5 % 0!8!!2 1!9 (do ot cofuse these Aáé with mior matrices). If A ad B are block matrices that are partitioed ito the same umber of blocks such that each of the correspodig blocks is of the same size, the it is clear that (i a obvious otatio)! A 11 + B 11! A 1 + B 1 A + B =! " "!!. " A m1 + B m1! A m + B m % I additio, if C ad D are block matrices such that the umber of colums i each Cáé is equal to the umber of rows i each DéÉ, the the product of C ad D is also a block matrix CD where (CD)áÉ = Íé Cáé DéÉ. Thus block matrices

36 4.5 BLOCK MATRICES 205 are multiplied as if each block were just a sigle elemet of each matrix i the product. I other words, each (CD)áÉ is a matrix that is the sum of a product of matrices. The proof of this fact is a exercise i matrix multiplicatio, ad is left to the reader (see Exercise 4.5.1). Theorem 4.14 If A! M(F) is a block triagular matrix of the form! A 11 A 12 A 13! A 1k 0 A 22 A 23! A 2k " " " " " 0 0 0! A kk % where each Aáá is a square matrix ad the 0 s are zero matrices of appropriate size, the k det A =! det A ii!!. i=1 Proof What is probably the simplest proof of this theorem is outlied i Exercise However, the proof that follows serves as a good illustratio of the meaig of the terms i the defiitio of the determiat. We first ote that oly the diagoal matrices are required to be square matrices. Because each Aii is square, we ca simply prove the theorem for the case k = 2, ad the geeral case will the follow by iductio. We thus let A = (aáé)! M(F) be of the form! B C " 0 D% where B = (báé)! Mr(F), D = (dáé)! Ms(F), C = (cáé)! Mrxs(F) ad r + s =. Note that for 1 i, j r we have aáé = báé, for 1 i, j s we have ai+r j+r = dáé, ad if i > r ad j r the aáé = 0. From the defiitio of determiat we have det A =! " S (sg" )a 1!"1!!!a r!" r a r+1!" (r+1)!!!a!"!!. By defiitio, each ß! Sñ is just a rearragemet (i.e., permutatio) of the elemets i Sñ. This meas that for each ß! Sñ with the property that ßi > r for some i r, there must be some iæ > r such that ßiæ r. The for this iæ we have aiæßiæ = 0, ad hece each term i det A that cotais oe of these factors is zero. Therefore each ozero term i the above sum must be over oly those permutatios ß such that ßi > r if i > r (i.e., the block D), ad ßi r if i r (i.e., the block B).

37 206 DETERMINANTS To separate the actio of the allowed ß o the blocks B ad D, we defie the permutatios å! Sr ad! Ss by åi = ßi for 1 i r, ad i = ß(i + r) - r for 1 i s. I other words, each of the allowed ß is just some rearragemet å of the values of i for 1 i r alog with some rearragemet of the values of i for r < i r + s =, ad there is o mixig betwee these blocks. The permutatios å ad are thus idepedet, ad sice sg ß is defied by the umber of traspositios ivolved, this umber simply separates ito the umber of traspositios i å plus the umber of traspositios i. Therefore sg ß = (sg å)(sg ). The result of this is that all ozero terms i det A are of the form (sg!)b 1!!1!!!b r!!r (sg ")d 1!"1!!!d s!"s!!. Furthermore, every term of this form is icluded i the expressio for det A, ad hece det A = (det B)(det D). There is aother way to prove this theorem that is based o our earlier formula D(A) = (det A)D(I) (2) where D is ay multiliear ad alteratig fuctio o the set of all x matrices (see the discussio followig the proof of Theorem 4.9). To show this, cosider the block triagular matrix! A B " 0 C% where A ad C are square matrices. Suppose we defie the fuctio D(A,!B,!C) = A B 0 C!!. If we cosider the matrices A ad B to be fixed, the this is clearly a multiliear ad alteratig fuctio of the rows of C. Applyig equatio (2), this may be writte as D(A, B, C) = (det C) D(A, B, I) where D(A,!B,!I) = A B 0 I!!.

38 4.5 BLOCK MATRICES 207 But usig Theorem 4.4(c) it should be obvious that we ca subtract suitable multiples of the rows of I from the matrix B so that D(A, B, I) = D(A, 0, I). Applyig the same reasoig, we observe that D(A, 0, I) is a multiliear ad alteratig fuctio of the rows of A, ad hece (2) agai yields D(A, 0, I) = (det A) D(I, 0, I). Puttig all of this together alog with the obvious fact that D(I, 0, I) = 1, we obtai D(A,!B,!C) = (detc)d(a,!b,!i) which agrees with Theorem Example 4.7 Cosider the matrix = (detc)d(a,!0,!i) = (detc)(det!a)d(i,!0,!i) = (detc)(det A) " 1!1!!2!!3% 2!!2!!0!!2 A =!!. 4!!1!1!1 1!!2!!3!!0 By the additio of suitable multiples of oe row to aother, it is easy to rowreduce A to the form " 1!1!2!3 % 0!4!4!4 B = 0 0!4!8 0 0!4!0 with det B = det A. Sice B is i block triagular form we have det A = det B = 1!1 0!!4!4!8 = 4(32) = 128!!.!!4!!0

39 208 DETERMINANTS Exercises 1. Prove the multiplicatio formula give i the text (just prior to Theorem 4.14) for block matrices. 2. Suppose A! M(F), D! Mm(F), U! Mxm(F) ad V! Mmx(F), ad cosider the ( + m) x ( + m) matrix! M = A U!!. " V D% If Aî exists, show that " A!1 0 %!VA!1 " A V I m U % " = I A!1 U % D 0!VA!1 U + D ad hece that A V U D = (det A)det(D!VA!1 U)!!. 3. Let A be a block triagular matrix of the form! A = B C " 0 D% where B ad D are square matrices. Prove det A = (det B)(det D) by usig elemetary row operatios o A to create a block triagular matrix! B!A =! C! " 0 D! % where Bÿ ad Dÿ are upper-triagular. 4. Show! A " C T B D%! = AT C T " B T D T!!. % 4.6 THE CAUCHY-BINET THEOREM This sectio deals with a geeralizatio of Theorem 4.8 that is ot used aywhere i this book except i Chapter 8. Because of this, the reader should feel

40 4.6 THE CAUCHY-BINET THEOREM 209 free to skip this sectio ow ad come back to it if ad whe it is eeded. Let us first poit out that while Theorem 4.8 dealt with the product of square matrices, it is evertheless possible to formulate a similar result eve i the case where A! Mxp(F) ad B! Mpx(F) (so that the product AB! M(F) is square ad det AB is defied). This result is the mai subject of this sectio, ad is kow as the Cauchy-Biet theorem. Before proceedig with our discussio, we briefly preset some ew otatio that will simplify some of our formulas sigificatly, although at first it may seem that all we are doig is complicatig thigs eedlessly. Suppose that we have a matrix A with m rows ad colums. The we ca easily form a ew matrix B by cosiderig oly those elemets of A belogig to, say, rows 2 ad 3 ad colums 1, 3 ad 4. We shall deote this submatrix B of A by B = A[2, 3\1, 3, 4]. What we ow wish to do is make this defiitio precise. To begi with, let k ad be positive itegers with 1 k. We let MAP(k, ) deote the set of all mappigs from the set kõ = (1,..., k) to the set Õ = (1,..., ). For example, if = 5 ad k = 3, we ca defie å! MAP(3, 5) by å(1) = 2, å(2) = 5 ad å(3) = 3. Note that a arbitrary å! MAP(k, ) eed ot be ijective so that, for example, åæ! MAP(3, 5) could be defied by åæ(1) = 2, åæ(2) = 5 ad åæ(3) = 5. I fact, we will let INC(k, ) deote the set of all strictly icreasig fuctios from the set kõ ito the set Õ. Thus if! INC(k, ), the (1) < (2) < ~ ~ ~ < (k). We also deote the mappig by simply the k-tuple of umbers = ((1),..., (k)). Note that this k-tuple cosists of k distict itegers i icreasig order. Now cosider the set of all possible permutatios of the k itegers withi each k-tuple for every! INC(k, ). This yields the set INJ(k, ) cosistig of all ijective mappigs from the set kõ ito the set Õ. I other words, if! INJ(k, ), the = ((1),..., (k)) is a k-tuple of k distict itegers i ay (i.e., ot ecessarily icreasig) order. I the particular case that k =, we see that the set INJ(, ) is just the set of all permutatios of the itegers 1,...,. The set INJ(, ) will be deoted by PER(). (Note that the set INJ(, ) is the same as the set S, but without the additioal group structure.) Now suppose that A = (aáé)! Mmx(F), let å = (iè,..., ié)! INC(k, m) ad let = (jè,..., jt)! INC(t, ). The the matrix B! Mkxt(F) whose (r, s)th etry is ai j (where 1 r k ad 1 s t) is called the submatrix of A lyig i rows å ad colums. We will deote this submatrix by A[å\]. Similarly, we let A(å\]! M(m-k)xt(F) deote the submatrix of A whose rows are precisely those complemetary to å, ad whose colums are agai give by. It should be clear that we ca aalogously defie the matrices A(å\)! M(m-k)x(- t)(f) ad A[å\)! Mkx(- t)(f). Fortuately, these ideas are more difficult to state carefully tha they are to uderstad. Hopefully the ext example should clarify everythig.