Linear Algebra Existence of the determinant. Expansion according to a row.

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1 Lir Algbr Existc of th dtrmit. Expsio ccordig to row. W dfi th dtrmit for 1 1 mtrics s dt([]) = (1) It is sy chck tht it stisfis D1)-D3). For y othr w dfi th dtrmit s follows. Assumig th dtrmit is dfid lrdy for ( 1) ( 1) mtrics, for y A R dfi dt(a) usig xpsio ccordig to th i-th row: dt(a) = j=1 ( 1) i+j i j dt(a i j ) (2) whr A i j is ( 1) ( 1) mtrix tht is mtrix A without th i-th row d j-th colum. For xmpl (usig xmpl from th txtbook t pg 165), if Notic tht th formul for th dtrmit of 2 2 mtrix, which ws prstd i th lctur: is comptibl with xpsio ccordig to th first row: dt(a) = dt ([ ]) = (3) ( 1) dt(a 1 1 ) + ( 1) dt(a 1,2 ) = 1 1 dt([ 2 2 ]) 2 1 dt([ 1,2 ]) = (4) whr t th lst stp w usd th dfiitio of th dtrmit for 1 1 mtrics. This dfiitio is lso comptibl with xpsio ccordig to th scod row: ( 1) dt(a 2 1 ) + ( 1) dt(a 2,2 ) = 2 1 dt([ 1 2 ]) dt([ 1,1 ]) = (5) I th lctur w hv prov tht formul (3) stisfis D1)-D3), so it givs vlid dfiitio of th dtrmit for 2 2 mtrics. Our clcultio shows tht o c us xpsio ccordig to row to clcult th dtrmit. To udrstd how xpsio ccordig to row works, lt s us xmpl from th txtbook for 3 3 mtrix d xpsio ccordig to th first row: 1

2 Lt us ow prov th dfiitio of th dtrmit usig th xpsio ccordig to row is vlid. So ssum tht th dtrmit is dfid for ( 1) ( 1) mtrics d for thos mtrics it stisfis D1) D3) (d s cosquc DP1) DP5)). W will prov tht o mttr which row is usd, xpsio (2) stisfis D1) D3). D1): lirity wrt y colum. Cosidr k-th colum. Cosidr trm i th sum (2): ( 1) i+j i j dt(a i j ) For j k, th i j dos ot dpd o th k-th colum d dt(a i j ) dpds lirly o th k-th colum (s th dtrmit for ( 1) ( 1) mtrics stisfis D1)). If j = k th i j dpds lirly o th k-th colum d dt(a i j ) dos ot dpd o th k-th colum. I y cs our trm dpds lirly o th k-th colum. Sic dt(a) is sum of such trms, it dpds lirly o th k-th colum. D2): If two djct colums r qul, th th dtrmit is qul to 0. Suppos two djct colums of A r qul, mly k = k+1. Agi cosidr trm i th sum (2): ( 1) i+j i j dt(a i j ) If j is ot qul to k d k + 1, th th mtrix A i j hs two djct colums qul, d hc its dtrmit is qul to 0. Thus th trm corrspodig to idx j ot qul to k d k + 1 givs zro cotributio to dt(a). Th othr two trms c b writt s ( 1) i+k i k dt(a i k ) + ( 1) i+k+1 i k+1 dt(a i k+1 ) Th two mtrics A i k d A i k+1 r qul bcus of our ssumptio tht th k-th colum of A is qul to th (k + 1)-th colum. Similrly i k = i k+1. Hc ths two trms ccl s thy hv th opposit sigs. D3): dt(i) = 1 Lt A b th idtity mtrix. Th i j = 0 ulss i = j, i which cs i i = 1. Thus th oly trm is th sum which givs o-zro cotributio is ( 1) i+i i i dt(a i i ) = dt(a i i ) As A i i is th ( 1) ( 1) idtity mtrix, its dtrmit is qul to 1, s th dtrmit for ( 1) ( 1) mtrics stisfis D3). Thus o c us xpsio to y row to dfi th dtrmit of mtrix d such dtrmit stisfis D1) D3). As thr is oly o fuctio stisfyig D1) D3) (prov i th lctur) thos xpsios hv to giv th sm rsults. W hv prov tht if th dtrmit is dfid for ( 1) ( 1) mtrics, th w my dfi it for mtrics. Formlly it is dfiitio usig mthmticl iductio, which my b xplid s follows. As th dtrmit is dfid for 1 1 mtrics, dfi it for 2 2 mtrics usig th xpsio ccordig to row, th dfi it for 3 3 mtrics usig th xpsio ccordig to row d its dfiitio for 2 2 mtrics. Th dfi it for 4 4 mtrics usig xpsio ccordig to row d its dfiitio for 3 3 mtrics. Th dfi it for 5 5 mtrics... 2

3 2 Expsio ccordig to colum. Dtrmit of A T. W hv prov tht th dtrmit stisfyig D1) D5) xists d w hv lso prov (i th lctur) tht it is uiqu. W hv two lgorithms to fid th dtrmit. O usig xpsio ccordig to row d othr usig colum oprtios (or row oprtios o A T ). W will prov ow tht o could lso us xpsio ccordig to colum. Lt A R. Expsio ccordig to th j-th colum is dfid s Thorm 2.1 Lt A R dt c j(a) = i=1 ( 1) i+j i j dt(a i j ) (6) dt c j(a) = dt A (7) Proof For = 1 thr is o xpsio. For = 2, th fct tht dt c 1(A) = dt c 2(A) = dt A is sy to show. For proof by iductio, ssum tht if A R ( 1) ( 1), th (7) is stisfid. W will prov tht th (7) is stisfid for y j d for y A R. To prov it, w will show tht for y j, th fuctio dt c j ctig o mtrics stisfis D1), D2), D3) d tht implis tht it hs to b qul to th dtrmit. Proofs tht D1) d D3) r stisfid by dt c j r similr to thos for xpsio by row, so w will skip thm. Lt us focus o D2) which sys tht if two djct colums r th sm, th dt c j(a) = 0. Lt k = k+1. For xpsio ccordig th j-th colums, whr j is ot qul to k or k + 1, ch trm dt(a i j ) hs two djct colums tht r qul, so dt(a i j ) = 0. Thus th sum is qul to zro. Cosidr j = k. Usig th iductio ssumptio, s A i k R ( 1) ( 1), w hv dt c k (A i k) = dt(a i k ). I fct w r prformig othr xpsio ccordig to th k-th colum of A i k, which corrspods to th (k + 1)-th colum of A: dt c k (A) = ( 1) i+k i k dt(a i k ) = i=1 i=1 = ( 1) i+k i k i=1 l=1,l i l=1,l i ( 1) p l+k l k+1 dt((a i j ) pl k) ( 1) i+k ( 1) p l+k i k l k+1 dt((a i j ) pl k) whr p l = l if l < i d p l = l 1 if l > i. Ech mtrix (A i j ) pl k is th mtrix A without two rows d without two colums k, k + 1. Cosidr two rows i 1, i 2 whr i 1 < i 2. Lt A (i1,i 2 ) (k,k+1) dot mtrix A without rows i 1, i 2 d without colums k, k + 1. Exprssio dt(a (i1,i 2 ) (k,k+1)) pprs i th sum (8) xctly twic, s dt((a i1 k) i2 1 k) d s dt((a i2 k) i1 k). As rsult oc it is multiplid by d th scod tim it is multiplid by ( 1) i 1+k ( 1) i 2 1+k i1 k i2 k+1 = ( 1) i 1+i 2 1 i1 k i2 k (8) ( 1) i 2+k ( 1) i 1+k i2 k i1 k+1 = ( 1) i 1+i 2 1 i1 k i2 k As rsult thos two trms sum up to 0. This implis tht th whol sum i (8) is qul to 0. Th proof for th cs of xpsio ccordig to th j = k + 1 colum is similr. This implis tht D2) is stisfid for. Togthr with D1),D3) (ot show hr) it ms tht if (7) is tru for y A R ( 1) ( 1), th D1), D2), D3) stisfid by fuctio dt c j for mtrics. As thr is oly o fuctio stisfyig D1),D2),D3), it hs to b tht for A R (7) is stisfid. Usig mthmticl iductio, w hv prov tht (7) for y squr mtrix A R. This llows us to cosidr othr tool for fidig th dtrmit, but lso llows us to prov tht: Thorm 2.2 Lt A R, th dt(a T ) = dt(a) (9) 3

4 Proof Proof is by iductio. For A R 1 1 it is trivil rsult, s th A T = [] T = [] = A Lt us ssum tht (9) is stisfid for y ( 1) ( 1) mtrix. W shll prov tht it must b stisfid for y mtrix. Lt Dfi B s A = b 1 1 b b b B = 2 1 b b 2 = A T = b 1 b 2... b Lt us prform xpsio ccordig to th first row of B = A T : Notic tht dt(a T ) = dt(b) = j=1 B 1 j = (A T ) 1 j = (A j 1 ) T ( 1) 1+j b 1 j dt(b 1 j ) (10) Ad s (A j 1 ) is ( 1) ( 1) mtrix, w ssumd tht (9) is stisfid for it, so Additiolly b 1 j = j 1 d thus (10) is qul to dt((a j 1 ) T ) = dt(a j 1 ) j=1 ( 1) j+1 j 1 dt(a j 1 ) which is th xpsio ccordig to th 1st colum for A. Thorm 2.1 sys it hs to b qul to dt(a). As rsult s th rows of A r colums of A, w obti tht D1) D2) d DP1) DP5) r tru for colums s wll. Thus o c do row oprtios to fid th dtrmit. 3 Dtrmit of th product of two mtrics Thorm 3.1 Lt A, B R, th dt(a B) = dt(a) dt(b) Proof W will us, wht ws prov i clss, tht for y mtrix A R, w hv Lt C = A B dt(a) = dt = dt =... i1 1 i i dt i 1 =1 i 2 =1 i =1 i1 i2... i (11) 4

5 Lt us prform som clcultios for = 3 first. C = 1 b b b b b b b b b 3 3 s rsult, usig lirity of th dtrmit with rspct to th ch colum, w obti: dt(c) = i 1 =1 b i1 1 dt i1 1 b b b b b b 3 3 = b i1 1 b i2,2 dt i 1 =1 i 2 =1 i1 i2 1 b b b 3 3 = b i1 1 b i2,2 b i3,3 dt i 1 =1 i 2 =1 i 3 =1 i1 i2 i3 = b i1 1 b i2,2 b i3,3 dt i 1 =1 i 2 =1 i 3 =1 i1 i2 i3 A similr clcultio for mtrics shows tht dt(c) =... b i1 1 b i2,2... b i, dt i 1 =1 i 2 =1 i 3 =1 i1 i2... i (12) Now otic tht if y two colums of i1 i2... i r qul th dtrmit of this mtrix is zro. So th oly o-zro cotributio coms from thos trms, whr this mtrix hs th sm colums s mtrix, but i diffrt ordr. By prformig colum itrchgs, o c trsform this mtrix to mtrix A. As rsult dt i1 i2... i = ( 1) k dt = ( 1) k dt(a) (13) whr k is th umbr of colums itrchgs dd. Notic tht th sm colums itrchgs will trsform mtrix i1 i2... i to mtrix = I Thus w obti: dt i1 i2... i = ( 1) k dt(i) = ( 1) k 1 = ( 1) k Usig th bov, (13) o c rwrit (12) s dt(c) = i 1 =1 i 2 =1 = dt(a)... i 1 =1 i 3 =1 i 2 =1 b i1 1 b i2,2... b i, dt i1 i2... i dt(a)... b i1 1 b i2,2... b i, dt i 3 =1 i1 i2... i 5

6 which, usig (11) for mtrix B, yilds dt(c) = dt A dt B 6