Lecture10 - Lecture 10[Interlude DetA(H =(Det A Tr(A1 H if A Mn(R is invertible and H Mn(R Proof Write A= a1 an with columns a1 an Rn In particular

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Lecture 10 [Interlude : Det 0 A ( H ) = (Det A ) Tr( A - 1 H ) if A M n ( R ) is invertible and H M n ( R ). Proof. Write A = ± a 1 ... a n ² with columns a 1 ,...,a n R n . In particular, A = ± e 1 ... e n ² for the standard basis e 1 ,...,e n of R n . Since Det A is multilinear in a 1 ,...,a n , Det is differentiable and Det 0 A ( H ) = n X j =1 Det ± a 1 ... h j ... a n ² . Special Case : A = I . Det 0 I ( H ) = n X j =1 Det ± e 1 ... h j ... e n ² = n X j =1 h jj = Tr H. Any invertible A : Note that if U M n ( R ) then Det A · Det( A - 1 U ) = Det U . That is, if we define λ : M n ( R ) M n ( R ) : U 7→ A - 1 U then Det = (Det A ) · Det λ. 1
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Apply the chain rule: as λ is linear, so Det 0 A ( H ) = (Det A ) · Det 0 λ ( H ) ( λ 0 A ( H )) = (Det A ) · Det 0 I ( λ ( H )) = (Det A ) · Tr( A - 1 H ) Remark: In general, if A M n ( R ) then Det 0 A ( H ) = Tr( ˜ AH ) where ˜ A is the ‘adjugate’ of A (the matrix whose ( i,j ) entry is ( - 1) i + j A ji and A ji is the ( j,i ) cofactor of
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This note was uploaded on 07/08/2011 for the course MAT 6932 taught by Professor Staff during the Summer '10 term at University of Florida.

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Lecture10 - Lecture 10[Interlude DetA(H =(Det A Tr(A1 H if A Mn(R is invertible and H Mn(R Proof Write A= a1 an with columns a1 an Rn In particular

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