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lecture11 - Lecture 11 Interlude Theorem 1 DetA(H = T r(AH...

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Lecture 11 Interlude: Theorem 1. Det 0 A ( H ) = Tr ( e AH ) Proof. It is enough to see both sides agree when H is a standard basis matrix E ij . Left Hand Side: Det 0 A ( E ij ) = d dt Det ( A + tE ij ) | t =0 = d dt ( DetA + ( - 1) i + j tA ij ) | t =0 (where A ij is the minor of A) = ( - 1) i + j A ij . Right Hand Side: ( BE ij ) pq = n r =1 B pr ( E ij ) rq = n r =1 B pr δ ir δ jq = B pi δ jq Tr ( BE ij ) = n p =1 B pi δ jp = B ji . In particular, Tr ( e AE ij ) = e A ji = ( - 1) i + j A ij . Lemma 1. Let F : U R where U X and X is a normed space. Suppose F is differentiable at a U and has a local maximum value at a then F 0 a ( h ) = 0 for all h . (Exercise) 1
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Proof. Suppose BWOC there exists x U such that F 0 a ( x ) 6 = 0. WLOG we can take F 0 a ( x ) > 0 (if F 0 a ( x ) < 0 look at - x ). Since a is a local maximum, there exists δ 1 such that k y - a k < δ 1 f ( y ) f ( a ) . Now let F 0 a ( x ) = s > 0. Let ε > 0 such that ε < s k x k . Now since F is differentiable at a , there exists δ 2 > 0 such that if k h k < δ 2 , then | F ( a + h ) - F ( a ) - F 0 a ( h ) | < ε k h k .
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