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1.Let $A \in GL_n(\mathbb{C})$. Show that $\det(I+A)=1+\operatorname{tr}(A)+ \epsilon(A)$ where Modulas of epsilon(A) by norm of A=0 as A tends to 0,for any matrix norm. If I define J(A)= det(A) for A is in GLn(C),then J is differentiable at all such A and that,if H is in Mn(C),then J'(A)(H)=det(A)tr(A*H). where A* is inverse of A.

Can any one say how to write latex here?

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    @emili: I fixed the math display on the first few occurrences, just to illustrate what Mariano wrote in his comment. Please take the time to put the rest of the math displays in TeX. If you need a "displayed formula" (one set apart from the text), enclose it in double dollar signs.2011-02-03

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You can use $\det(e^{M}) = \exp(\text{tr} M)$ valid for diagonalizable matrices $M$. For your case this implies $\det(I+\lambda A) = \exp[ \text{tr} \log (I+ \lambda A)] = \exp[ \text{tr} ( \lambda A + \mathcal{O}(\lambda^2) )] = 1 + \lambda \,\text{tr} A + \mathcal{O}(\lambda^2).$

The Jacobian of $\det(A)$ can be obtained using this result $\det(A+ \lambda H) - \det(A) = \det(A) [ \det( I + \lambda A^{-1} H) -1] = \lambda \det(A) \,\text{tr} (A^{-1} H) + \mathcal{O}(\lambda^2)$ proving the result stated in the question.