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- Thread starter Bertrandkis
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Defennder

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morphism

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as we all know with eigenvalues A

the easiest way I have found to show that A^(-1)

Is to use matrix multiplcation to get A^(-1) on the right side and then get to your 1/lamda from there.

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Dick

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HallsofIvy

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For 3, it's not too hard to prove that |det(A)| is the absolute value of the constant term of the characteristic polynomial of A. Why does this help?

As said the first time you posted this you don't have to do any evaluation or computation. If [itex]\lambda_1[/itex], [itex]\lambda_2[/itex], ..., [itex]\lambda_n[/itex] are are the eigenvalues of A, then its characteristic polynomial is [itex](x-\lambda_1)(x- \lambda_2)\cdot\cdot\cdot (x- \lambda_n)[/itex]. Multiplying that out, the constant term is clearly [itex]\lambda_1\lambda_2\cdot\cdot\cdot\lambda_n|[/itex]. If you have set up the characteristic equation as [itex]det(A- \lambda I)[/itex]= 0, as I learned, it is obvious that taking [/itex]\lambda= 0[/itex] gives just the determinant. If, as some people learn, you set it up as [itex]det(\lambda I- A)[/itex] it gives the negative of the determinant.How do you start to prove that? Evaluating determinants by co-factor expansions for nxn matrices become very messy.

In either case, it is correct to say that the absolute value of the determinant is equal to the absolute value of the constant term of the characteristic equation.

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