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Let A $\in$ Mat$_n (\mathbb{F})$ and let $f(x) = a_n x^n+\cdots+a_1 x+a_0$ be the characteristic polynomial of A. Prove that A is singular if and only if $a_{0} \neq 0$.

Any hint or technique.

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I presume you mean iff $a_0 = 0$ since otherwise $\det A = (-1)^n a_0 \neq 0$. –  copper.hat Dec 3 '12 at 17:29
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The characteristic polynomial of $A$ is $f(x) = \det (xI-A)$. Since $\det cA = c^n \det A$ for a constant $c$, we have $f(0) = a_0 = \det (-A) = (-1)^n \det A$.

Hence $A$ is singular iff $\det A = 0$ iff $a_0 = 0$ .

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Supposing you believe that a matrix satisfies its characteristic polynomial:

Then $p(A)=0$ implies $p(A)-a_0I=a_0I$. But obviously $A$ can be factored out of $p(A)-a_0I$, so you have something that looks like $A*q(A)=a_0I$. (Note that $q(x)$ is nonzero.)

If $a_0$ is nonzero, then you can divide through by $a_0$, and you have an inverse for $A$: $q(A)/a_0$, and hence $A$ is nonsingular.

If $a_0=0$, then you have an entire matrix $q(A)$ such that $Aq(A)=0$, so any of its nonzero rows will serve as evidence $A$ is singular.

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$A$ being singular is equivalent to $0$ being an eigenvalue of $A$. Since the eigenvalues of $A$ are precisely the roots of its characteristic polynomial, $A$ is singular iff $f(0) = a_0 = 0$.

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I already proved that A is singular iff 0 is the eigenvalue of A. What is the next step? –  ghet Dec 3 '12 at 17:57
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Let $P_\lambda$ denote the characteristic polynomial of $A$. Then $P_\lambda{(0)} = det(A-0I)= det(A)= a_0$.

$A$ is singular iff $a_0=det(A)= 0$.

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