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Let A be a real square matrix and $I$ the identity matrix of the same size. Let n > 0 be the smallest natural number for which $a_0, a_1, \ldots, a_{n-1} \in \mathbb{R}$ exist such that the following is true:

$A^n+a_{n-1}A^{n-1}+ \ldots +a_1A +a_0I = O$

Prove that A is invertible if and only if $a_0 \ne 0$ .

3 Answers 3

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Hint

If $\lambda$ is an eigenvalue of $A$ then:

$$\lambda^n+a_{n-1}\lambda^{n-1}+...+\lambda a_1+a_0=0$$

If $\lambda =0$ then $a_0=0$ and also if $a_0=0$ then $0$ is a root.

What can you conclude about $A$ if it has a eigenvalue equal to $0$? Can you finish?

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    Why may I replace A with $\lambda$? This could be quite usefull for other proofs. – 2017-01-10
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    1) $Ax=\lambda x \rightarrow A^2x=\lambda Ax=\lambda^2 x \rightarrow A^3x=\lambda Ax=\lambda^3 x ...$. Can you see? – 2017-01-10
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    But A is a matrix and lambda is not a matrix to my understanding – 2017-01-10
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What you wrote is the minimal polynomial $R(t)$ of $A$ evaluated at $A$. One knows that $$ R(t)^k=\det(A-tI) $$ for some $k\geq1$ and thus $a_0^k=\det(A)$. One of these quantities is $0$ if and only if the other is and the invertibility of $A$ is equivalent to $\det(A)\neq0$

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If $a_0 \ne 0$, then $A^n+a_{n-1}A^{n-1}+ \cdots +a_1A +a_0I = 0$ implies $$\dfrac{1}{-a_0}(A^{n-1}+a_{n-1}A^{n-2}+ \cdots +a_1)A = I$$ and so $A$ is invertible.

If $a_0 = 0$, then $(A^{n-1}+a_{n-1}A^{n-2}+ \cdots +a_1)A = 0$. If $A$ were invertible, then we'd have a polynomial equation for $A$ of smaller degree.