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Let $A$ be a real square matrix en let $I$ be the identity matrix with the same size. Let $n>0$ be the smallest natural number for which there exist $a_0,a_1,...a_{n-1} \in {R}$ such that $A^n+a_{n-1}A^{n-1}+....+a_1A+a_0I=O$.

Show that A is invertible if and only if $a_0\neq0 $.

So i need to proof

  1. $A$ invertible $\rightarrow$$a_0 \neq 0$
  2. $a_0 \neq0 \rightarrow$ $A$ invertible

I don't really know where to start with 1, and with 2 I tried to prove it with a contradiction but I don't think it's right either.

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    I think your problem statement is a bit unclear. It does sound like you have to prove the existence of $a_0,a_1,\ldots,a_{n-1}$ with $a_0\neq 0$ in the case where $A$ is invertible. I don't see any answer talking about it. However, it is not difficult. – 2017-01-08
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    @Batominovski my textbook states i don't have to prove the existence of those. The answers answered my question perfectly – 2017-01-08
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    Note that if degree $n$ is "the smallest natural number" possible, then you are asking about how the minimal polynomial for $A$ will look. See the proposed duplicate. – 2017-01-08

3 Answers 3

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Hint:

$$A^n+a_{n-1}A^{n-1}+\ldots+a_1A+a_0I=0\implies a_0I=-\left(A^n+a_{n-1}A^{n-1}+\ldots a_1A\right)\implies$$

$$a_0I=A\;\overbrace{\left(-A^{n-1}-a_{n-1}A^{n-2}-\ldots-a_1I\right)}^{=B}$$

Now, observe that $\;B\;$ is an excellent candidate to almost be $\;A^{-1}\;$ , but this will happen iff ...

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Let $A$ be an invertible matrix. If $a_0 = 0$ then $$A^n + a_{n-1}A^{n-1} + \ldots + a_1A = O$$ but then $$A^{n-1} + a_{n-2}A^{n-1} + \ldots + a_1I = A^{-1}(A^n + a_{n-1}A^{n-1} + \ldots + a_1A) = O$$

but that contradicts the assumption that $n$ was minimal. Therefore $a_0$ must be nonzero.

Conversely, if $a_0\neq 0$ then

$$\dfrac{-1}{a_0}(A^{n-1} + a_{n-2}A^{n-1} + \ldots + a_1I)A = I$$

so $$\dfrac{-1}{a_0}(A^{n-1} + a_{n-2}A^{n-1} + \ldots + a_1I) = A^{-1}$$

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    Thank you! But why does the $a_{n-1}$ change in to$ a_{n-2}$ and not the $A^{n-1} $into$ A^{n-2}$? – 2017-01-08
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    @Zoey nothing changed into anything, I just either multiplied by $A^{-1}$ or extracted a common factor $A$. – 2017-01-08
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Forward part $A$ is invertible implies $a_0\neq 0$. Lets look at the contrapositive. If $a_0=0$ then $$A(A^{n-1}+a_{n-1}A^{n-2}+...+a_1I)=0$$ Now if A is invertible then $A^{n-1}+a_{n-1}A^{n-2}+...+a_1I=0$ which contradicts the fact that n is the smallest number which annihilates A.
Backward implication if $a\neq 0$ then we can find the inverse of $A$ as $\frac{-1}{a_0}(A^{n-1}+a_{n-1}A^{n-2}+...+a_1I)$