I stumbled upon a question in my textbook saying to use the Cayley-Hamilton theory to deduce that if an $ n \times n $ square matrix $M$ has a non-zero determinant, $M$ is invertible.
I understand how to find the inverse using this theory, but I don't see how you can prove that the inverse exists.
Thanks for any input (:
On
We are given that $M$ has characteristic polynomial $$ x^n + a_{n-1}x^n{-1} + \cdots + a_1x + a_0 $$ with $a_0 \neq 0$. The formula for our candidate for the inverse is given by $$ Q = - \frac 1{a_0}(M^{n-1} + a_{n-1}M^{n-2} + \cdots + a_2 M+ a_1 I) $$ To prove that $Q$ is actually the inverse to $M$, it suffices to note that $$ MQ = M \cdot - \frac 1{a_0}(M^{n-1} + a_{n-1}M^{n-2} + \cdots + a_2 M+ a_1 I) = \\ - \frac 1{a_0}(M^{n} + a_{n-1}M^{n-1} + \cdots + a_2 M^2+ a_1 M) = \\ - \frac 1{a_0}(- a_0 I) = I $$
By Cayley-Hamilton, $M$ satisfies its characteristic polynomial:
$$ M^n + a_{n - 1}M^{n - 1} + \dots + a_1M + a_0I = 0$$
where $a_0 = (-1)^n\det M$. If $\det M \ne 0$ then you can subtract $a_0I$ from both sides and divide by $a_0$ to get
$$I = -a_0^{-1}M\left( M^{n-1} + a_{n - 1}M^{n - 2} + \dots + a_1 \right).$$
Thus
$$M^{-1} = -a_0^{-1}\left( M^{n-1} + a_{n - 1}M^{n - 2} + \dots + a_1 \right).$$