Let A be an n by n matrix with characteristic polynomial $f(t)=(-1)^n t^n+a_{n-1}t^{n-1}+…+a_1 t+a_0$. Prove that $f(0)=a_0=det(A)$. deduce that A is invertible if and only if $a_0 \neq 0$.
To prove $f(0)=a_0=det(A)$ it is obvious that by definition of characteristic polynomial we can set $det(A)=f(t)$ and plug in to get the result.
$\Rightarrow$ Since we are given $f(0)=det(A)=a_0$. If $A^{-1}$=$\frac{1}{det(A)}adj(A)$, this is possible only if $det(A) \neq 0$, which implies $a_0 \neq 0$
$\Leftarrow$ Suppose $a_0 \neq 0$, then $det(A) \neq 0$. Hence the matrix A is invertible.
Can someone check the logic flow? Appreciate it!
Best Answer
I get the impression that the intended answer to this question is more simple: as you noted, $a_0 = f(0) = \det(A - 0I) = \det(A)$. Since a square matrix $A$ is invertible if and only if $\det(A) \neq 0$, we have $A$ is invertible if and only if $a_0 \neq 0$.
This is basically no different from your argument, except that you justify $A^{-1}$ not existing based on the formula $A^{-1} = \frac{1}{\det(A)}\operatorname{adj}(A)$, which only makes sense if $\det(A) \neq 0$. There's a subtle issue with this logic. Specifically, $A$ being invertible, and its inverse, are not defined in terms of the above formula for $A^{-1}$.
This formula is a theorem proven about inverses, and as you've pointed out, it only really makes sense when $\det(A) \neq 0$. However, there's nothing to say, prima facie, that this formula for inverses doesn't just fail whenever $\det(A) = 0$. That is, it might be the case that this formula provides the inverse of $A$ whenever $\det(A) \neq 0$, but some other method needs to be employed in the case where $\det(A) = 0$.
To take an analogous example from power series, given a power series $\sum_{n=0}^\infty a_n x^n$, we can compute the radius of convergence to be $$\lim_{n \to \infty} \frac{a_n}{a_{n+1}},$$ wherever the above limit exists and is finite. If the limit doesn't exist, then there is still a radius of convergence, but it cannot be found using this formula. It would therefore be incorrect to assume, based on this formula, that the limit must exist whenever the radius of convergence exists.
All that said, it is, of course, true that $\det(A) \neq 0$ whenever $A$ is invertible; it's just a separate result.