# Thread: characteristic polynomial and invertible matrices

1. ## characteristic polynomial and invertible matrices

Given $\displaystyle A$, an $\displaystyle n \times n$ matrix, with the following characteristic polynomial

$\displaystyle f(t) = (-1)^n t^n +a_{n-1}t^{n-1}+ \ \dotso \ + a_1t+a_0$ show that $\displaystyle A$ is invertible if only if $\displaystyle a_0\neq 0$

what I have is f(t) can be written as $\displaystyle det(A-t \cdot I)$ where $\displaystyle I$ is the identity matrix, so I would have:

$\displaystyle det \left( \begin{bmatrix} 0 & 0 & \dotso & 0 & -a_0 \\ 1 & 0 & \dotso & 0 & -a_1\\ 0 & 1 & \dotso & 0 & -a_2\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \dotso & 1 & -a_{n-1} \end{bmatrix} - t\times \begin{bmatrix} 1 & 0 & \dotso & 0 & 0 \\ 0 & 1 & \dotso & 0 & 0\\ 0 & 0 & \dotso & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \dotso & 0 & 1 \end{bmatrix}\right)$

now from the definition of invertible there must exist a matrix $\displaystyle B$ which is $\displaystyle n \times n$ matrix such that $\displaystyle A\cdot B = I$

but how do I find such a matrix, and what is the connections with $\displaystyle a_0$, beyond the fact that it can't be 0?

2. Originally Posted by lllll
Given $\displaystyle A$, an $\displaystyle n \times n$ matrix, with the following characteristic polynomial

$\displaystyle f(t) = (-1)^n t^n +a_{n-1}t^{n-1}+ \ \dotso \ + a_1t+a_0$ show that $\displaystyle A$ is invertible if only if $\displaystyle a_0\neq 0$

what I have is f(t) can be written as $\displaystyle det(A-t \cdot I)$ where $\displaystyle I$ is the identity matrix, so I would have:

$\displaystyle det \left( \begin{bmatrix} 0 & 0 & \dotso & 0 & -a_0 \\ 1 & 0 & \dotso & 0 & -a_1\\ 0 & 1 & \dotso & 0 & -a_2\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \dotso & 1 & -a_{n-1} \end{bmatrix} - t\times \begin{bmatrix} 1 & 0 & \dotso & 0 & 0 \\ 0 & 1 & \dotso & 0 & 0\\ 0 & 0 & \dotso & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \dotso & 0 & 1 \end{bmatrix}\right)$

now from the definition of invertible there must exist a matrix $\displaystyle B$ which is $\displaystyle n \times n$ matrix such that $\displaystyle A\cdot B = I$

but how do I find such a matrix, and what is the connections with $\displaystyle a_0$, beyond the fact that it can't be 0?
The value of $\displaystyle a_0$ is the product of the eigenvalues.
Since it is non-zero it means the eigenvalues are not zero.
Thus, the matrix must be invertible.

3. I would have pointed out that every matrix is either diagonalizable, with its eigenvalues on the main diagonal, or has a Jordan form, again with its eigenvaues on the main diagonal, both of which have determinant equal to the products of the eigenvalues. Since the determinant is the product of the eigenvalues, a matrix is invertible if and only if it does not have 0 as an eigenvalue: if $\displaystyle a_0$ is non-zero. But ThePerfectHacker's method is much better.