# Thread: characteristic polynomial and invertible matrices

1. ## characteristic polynomial and invertible matrices

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

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

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

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

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

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

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

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

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

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

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

but how do I find such a matrix, and what is the connections with $a_0$, beyond the fact that it can't be 0?
The value of $a_0$ 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 $a_0$ is non-zero. But ThePerfectHacker's method is much better.