Calculating the signature of matrix A?

Question

given is a symmetrical bilinearform s that has the following matrix:

$A = M_\beta(s) = \begin{pmatrix} -3&0&-1\\0&-3&0\\-1&0&-1\end{pmatrix}$

I have to calculate the signature of s and tell, whether s is positive-definite, negative-definite or indefinite.

I take a look at $det(A-t\cdot E_n)$ and get three eigenvalues $< 0$. This means:

Signature $\sigma(s) = (3,0,0)$ and the matrix is positive definite, right?

Answer 1

Another way following "the path of Sylvester's Theorem":

$$\begin{pmatrix} -3&0&-1\\0&-3&0\\-1&0&-1\end{pmatrix}\stackrel{R_1\leftrightarrow R_3}\rightarrow\begin{pmatrix} -1&0&-1\\0&-3&0\\-3&0&-1\end{pmatrix}\stackrel{C_1\leftrightarrow C_3}\rightarrow\begin{pmatrix} -1&0&-1\\0&-3&0\\-1&0&-3\end{pmatrix}\stackrel{R_3-R_1}\rightarrow$$

$$\begin{pmatrix} -1&0&-1\\0&-3&0\\0&0&-2\end{pmatrix}\stackrel{C_3-C_1}\rightarrow\begin{pmatrix} -1&0&0\\0&-3&0\\0&0&-2\end{pmatrix}$$

So we got a diagonal matrix congruent to the original one all of which diagonal elements are negative, and thus the signature is $\;(0,3)\;$ and the matrix is negative definite.