determine signature of matrix

what is the signature of this matrix:

$\begin{pmatrix} -3&0&-1 \\0&-3&0 \\ -1&0&-1 \end{pmatrix}$ ?

I tried calculating them without eigenvalues; this should be done via using Gauß, right?

However, I don't see a way to get the signature by using simultaneously raw and line-transformations.. ?

Solution 1:

You're on the right track, simultaneous row and column transformations are the way to go. So, as a first step, we will use the first row to eliminate the entry in the lower left corner: $$\begin{pmatrix} -3&0&-1 \\0&-3&0 \\ -1&0&-1 \end{pmatrix} \to \begin{pmatrix} -3&0&-1 \\0&-3&0 \\ 0&0&-\frac{2}{3}\end{pmatrix}$$ The transformation I used was "replace row 3 by row 3 - 1/3*row 1". Next, we must perform the same transformation on columns: replace the third column by the third column minus one third of the fist column: $$\begin{pmatrix} -3&0&-1 \\0&-3&0 \\ 0&0&-\frac{2}{3}\end{pmatrix} \to \begin{pmatrix} -3&0&0 \\0&-3&0 \\ 0&0&-\frac{2}{3}\end{pmatrix}$$ Fortunately, we now already have a diagonal matrix. But even for more complicated matrices the procedure is basically the same.

Now you can simply read off the signature.

Of course there are other criteria to determine the signature, using determinants of minors of the matrix, etc. But this method is, in my opinion, the most straightforward.

Edit: Here is a small example of the calculation of a suitable transformation matrix.

I'll start with $A=\begin{pmatrix}1&2\\2&3\end{pmatrix}\in\mathbb{Q}^{2\times 2}$ and I want to find an invertible matrix $T$ over $\mathbb{Q}$ such that $TAT^{tr}$ is diagonal.

I'll perform simultaneous row and column operations on $A$ and perform only the row operations on the $2\times 2$-identity matrix:

$$\begin{pmatrix}\begin{array}{cc|cc} 1&2 & 1&0 \\ 2&3 & 0&1 \end{array}\end{pmatrix} \to \begin{pmatrix}\begin{array}{cc|cc}1&2&1&0 \\ 0&-1 & -2&1 \end{array}\end{pmatrix} \to \begin{pmatrix}\begin{array}{cc|cc} 1&0 &1&0 \\ 0&-1 & -2&1 \end{array}\end{pmatrix}$$ So, my diagonal matrix is $\begin{pmatrix}1 & \\ & -1\end{pmatrix}$ and my transformation matrix $T$ is $\begin{pmatrix}1&0 \\ -2&1\end{pmatrix}$. Notice how I first performed a row operation, which I also performed on the right (identity) matrix. Then I performed the corresponding column operation, which I did not perform on the right side.

Solution 2:

$-3$ Is an eigenvalue, as $\text{rank }(A+3I) < 3$.

The sum of th eigenvalues is $\text{tr }A = -7$, and their product is $\det A = -6$. So if $a_{1,2}$ are the other eigenvalues, $a_1 a_2 = 2$ (so they has the same sign) and $a_1 + a_2 = -4$ so they are $<0$. So all the eigenvalues are $<0$.