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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.. ?

Vazrael
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  • Do you know how to relate definiteness to eigenvalues? I'm assuming you want to find the signature without finding the eigenvalues. – Git Gud Oct 24 '14 at 08:33

2 Answers2

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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.

Oliver Braun
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  • I know about it, I didn't read your answer carefully enough, I just noticed you saying "you're on the right track" and I didn't notice the OP mentioning simultaneous row and column operations because at first he just mentions 'Gauss', again, my fault for not reading everything. – Git Gud Oct 24 '14 at 08:51
  • Hi - thank you, this helped me a lot :) If I would do your row and column transformations simultaneously with unitary matrix $E_3$, what would be the 'transformed' $E_3$ matrix in the end? I mean: What sort of matrix would this be? Is this the normal-form? – Vazrael Oct 24 '14 at 09:27
  • You didn't specify the field over which you are working. E.g. over $\mathbb{R}$ the normal form of $A$ would be $-E_3$, because of the signature. If you do simultaneous row and column transformations and you perform one kind of those transformations on the identity matrix, you obtain the transformation matrix $T$, i.e. a matrix such that $TAT^{tr}$ is the diagonal matrix you obtain in the end of the calculation. Does this answer your question? I'll provide a small example in my answer. – Oliver Braun Oct 24 '14 at 09:44
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$-3$ 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 $<0$.

mookid
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