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?

How can a square matrix be positive definite when it has nonpositive diagonal entries?! — user1551, Sep 06 '14 at 13:06
Note that the eigenvalues of $I_n$ are all positive, and the corresponding bilinear form is positive definite. — Travis Willse, Sep 06 '14 at 13:10
@Vazrael Yes, that's right. Note that in practice people often simply write signature as a pair $(p, q)$ indicating "the number of positive and negative directions" when the bilinear form is understood from context to be nondegenerate. — Travis Willse, Sep 06 '14 at 13:27
@Vazrael If you meant "negative definite" then edit accordingly your question — Timbuc, Sep 06 '14 at 16:24
Does this answer your question? determine signature of matrix — Michael E2, Jun 30 '20 at 17:15

score 1 · Answer 1 · answered Sep 06 '14 at 16:39

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.

answered Sep 06 '14 at 16:39

Timbuc

34,191

What are the first two steps for? – celtschk Sep 06 '14 at 16:53
Well, the first step is to make things a little easier, and the second one must be there as to get congruent matrices we must do on columns exactly what we do on rows. – Timbuc Sep 06 '14 at 16:54
Given that the first two steps just turn the matrix upside down, I don't see how they make anything easier. – celtschk Sep 06 '14 at 16:55
For me it is easier to substract three times the first row from the last one , rather than one third of the first row from the last one. Matter of taste. – Timbuc Sep 06 '14 at 16:56
But is it also easier to subtract three times the first row from the last one, than it is to subtract three times the last row from the first one? – celtschk Sep 06 '14 at 16:57
I reduce matrices from top to bottom and not the other way around. I think most people does the same. – Timbuc Sep 06 '14 at 16:58

Calculating the signature of matrix A?

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