Signature of a bilinearform

Question

I have to determine the signature of the following matrix:

$$ A= \begin{pmatrix} 5&5&0\\ 5&-8&-2\\ 0&-2&0 \end{pmatrix} $$

The bilinear form $<,>: V \times V \rightarrow \mathbb{R}$ is defined by $(x,y)\rightarrow x^tAy$

a.) Determine the signature of the symmetrical bilinear form.

I'm not really sure how to do it. I first tried to calculate the upper triangular matrix:

$$ \begin{pmatrix} 5&5&0\\ 5&-8&-2\\ 0&-2&0 \end{pmatrix} \rightarrow \begin{pmatrix} 5&5&0\\ 0& 13& 2\\ 0& 0&4 \end{pmatrix}$$

Since all the eigenvalues are positve ($\lambda_1=5, \lambda_2 = 13$ and $\lambda_3 = 4)$ does it mean that the signature is also positive? Is (3,0) the signature? (3 positve and o negative eigenvalues)?

If not, how do I determine the signature of a matrix?

Answer 1

Added: "Repeated Completing the Square" gives the same outcome this time. The only requirement is Sylvester's Law of Inertia, the details could have been different. You begin with $5x^2 - 8 y^2 - 4 y z + 10 xy.$ We take care of two terms with $5(x+y)^2.$ But now we need $-13 y^2 - 4 y z . $ So we take $-13 (y + \frac{2z}{13})^2.$ At the end, we need to add back $\frac{4}{13} z^2.$

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & \frac{ 2 }{ 13 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - 13 & 0 \\ 0 & 0 & \frac{ 4 }{ 13 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & \frac{ 2 }{ 13 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 5 & 0 \\ 5 & - 8 & - 2 \\ 0 & - 2 & 0 \\ \end{array} \right) $$

So, there are two positive and one negative eigenvalue. The actual eigenvalues are approximately -10.0579, 0.2939, 6.7639.

The method, sometimes called "congruence diagonalization"

Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 5 & 5 & 0 \\ 5 & - 8 & - 2 \\ 0 & - 2 & 0 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 5 & 5 & 0 \\ 5 & - 8 & - 2 \\ 0 & - 2 & 0 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - 13 & - 2 \\ 0 & - 2 & 0 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - \frac{ 2 }{ 13 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - 1 & \frac{ 2 }{ 13 } \\ 0 & 1 & - \frac{ 2 }{ 13 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & \frac{ 2 }{ 13 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - 13 & 0 \\ 0 & 0 & \frac{ 4 }{ 13 } \\ \end{array} \right) $$

==============================================

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ \frac{ 2 }{ 13 } & - \frac{ 2 }{ 13 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 5 & 0 \\ 5 & - 8 & - 2 \\ 0 & - 2 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & \frac{ 2 }{ 13 } \\ 0 & 1 & - \frac{ 2 }{ 13 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - 13 & 0 \\ 0 & 0 & \frac{ 4 }{ 13 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & \frac{ 2 }{ 13 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - 13 & 0 \\ 0 & 0 & \frac{ 4 }{ 13 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & \frac{ 2 }{ 13 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 5 & 0 \\ 5 & - 8 & - 2 \\ 0 & - 2 & 0 \\ \end{array} \right) $$

Answer 2

By Sylvester since we have that $\det(A)=-20<0$ the signature can be

$$(-,-,-) \quad \text{or} \quad (+,+,-)$$

but since $\det(5)=5>0$ we have that the signature is $(+,+,-)$.