# Signature of a bilinearform

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?

• What is your definition of the signature? Commented Sep 3, 2018 at 13:59
• Since $\det(A)<0$ we can't have three positive eigenvalues.
– user
Commented Sep 3, 2018 at 14:02
• How did you obtain the last form? Commented Sep 3, 2018 at 14:27
• Elementary row operations don’t preserve eigenvalues (they don’t even preserve the determinant, although the affect it in known ways). If they did, then the only eigenvalue of every invertible matrix would be $1$.
– amd
Commented Sep 3, 2018 at 21:02

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
$$H = \left( \begin{array}{rrr} 5 & 5 & 0 \\ 5 & - 8 & - 2 \\ 0 & - 2 & 0 \\ \end{array} \right)$$ $$D_0 = H$$ $$E_j^T D_{j-1} E_j = D_j$$ $$P_{j-1} E_j = P_j$$ $$E_j^{-1} Q_{j-1} = Q_j$$ $$P_j Q_j = Q_j P_j = I$$ $$P_j^T H P_j = D_j$$ $$Q_j^T D_j Q_j = H$$

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

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

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

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

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

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

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

• For these kind of OP we always give our complementary answers! That’s nice :)
– user
Commented Sep 3, 2018 at 18:24

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 $(+,+,-)$.