# Signature and Orthogonal Basis of Symmetric Bilinear Form

I am trying to find the signature and a orthogonal basis for the symmetric bilinear form $$(x,y) \mapsto x^tAy$$ where $$A$$ is $$\begin{bmatrix} 5 & 5 & 0 \\ 5&-8&-2 \\ 0&-2&0 \end{bmatrix}$$. Furthermore, $$BLF: V \times V \to R$$, with $$V := R^3$$ (column vectors). The problem is that I get into really ugly numbers when I try to find the eigenvalues - thus I guess there is an easier way to proceed. Here is what I got so far:

• I applied the BLF to all standard basis vectors of $$R^3$$ and got the same matrix A as the representation matrix of the BLF (?).
• I calculated the characteristic polynomial and got: $$-t^3-3t^2+69t-20$$. Which implies really ugly eigenvalues (0.29398.., 6.7639.., -10.057..)
• I concluded that the signature is $$(p, n, z) = (2, 1, 0)$$.

How can I find a orthogonal basis from here? Should I apply Gram-Schmidt Orthogonalisation to a eigenvector to get 3 orthogonal vectors (the ugly eigenvalues make me think that there is another way)? Or apply Gram-Schmidt to one of the standard basis vectors of V? I am really confused and would appreciate any help.

• I put an answer, but I like your idea of Gram-Schmidt; the one detail is that you do not make any attempt to normalize the vectors. Take your first vector $x_1$ and write it on the left of the matrix $A.$ Might as well take $x_1=(1,0,0)$ as it really does not matter. Then find the row vector $x_1 A.$ For a vector to be perpendicular to that, all you need is $(x_1A) \cdot x_2 = 0$ Finally, row vectors $x_1,x_2,$ you need $x_3$ to have dot product zero with both $x_1A$ and $x_2A$ which can be found (in this dimension) by cross product. Commented Aug 29, 2022 at 21:51
• @WillJagy, I applied the Gram-Schmidt process and got the (column) vectors $(1,0,0), (-1,1,0), (2/13,-2/13, 1)$ which are all perpendicular to each other. Can I now just state that this is a orthogonal basis and that's it? Thank you again for your time Commented Aug 30, 2022 at 11:04

Notice that they say orthogonal basis, not "orthonormal," so all you need for your symmetric $$H$$ is a nonsingular matrix $$P$$ such that $$P^T H P = D$$ is diagonal. If you can't figure out how to do this by repeatedly completing the square, there is an algorithm. Starting with integer matrix $$H$$ it constructs rational matrix $$P,$$ one step at a time. I display all the matrices, one does not need to write quite that much by hand...

You mention Gram Schmidt. You should try that, except you are not going to normalize "lengths," you are just going to make them perpendicular to each other according to the given bilinear form

You do not need eigenvalues. There is no requirement for $$P$$ to be orthogonal.

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

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

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

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

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

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

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

$$\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$$

$$P^T H P = D$$

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

$$Q^T D Q = H$$

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

$$\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$$

• Thank you very much for your explanation and your time! So, I guess that the orthogonal basis vectors are $b_1 =(5, 0,0), b_2=(0, -13, 0), b_3=(0,0,4/13)$? Shouldn't then for example ${b_1}^t A b_2 = 0$? Commented Aug 30, 2022 at 10:30