symmeric bilinear form defined by the following matrix

Question

what is the signature of the symmeric bilinear form defined by the following matrix

$\left(\begin{matrix} 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -5\\ 0 & 0& 0 & -5 & 0\\ 0 & 0 & -5 & 0 & 0\\ 0 & -5 & 0 & 0 & 0\\ \end{matrix}\right)$

i was thinking that sigantutre =$2p-r $ where p is a numberof positive entries and r is the ranks of the given matrix

so her p=1 and r= 5 and signature $=2.1-5=-3$

Answer 1

By Sylvester, since $\det(1)=1$ and $\det(M_{5\times5})>0$ the signature can be

1. $+++++$

2. $+++--$

3. $+----$

Case 1 can be excluded since

$$\det(M_{2\times2})=\begin{vmatrix}0&-5\\-5&0\end{vmatrix}=-25 <0$$

thus we have at least one negative eigenvalue. Moreover

$$\det(M_{3\times3})=\begin{vmatrix}1&0&0\\0&0&-5\\0&-5&0\end{vmatrix}=-25 <0$$

has signature $++-$ then also case 3 can be excluded.

Then the signature is $$n_+=3 \quad n_-=2 \quad n_0=0$$

Answer 2

This is congruence diagonalization of your matrix, I am calling it $H.$ The entires of $D$ are not the eigenvalues, but they have the exact same counts of positive, negative, zero. As you can see, three positive, two negative. Oh, as I arranged to have $\det P = \pm 1,$ the determinant is the same as the determinant of $D.$

Using the letters in the question, $p = 3, r = 5,$ $2p-r = 6-5 = 1$

$$ P^T H P = D $$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & 0 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & - 5 \\ 0 & 0 & 0 & - 5 & 0 \\ 0 & 0 & - 5 & 0 & 0 \\ 0 & - 5 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & - 10 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & - 10 & 0 \\ 0 & 0 & 0 & 0 & \frac{ 5 }{ 2 } \\ \end{array} \right) $$

Begins with $D_0 = H,$ after which we find useful elementary matrices $E_j \; :$

$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & - 5 \\ 0 & 0 & 0 & - 5 & 0 \\ 0 & 0 & - 5 & 0 & 0 \\ 0 & - 5 & 0 & 0 & 0 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & - 1 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & - 10 & 0 & 0 & - 5 \\ 0 & 0 & 0 & - 5 & 0 \\ 0 & 0 & - 5 & 0 & 0 \\ 0 & - 5 & 0 & 0 & 0 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & \frac{ 1 }{ 2 } \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 1 }{ 2 } & 0 & 0 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & - 1 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & - 10 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 5 & 0 \\ 0 & 0 & - 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{ 5 }{ 2 } \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 1 }{ 2 } & 0 & 0 & \frac{ 1 }{ 2 } \\ 0 & - 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & - 10 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & 0 & - 5 \\ 0 & 0 & 0 & - 5 & 0 \\ \end{array} \right) $$

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$$ E_{4} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 1 }{ 2 } & 0 & 0 & \frac{ 1 }{ 2 } \\ 0 & - 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & - 1 & 0 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & - 10 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & - 10 & - 5 \\ 0 & 0 & 0 & - 5 & 0 \\ \end{array} \right) $$

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$$ E_{5} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{5} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ \end{array} \right) , \; \; \; Q_{5} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 1 }{ 2 } & 0 & 0 & \frac{ 1 }{ 2 } \\ 0 & - 1 & 0 & 0 & 1 \\ 0 & 0 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 & - 1 & 0 \\ \end{array} \right) , \; \; \; D_{5} = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & - 10 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & - 10 & 0 \\ 0 & 0 & 0 & 0 & \frac{ 5 }{ 2 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & 0 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & - 5 \\ 0 & 0 & 0 & - 5 & 0 \\ 0 & 0 & - 5 & 0 & 0 \\ 0 & - 5 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & - 10 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & - 10 & 0 \\ 0 & 0 & 0 & 0 & \frac{ 5 }{ 2 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 1 }{ 2 } & - 1 & 0 & 0 \\ 0 & 0 & 0 & \frac{ 1 }{ 2 } & 1 \\ 0 & 0 & 0 & \frac{ 1 }{ 2 } & - 1 \\ 0 & \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & - 10 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & - 10 & 0 \\ 0 & 0 & 0 & 0 & \frac{ 5 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 1 }{ 2 } & 0 & 0 & \frac{ 1 }{ 2 } \\ 0 & - 1 & 0 & 0 & 1 \\ 0 & 0 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 & - 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & - 5 \\ 0 & 0 & 0 & - 5 & 0 \\ 0 & 0 & - 5 & 0 & 0 \\ 0 & - 5 & 0 & 0 & 0 \\ \end{array} \right) $$