How to convert this symmetric matrix to normal form?

Question

I have to find the signature of the symmetric matrix $\begin{pmatrix}1 & 0 & 0 &0\\0 &0 &1 &0\\0&1&0 &0\\0& 0 &0 & 1\end{pmatrix}$. Now to find the signature I want to reduce this matrix to normal form by doing congruence operation.But I am stuck as if we do $R_{23}$ then we have to do $C_{23}$.Then our purpose is not satisfied as we get the same matrix.We should be able to transform this matrix to $\begin{pmatrix} I_k & &\\ &-I_{r-k} &\\ & & O\end{pmatrix}$.

Answer 1

The trick in situations like this is to first use off-diagonal non-zero entries to produce a non-zero on the diagonal. In your case, adding the third row to the second (and then likewise the third column to the second to preserve symmetry) gives $$ \begin{pmatrix} 1&0&0&0 \\ 0&2&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{pmatrix} $$ Now you can continue as usual, i.e. by subtracting half of the second row from the third (and likewise for columns): $$ \begin{pmatrix} 1&0&0&0 \\ 0&2&0&0 \\ 0&0&-1/2&0 \\ 0&0&0&1 \end{pmatrix} $$ Normalizing (and reordering) the entries gives the signature: $(3,1)$.

Answer 2

Hint $$ \left( {\matrix{ 0 & 1 \cr 1 & 0 \cr } } \right) = \left( {\matrix{ { - 1} & 1 \cr 1 & 1 \cr } } \right)^{ - 1} \left( {\matrix{ { - 1} & 0 \cr 0 & 1 \cr } } \right)\left( {\matrix{ { - 1} & 1 \cr 1 & 1 \cr } } \right) $$