Given the matrix
$$\begin{bmatrix}1 && 2 && 3 \\ 2 && 0 && 1 \\ 3 && 1 && -1\end{bmatrix}$$
Find a Diagonal Matrix congruent to it.
I know that the Matrix congruency relation over $n\times n$ symmetric matrices over the real field has the canonical forms as diagonal matrices containing diagonal elements as $1,-1,0$ up to ordering.
My first attempt was to find the explicit orthogonal transformation that can convert the given matrix to a diagonal matrix but that proved to be cumbersome manually (the characteristic polynomial wasn't simple to solve).
Then I figured that $\text{tr}(A)=0$ and $\det(A)=15$. Since determinant is positive (and non-zero) either 1 or 3 eigenvalues are positive. However, $\text{tr}(A)=0 \implies$ some eigenvalues are negative definitely. Therefore, the signature of the form is $( + – -)$. Hence, one congruent diagonal matrix is simply the following:
$$\begin{bmatrix}1 && 0 && 0 \\ 0 && -1 && 0 \\ 0 && 0 && -1\end{bmatrix}$$
Is this reasoning correct? Moreover, is there a simple way to find an explicit transformation leading to a diagonal matrix?
Best Answer
By completing squares you can see that the form is $$ (x+2y+3z)^2 - (2y+z)^2 -(3z)^2 $$ from which you can work out the change of basis needed.