Signature of quadratic form

Question

Give is the quadratic form:

$q(x,y,z,t) = x^2 + (4 + λ)y^2 + (1 + 4λ)z^2 + λt^2 + 4xy + 2xz + 4(1 − λ)yz + 2λyt + (1 − 4λ)zt $

I should compute the signature of $q$.

I load it into a matrix:

$q(x,y,z,t) =\begin{bmatrix} x&y&z&t \end{bmatrix}\begin{bmatrix} 1&2&1&0\\2&4+\lambda&2(1-\lambda)& \lambda\\1&2(1-\lambda)&1+4\lambda&1/2(1-4\lambda)\\0&\lambda&1/2(1-4\lambda)&\lambda \end{bmatrix}\begin{bmatrix} x\\y\\z\\t \end{bmatrix}$

Now I'm stuck.

How do I move on from here? Simplifying is a mess, is there another way?

Answer 1

Note that you can write

\begin{equation} q(x,y,z,t) = (x+2y+z)^2 + \frac{1}{4}(z+t)^2 - \frac{1}{4}(z-t)^2 + \lambda (y-2z+t)^2 \end{equation}

The answer should follow immediately from this.

Answer 2

Diagonalizing the matrix isn’t all that much of a mess.

Going down the first column, subtract twice the first row/column from the second, and subtract the first row/column from the third: $$\begin{bmatrix}1&0&0&0 \\ 0&\lambda&-2\lambda&\lambda \\ 0&-2\lambda&4\lambda&\frac12(1-4\lambda) \\ 0&\lambda&\frac12(1-4\lambda)&\lambda\end{bmatrix}.$$ Moving down the next column, add two times row/column two to three and subtract it from the last one: $$\begin{bmatrix}1&0&0&0 \\ 0&\lambda&0&0 \\ 0&0&0&\frac12 \\ 0&0&\frac12&0\end{bmatrix}.$$ You have to be a bit careful with the last two rows/columns, since you can’t just swap them as you might when performing Gaussian elimination. Instead, you need to add the fourth row/column to the third to create a non-zero diagonal entry and then proceed as before. After a few more operations you end up with $$\begin{bmatrix}1&0&0&0 \\ 0&\lambda&0&0 \\ 0&0&1&0 \\ 0&0&0&-\frac14\end{bmatrix}.$$ If you also want the basis that diagonalizes this matrix, apply the same row, but not column, operations to the identity matrix.