I got this bilinear form: $f:\mathbb{R}^2\times\mathbb{R}^2\longrightarrow\mathbb{R}$
$$f((x_1,x_2),(y_1,y_2))=-x_1y_1-16x_2y_2+4x_1y_2+4x_2y_1$$
I am asked to find the basis that diagonalizes the matrix.
Using the standard method (eigenvalues/eigenvectors), I found that $P=\begin{pmatrix}4 & 1\\1 & -4\end{pmatrix}$ diagonalizes $f$ into $diag\{0,-17\}$.
However, this is a fairly tedious process (if you want to verify the result) and I am sure there is a much simpler one (both intuitively and the fact that this is supposed to be a <15-minute problem). So I expressed the bilinear form as a quadratic form:
$$q(x,y)=-x^2-16y^2+8xy$$
$$[q]_E = \begin{pmatrix}-1 & 4\\4 & -16\end{pmatrix}$$
I noticed that $-x^2-16y^2+8xy=-(x-4y)^2$, so it should be very easy to get the basis using Lagrange's method. I get that:
$$D=\begin{pmatrix}-1 & 0\\0 & 0\end{pmatrix}$$
And:
$$x'=x-4y$$
$$y'=?$$
Here I am stuck. $y'$ can't be $0$ because it would make the diagonalizing matrix singular. I tried with an arbitrary number, but it didn't work out. Also tried the congruence by elementary steps method (not sure what the official name is—it's the algorithm where you perform elementary row/column operations on the matrix until you reach the diagonal, and then do just the row operations on $I$), and the result was similar. What am I missing?
Best Answer
This is what I get if I apply the "congruence by elementary steps method" (also don't know what it is called ;)). First step: $$ \begin{bmatrix} -1 & 0 & :: & 1 & 0\\ 4 & 0 & :: & 0 & 1 \end{bmatrix} $$ Second step: $$ \begin{bmatrix} -1 & 0 & :: & 1 & 0\\ 0 & 0 & :: & 4 & 1 \end{bmatrix} $$ The matrix on the right is a change of basis matrix, right? Remember congruence and similarity is not the same. Congruent matrices have same numbers of positive, negative, and zero eigenvalues, whereas similarity ensures the same eigenvalues.