# Simultaneous diagonalisation of quadratic forms

Is there a linear transformation that simultaneously reduces the pair of real quadratic forms $$x^2-y^2$$ and $$2xy$$ to diagonal forms?

My attempt I know that neither of these forms are positive definite, but if P=$$\left[ \begin{array}{ c c } 2 & 0 \\ 1 & 0 \end{array} \right]$$ so that $$x=2X$$ and $$y=X$$ then both are diagonal. Is this correct?

• The given matrix $P$ is not invertible and so cannot be used for diagonalization. – Travis Dec 18 '18 at 15:50

First note that the given matrix $$P$$ is not invertible, so it cannot be used for diagonalization of a nondegenerate quadratic form.
The matrix representations of the two quadratic forms with respect to the standard basis are $$Q = \pmatrix{1&\\&-1}, \qquad Q' = \pmatrix{&1\\1&}.$$
Computing for a general change-of-basis matrix $$P = \pmatrix{p_{ij}}$$ the matrix representations $$P^{\top} Q P, P^{\top} Q' P$$ of the quadratic forms w.r.t. a general basis have respective off-diagonal components $$a = p_{11} p_{12} - p_{21} p_{22}, \qquad b = p_{11} p_{22} + p_{12} p_{21} ,$$ and by definition an invertible matrix $$P$$ simultaneously diagonalizes $$Q, Q'$$ iff $$a = b = 0$$.
For a diagonal matrix, $$0 = a^2 + b^2 = (p_{11}^2 + p_{21}^2)(p_{12}^2 + p_{22}^2) ,$$ but for real $$P$$ this only vanishes if one of its columns has magnitude zero: Thus, the quadratic forms are not simultaneously diagonalizable over $$\Bbb R$$. On the other hand, over $$\Bbb C$$ this equation doesn't force degeneracy of $$P$$, only that $$p_{21} = \pm i p_{11}$$ or $$p_{12} = \pm i p_{21}$$. Substituting quickly leads to the solutions $$P = \pmatrix{\lambda&\pm i\mu\\\pm i\lambda&\mu}$$, so the quadratic forms are simultaneously diagonalizable over $$\Bbb C$$.
You need an invertible matrix for diagonalization, so your $$P$$ doesn’t work. It collapses everything onto a single line.
You can approach this problem geometrically. Restricted to the reals, $$x^2-y^2=a$$ is a family of hyperbolas with common asymptotes $$x=y$$ and $$x=-y$$, and $$2xy=b$$ is a family of hyperbolas with common asmyptotes $$x=0$$ and $$y=0$$. Simultaneous diagonalization makes these asymptotes coincide, but $$x=0$$ intersects every hyperbola of the first family at two finite points, so there’s no linear transformation of the plane that can turn this line into an asymptote of those hyperbolas.