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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?

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

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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.

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