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I'm trying to do Exercise 3.6 at the end of this pdf (in $\Bbb CP^2$):

  1. Show that the two quadratic forms $$x^2+y^2-z^2, \quad x^2+y^2-y z$$ cannot be simultaneously diagonalized.

Attempt 1:

Their matrices don't commute: $$\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right) \cdot\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{1}{2} \\ 0 & -\frac{1}{2} & 0\end{array}\right)\ne\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{1}{2} \\ 0 & -\frac{1}{2} & 0\end{array}\right) \cdot\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right)$$ but simultaneously diagonalizable matrices commute, so they cannot be simultaneously diagonalized.

Then I realize I am wrong. To diagonalize quadratic forms we want $S^TAS$ and $S^TBS$ to be diagonal matrices, but in the linked is another question: $S^{-1}AS$ and $S^{-1}BS$ are diagonal matrices.

Attempt 2:

I try to relate this exercise to a proposition in the PDF:

The non-diagonalizable case is where the intersection is at an odd number of points: enter image description here

The two conics $$x^2+y^2-z^2=0, \quad x^2+y^2-y z=0$$ intersect at two points $[1,\pm i,0]$ with multiplicity $1$, and one point $[0,1,1]$ with multiplicity $2$. So the intersection is at an odd number of points, so they cannot be simultaneously diagonalized.

But how do we prove this proposition?

Attempt 3:

For the first quadratic form $x^2+y^2-z^2$, let $z'=iz$, then $x^2+y^2-z^2=x^2+y^2+z'^2$, then the second quadratic form $x^2+y^2-yz=x^2+y^2+iyz'$, so it has matrix $\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & \frac{i}{2} \\ 0 & \frac{i}{2} & 0\end{array}\right)$.

The Jordan normal form of $\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & \frac{i}{2} \\ 0 & \frac{i}{2} & 0\end{array}\right)$ is $J=\left(\begin{array}{lll}\frac{1}{2} & 1 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & 1\end{array}\right)$ WA computation

this is non-diagonalizable over $\Bbb C$, since it has a Jordan block of size $2$.

Is it true in general that, if $J$ is a $3\times3$ matrix with a Jordan block of size $>1$, and $I_3$ is the identity matrix, then the two corresponding conics has odd number of intersections?

hbghlyj
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  • Be careful. Diagonalizing quadratic forms is different from diagonalizing linear maps. – Ted Shifrin Jun 13 '24 at 00:02
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    Given your symmetric matrices $A,B$ which are both invertible. Define $C= A^{-1} B.$ A necessary and sufficient condition to be able to construct $P$ such that both $P^T AP$ and $P^T BP$ are diagonal is that $C$ be diagonalizable, that is $R^{-1}CR$ is diagonal. Horn and Johnson 1985, paperback edition, page 229 has nice table of results. The construction for case II is pages 231-232. That title is Matrix Analysis. Their 1991 book is Topics in Matrix Analysis. – Will Jagy Jun 13 '24 at 00:49
  • Look at that, your $A^{-1} B$ is not diagonalizable. – Will Jagy Jun 13 '24 at 00:54

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