Is there a basis for which the quadratic forms are both diagonalized?

Question

The question asks if there is a basis for which both of the following quadratic forms are in a diagonalized form:

$$ q_1(x_1,x_2) = x_1^2 + x_1x_2 - x_2^2 $$ $$ q_2(x_1,x_2) = x_1^2 - 2x_1x_2 $$

What i tried:

I tried to see if one of the quadratic forms is positive definite, but I got that both are not.

Yet, it's not enough to say that none are positive definite in order to prove that there isn't a basis for which both are diagonalized, or at least I am not familiar with such a sentence.

So how can I determine this ?

Thank you.

Answer 1

The answer is no.

In fact if there exists a base of $\mathbb{R}^2$ such that the symmetric matrixes $A,B$ associated to the two quadratic forms are diagonalizable in that base, then there exists an orthogonal change base matrix $P$ such that

$P^TAP=I_1$

and

$P^TBP=I_2$

where $I_1, I_2$ are diagonal matrixes. Then we get

$AB=P^TI_1I_2 P=P^TI_2 I_1 P=BA$

because $P$ is orthogonal and so $P^{T}=P^{-1}$.

This means that you would have $AB=BA$.

In your case

$A=[[1, \frac{1}{2}], [\frac{1}{2}, -1]]$

$B=[[1, -1],[-1, 0]]$

and you can observe they not commute.

A necessary condition to get two matrixes are simultaneously diagonalizable is that they commute.

When the field is algebraically closed, then this condition is also sufficient.

Answer 2

Let's translate the problem into the language of matrices. Write $q_1(x) = x^T A x$ and $q_2(x) = x^T B x$ with $A,B$ symmetric and assume that $q_1,q_2$ are simultaneously diagonalizable. Then we can find an invertible $P$ with

$$ P^TAP = D_1, P^TBP = D_2 $$

where $D_1,D_2$ are diagonal. Let's assume that $A,B$ (and so $D_1,D_2$) are invertible. Then

$$ A = \left( P^T \right)^{-1} D_1 P^{-1}, B = \left( P^T \right)^{-1} D_2 P^{-1} $$

so

$$ A^{-1} B = P D_1^{-1} P^T \left( P^T \right)^{-1} D_2 P^{-1} = P \left( D_1^{-1} D_2 \right) P^{-1} $$

which implies that $A^{-1}B$ is similar to a diagonal matrix $D_1^{-1} D_2$. So a necessary condition for $q_1,q_2$ to be simultaneously diagonalizable is for $A^{-1}B$ to be a diagonalizable matrix.

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In your case,

$$ A = \begin{pmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & -1 \end{pmatrix}, B = \begin{pmatrix} 1 & -1 \\ -1 & 0 \end{pmatrix} $$

so

$$ A^{-1} B = \frac{1}{5} \begin{pmatrix} 2 & -4 \\ 6 & -2 \end{pmatrix}. $$

This matrix $5 A^{-1} B$ has characteristic polynomial $X^2 + 20$ which doesn't have real roots so it is not diagonalizable and hence your quadratic forms can't be simultaneously diagonalized over $\mathbb{R}$.