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I need to find a base in which both of these forms are diagonal. Provide this form for both forms.

$f(x,y,z)=z^2+2xz+3x^2+y^2-2xy$

$g(x,y,z)=5x^2-2xy+y^2+4xz+2z^2$

I know how to bind a basis, where $f$ or $g$ is diagonalized, but not where both of them are. Don't know if it's useful, but that's what I got:

$f$ is $Diag(1,1,1)$ in $\mathcal{A}=(\frac{1}{\sqrt{3}}(1,0,0), \frac{1}{\sqrt{2}}(0,1,1),\frac{1}{\sqrt{6}}(2,3,-3))$

$g$ is $Diag(1,1,1)$ in $\mathcal{B}=(\frac{1}{\sqrt{5}}(1,0,0), \frac{1}{\sqrt{6}}(0,2,1),\frac{1}{\sqrt{30}}(-3,-5,5))$

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  • $\begingroup$ I don't think the question asks to diagonalize both $f$ and $g$ simultaneously. $\endgroup$ Commented Jun 17 at 19:34
  • $\begingroup$ "Find a base in which both of these forms are diagonal. Provide this form for both forms." thats what I have to do $\endgroup$
    – zaba12
    Commented Jun 17 at 19:36
  • $\begingroup$ Edit the question to include that exact text. $\endgroup$ Commented Jun 17 at 19:39
  • $\begingroup$ done :) hope it helps $\endgroup$
    – zaba12
    Commented Jun 17 at 19:42
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    $\begingroup$ are you sure you wrote down the forms correctly? $f=(x,y,z)A(x,y,z)^T$ and $g=(x,y,z)B(x,y,z)^T$ where $A=\begin{pmatrix} 3 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 5 & -1 & 4 \\ -1 & 1 & 0 \\ 4 & 0 & 2 \end{pmatrix}$, but $AB\neq BA$, so these two matrices cannot be simultaneously diagonalized. $\endgroup$
    – AnCar
    Commented Jun 17 at 19:59

1 Answer 1

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To simultaneously diagonalize two quadratic form there are several (essentially equivalent) methods. I will proceed as follows. You define the matrices $$ A = \begin{pmatrix} 3 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix},\qquad B = \begin{pmatrix} 5 & -1 & 2 \\ -1 & 1 & 0 \\ 2 & 0 & 2 \end{pmatrix}\, $$ with $f(x,y,z) = (x,y,z)^T A(x,y,z)$ and $g(x,y,z) = (x,y,z)^T B(x,y,z)\,.$

We solve the generalized eigenvalue problem $$ A v = \lambda B v\,.$$ And obtain the eigenvalues and eigenvectors $$\begin{align} \lambda_1 &= 1, & v_1&= \begin{pmatrix} 0&1&0\end{pmatrix}^T,\\ \lambda_2 &= \frac12, & v_2 &= \frac{1}{2}\begin{pmatrix} 1&1&0\end{pmatrix}^T,\\ \lambda_3 &= \frac12, & v_3 &= \frac{1}{2}\begin{pmatrix} -1&-1&2\end{pmatrix}^T\,; \end{align} $$ here, the normalization was chosen such that $$v_j^T B v_l = \delta_{jl}.$$ Moreover, we have (from the eigenvalue condition) that $$ v_j^T A v_l = \lambda_j \delta_{jl}\,.$$ Thus, the quadratic forms are diagonalized simultaneously in the basis $v_j$.

In terms of quadratic forms, we can introduce the new variables $x',y',z'$ with $$ x= \frac12(y'-z'), \quad y= x'+\frac12 (y'-z'), \quad z =z'\,.$$ In terms of these variables, the quadratic forms are given by $$f = x'^2 + \frac12 y'^2 + \frac12 z'^2, \qquad g= x'^2 + y'^2 +z'^2\,. $$

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  • $\begingroup$ wait so this works even if $A$ and $B$ do not commute? $\endgroup$
    – AnCar
    Commented Jun 17 at 20:40
  • $\begingroup$ @AnCar: yes, you can diagonalize two quadratic forms $A,B$ simultaneously (there are some degenerate cases which you can avoid by assuming that one is positive definite); e.g., there is an invertible matrix $T$ such that $T^T A T $ and $T^T B T$ are diagonal. $\endgroup$
    – Fabian
    Commented Jun 17 at 20:44
  • $\begingroup$ I guess you mixing it up with similarity transforms: if you want $T^{-1} A T$ and $T^{-1} B T $ simultaneously diagonal, you need that $A$ and $B$ commute. $\endgroup$
    – Fabian
    Commented Jun 17 at 20:45
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    $\begingroup$ thanks, I learned something new. I guess this means that you can also diagonalize a psd matrix without a unitary matrix. This is what I was missing I think. $\endgroup$
    – AnCar
    Commented Jun 17 at 20:46

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