Simultaneous orthogonal diagonalization of two matrices

Let $A=\begin{pmatrix} 1 & -2\\ -2 & 5 \end{pmatrix}$ and $B=\begin{pmatrix} -3 & 6\\ 6 & -10 \end{pmatrix}$. Obviously $A$ is positive-definite and thus we can simultaneously diagonalize $A$ and $B$. It's easy to check that $P=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}$ is invertible and diagonalizes $A$ and $B$ such that $P^t A P=I$ and $P^t B P=\text{diag}(-3,2)$. But is there an orthogonal matrix $Q$ such that $Q^t A Q$ and $Q^t B Q$ are diagonal? How can I possibly determine whether such a matrix exists?

• they would need to have the same eigenvectors. Not the same eigenvalues, or even the same ratio of eigenvalues. Only the directions of the eigenvectors matters. For two by two one can find this out with some square roots at worst. Commented Aug 29, 2015 at 20:30
• Oh, neither one need be positive definite for there to be a matrix $P$ such as you found, there is an algorithm as long as one of them is invertible. Commented Aug 29, 2015 at 20:31
• @WillJagy - can you please elaborate - why they must have the same eigenvectors? Is it a necessary and sufficient condition? Thank you. Commented Aug 29, 2015 at 20:42
• @WillJagy - probably, but I'm not aware of a necessary condition for simultaneous diagonalization (I only know that the sufficient condition is one of the matrices being positive-definite). Commented Aug 29, 2015 at 20:43
• in the first edition of Horn and Johnson, Matrix Analysis, it is in table 4.5.15T on page 229. The second edition (2013) does it some other way, I prefer the first. Theorem 4.5.15 is stated on page 228, and case II(b) is expanded on page 231 mostly. Commented Aug 29, 2015 at 20:49

Simultaneous congruence via an orthogonal matrix implies in particular simultaneous orthogonal diagonalization and in particular, simultaneous diagonalization. Assume that there exists an orthogonal matrix $Q$ such that $Q^tAQ = \mathrm{diag}(\lambda_1, \lambda_2)$ and $Q^tBQ = \mathrm{diag}(\mu_1, \mu_2)$. Since $Q^t = Q^{-1}$, we must have that $\lambda_1,\lambda_2$ are the eigenvalues of $A$ and the columns of $Q$ are an orthonormal basis of eigenvectors (with respect to the standard inner product on $\mathbb{R}^2$) of $A$ corresponding to $\lambda_i$. Similarly, $\mu_1,\mu_2$ are the eigenvalues of $B$ and the columns of $Q$ must form an orthonormal basis of eigenvectors of $B$ corresponding to $\mu_i$. The matrices $A$ and $B$ will be simultaneously congruent to diagonal matrices via an orthogonal matrix if and only if you can find an orthonormal basis of $\mathbb{R}^2$ that consists of eigenvectors both of $A$ and $B$.

The characteristic polynomial of $A$ is $\chi_A(t) = t^2 - 6t + 9$ and so $\{\lambda_1, \lambda_2 \} = \{3 - 2\sqrt{2}, 3 + 2\sqrt{2}\}$. The corresponding eigenspaces are

$$V_{3 - 2\sqrt{2}} = \mathrm{span}\{(1 + \sqrt{2}, 1)^t\}, V_{3 + 2\sqrt{2}} = \mathrm{span}\{(1 - \sqrt{2}, 1)^t\}$$.

Since

$$\begin{pmatrix} -3 & 6\\ 6 & -10 \end{pmatrix} \begin{pmatrix} 1 + \sqrt{2} \\ 1 \end{pmatrix} = \begin{pmatrix} 3 + 6\sqrt{2} \\ -4 + 6\sqrt{2} \end{pmatrix} \neq (-4 + 6\sqrt{2}) \begin{pmatrix} 1 + \sqrt{2} \\ 1 \end{pmatrix}$$

we see that an eigenvector of $A$ won't be an eigenvector of $B$ and thus simultaneous congurence via an orthogonal matrix is impossible.

Alternatively, you can prove that two symmetric real matrices are orthogonally diagonalizable simultaneously if and only if they commute. Since $AB \neq BA$, this shows that $A$ and $B$ can't be diagonalized simultaneously.

• good; statement and proof on commuting available (in pages shown in the preview, 56-58) in Bellman, books.google.com/… Commented Aug 30, 2015 at 19:01