Exercise 6.4.15 in Shifrin and Adams' Linear Algebra: a Geometric Approach says

Suppose $A$ and $B$ are symmetric and $AB=BA$. Prove there is an orthogonal matrix $Q$ so that both $Q^{-1}AQ$ and $Q^{-1}BQ$ are diagonal. (Hint: Let $\lambda$ be an eigenvalues of $A$. Use the Spectral Theorem to show that there is an orthonormal basis for E$(\lambda)$ consisting of eigenvectors of $B$.)

We know that if those two products of matrices were diagonal then $Q$ would be made of eigenvectors. Following the first part of the hint we know that $A$x$=\lambda$x. Can someone please solve this?

  • $\begingroup$ Sorry, but what is $\mathbf{E}(A)$ meant to indicate? $\endgroup$
    – Axesilo
    May 10 '16 at 3:25
  • $\begingroup$ @DA29731: It's probably a (bad) notation for the eigenspace of $A$ associated with $\lambda$. Hint: If $AB=BA$ then the eigenspaces of $A$ are $B$-invariant. $\endgroup$ May 10 '16 at 3:33
  • $\begingroup$ @jazzinsilhouette Why do you think the notation is bad? $\endgroup$ May 13 '16 at 16:57
  • $\begingroup$ @QuinnCulver: I think this notation is "ok" if we know that $A$ has exactly one eigenvalue $\lambda$. However, in general this won't be the case and I think a meaningful notation for a general setting is something like $E_A(\lambda)$ or $E(A, \lambda)$. If it's clear which matrix we're talking about $E(\lambda)$ is okay too. But $E(A)$ carries almost no information. $\endgroup$ May 13 '16 at 17:17
  • $\begingroup$ @jazzinsilhouette Yes, you are right. The problem is that the post contained a typo. The book cited says $\mathbf{E}(\lambda)$, not $\mathbf{E}(A)$. I'll correct it. $\endgroup$ May 13 '16 at 18:43

The spectral theorem in the textbook states that

Theorem. (Spectral Theorem). Let $A$ be a symmetric $n\times n$ matrix. Then

  1. The eigenvalues of $A$ are real.

  2. There is an orthonormal basis $\{{\bf q}_1,{\bf q}_2,\ldots,{\bf q}_n\}$ for $\mathbb{R}^n$ consisting of eigenvectors of $A$. That is, there is an orthogonal matrix $Q$ so that $Q^{-1}AQ=\Lambda$ is diagonal.

Now we start proving the hint of the problem. Let $A$ and $B$ be $n\times n$ matrices, $\lambda$ be an eigenvalue of $A$, and ${\bf E}(\lambda)$ be the $\lambda$-eigenspace of $A$. Let $\beta=\{{\bf q}_1,{\bf q}_2,\ldots,{\bf q}_k\}\subseteq\mathbb{R}^n$ be an orthonormal basis for ${\bf E}(\lambda)$. Since $AB=BA$, given ${\bf x}\in{\bf E}(\lambda)$, we have $$AB{\bf x}=BA{\bf x}=B\lambda{\bf x}=\lambda B{\bf x}.$$ It follows that $B{\bf x}\in{\bf E}(\lambda)$, that is, $B({\bf E}(\lambda))\subseteq{\bf E}(\lambda)$ and we have a linear transformation $\mu_B:{\bf E}(\lambda)\rightarrow{\bf E}(\lambda)$ defined by $\mu_B({\bf x})=B{\bf x}$ for all ${\bf x}\in{\bf E}(\lambda)$. Also, since $$B{\bf x}\cdot{\bf y} ={\bf x}^\top B^\top{\bf y} ={\bf x}^\top B{\bf y} ={\bf x}\cdot B{\bf y},\quad\forall {\bf x},{\bf y}\in{\bf E}(\lambda).$$ By observing the following equations \begin{align} \mu_B({\bf q}_j)=B{\bf q}_j=\sum_{i=1}^k(B{\bf q}_j\cdot{\bf q}_i){\bf q}_i,\quad\forall j=1,2,\ldots,k, \end{align} if we let $B'$ be the $k\times k$ matrix for $\mu_B$ w.r.t. the basis $\beta$, that is, $$B'=\begin{bmatrix} B{\bf q}_1\cdot{\bf q}_1& B{\bf q}_2\cdot{\bf q}_1& \cdots& B{\bf q}_k\cdot{\bf q}_1\\ B{\bf q}_1\cdot{\bf q}_2& B{\bf q}_2\cdot{\bf q}_2& \cdots& B{\bf q}_k\cdot{\bf q}_2\\ \vdots&\vdots&\ddots&\vdots\\ B{\bf q}_1\cdot{\bf q}_k& B{\bf q}_2\cdot{\bf q}_k& \cdots& B{\bf q}_k\cdot{\bf q}_k\\ \end{bmatrix}.$$ Then $B'$ is symmetric, and by the spectral theorem, there exists an orthonormal basis $\{{\bf q}'_1,{\bf q}'_2,\ldots,{\bf q}'_k\}$ for $\mathbb{R}^k$ consisting of eigenvectors of $B'$ corresponding to eigenvalues $\mu_1,\mu_2,\ldots,\mu_k$. Define $\gamma=\{{\bf q}''_1,{\bf q}''_2,\ldots,{\bf q}''_k\}\subseteq\mathbb{R}^n$ by $${\bf q}''_j=\sum_{i=1}^k({\bf q}'_j)_{i}{\bf q}_i,\quad\forall j=1,2,\ldots,k,$$ where $({\bf q}'_j)_i$ denote the $i$th entry of ${\bf q}'_j$. Then $\gamma$ is clearly an orthonormal basis for ${\bf E}(\lambda)$. Moreover, observe that \begin{align} B{\bf q}''_j\cdot{\bf q}_l =\sum_{i=1}^k({\bf q}'_j)_{i}(B{\bf q}_i\cdot{\bf q}_l) =(B'{\bf q}'_j)_l =(\mu_j{\bf q}'_j)_l =\mu_j({\bf q}'_j)_l, \end{align} which implies \begin{align} B{\bf q}''_j\cdot{\bf q}''_l &=\sum_{i=1}^k({\bf q}'_j)_{i}(B{\bf q}_i\cdot{\bf q}_l'') =\sum_{i=1}^k({\bf q}'_j)_{i}(B{\bf q}''_l\cdot{\bf q}_i)\\ &=\mu_l\sum_{i=1}^k({\bf q}'_j)_{i}({\bf q}'_l)_i =\mu_l{\bf q}'_j\cdot{\bf q}'_l =\left\{\begin{array}{ll} \mu_l&\mbox{if }j=l;\\0&\mbox{if }j\ne l. \end{array}\right. \end{align} Therefore $B{\bf q}''_j =\displaystyle\sum_{i=1}^k(B{\bf q}''_j\cdot{\bf q}''_i){\bf q}''_i =\mu_j{\bf q}''_j$, that is, $\gamma$ consists of eigenvectors of $B$, and we complete the proof of the hint. The rest of the problem will follow immediately.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.