# Showing that the eigenvalues of $B$ are all real.

Let $$A$$ be a real $$n \times n$$ symmetric matrix with distinct eigenvalues (that is, $$A$$ has no repeated eigenvalue). Let $$B$$ be a real $$n \times n$$ matrix that commutes with $$A.$$ Then show that the eigenvalues of $$B$$ are all real.

I couldn't quite able to do this problem. Could anyone please give me some small hint? I started with some like that $$:$$

Let $$\lambda$$ be an eigenvalue of $$B$$ corresponding to an eigenvector $$x$$ then we have $$\overline {\lambda}\ \|x\|^2 = \langle Bx, x \rangle = \langle x, B^t x \rangle.$$ But since $$A$$ is symmetric and $$B$$ commutes with $$A$$ hence so is $$B^t.$$ Using commutativity of $$A$$ and $$B^t$$ one can easily show that if $$x$$ is eigenvector of $$A$$ corresponding to some eigenvalue $$\mu$$ then $$B^t x$$ is also an eigenvector of $$A$$ corresponding to the same eigenvalue. But eigenspaces of $$A$$ are all one dimensional and hence this proves that $$x$$ is also an eigenvector of $$B^t.$$ Now if we can show that $$\lambda$$ is an eigenvalue of $$B^t$$ corresponding to the eigenvector $$x$$ we are through. This is where I got stuck.

Thanks a bunch.

Hint: By the spectral theorem, all eigenvalues of $$A$$ are real. Let $$x \in \Bbb R^n$$ be an eigenvector of $$A$$ with associated eigenvalue $$\lambda$$. Show that $$x$$ must also be an eigenvector of $$B$$. You might find it helpful to separately consider the cases where $$\lambda \neq 0$$ and $$\lambda = 0$$.
• BenGrossman$:$ please have a look at my edit. Sep 15 at 20:08
• @Rabin Your proof is mostly fine. However, it is not necessarily true that $B^t x$ is an eigenvector; it is also possible that $B^t x = 0$; this needs to be addressed. Also, there is no reason to go from $B$ to $B^t$; you could make the same argument using $B$ instead. Sep 15 at 20:12
• @Rabin If $x \in \Bbb R^n$ is an eigenvector of a real matrix $B$, then it must hold that the associated eigenvalue is real. Sep 15 at 20:13
If $$A$$ is real symmetric with distinct eigenvalues, it is diagonalizable with real eigenvalues and orthogonal eigenvectors. So $$A=U^TDU$$ where $$U\in\mathbb R^{n\times n}$$ and $$U^TU=I$$ and $$D$$ is diagonal with distinct real elements. Hence $$AB=BA \quad\Longleftrightarrow\quad U^TDUB=BU^TDU$$ and hence $$DUBU^T=UBU^TD$$ If $$D=\mathrm{diag}(d_1,\ldots,d_n)$$, $$L=UBU^T=(L_{ij})$$, then $$LD=DL$$, and for $$i\ne j$$, $$d_iLe_i=L(d_ie_i)LDe_i=DLe_i$$ and hence $$Le_i$$ is a multiple of $$e_i$$. (Here $$e_1,\ldots,e_n$$ is the standard basis of $$\mathbb R^n$$.) Say $$Le_i=\lambda_i e_i$$. This means that $$L=\mathrm{diag}(\lambda_1,\ldots,\lambda_n).$$ Hence $$B=U^TLU$$, If $$v_k\in\mathbb R^n$$ is the $$k-$$column of $$U^T$$, then $$Bv_k=U^TLUv_k=U^TLe_k=\lambda_k U^Te_k=\lambda_kv_k$$ and therefore $$\lambda_k=\langle v_k,Bv_k\rangle\in\mathbb R.$$