7
$\begingroup$

I read in G. Strang's Linear Algebra and its Applications that, if $A$ and $B$ are diagonalisable matrices of the form such that $AB=BA$, then their eigenvector matrices $S_1$ and $S_2$ (such that $A=S_1\Lambda_1S_1^{-1}$ and $B=S_2\Lambda_2 S_2^{-1}$) can be chosen to be equal: $S_1=S_2$.

How can it be proved?

I have found a proof here, but it is not clear to me how to see that $C$ is diagonalisable as $DCD^{-1}=Q$. Matrix $C$ obviously is the matrix with the coordinates of $B\mathbf{x}$ with respect to the basis $\{\mathbf{x}_1,...,\mathbf{x}_k\}$ of the eigenspaceof $V_\lambda (A)$, but I do not see how we can know that it is diagonalisable.

Thank you very much for any explanation of the linked proof or other proof!!!

EDIT: Corrected statement of the lemma I am interested in. See comments below by the users whom I thank for what they have noticed.

$\endgroup$
4
  • 1
    $\begingroup$ This isn't true. There are infinite ways to diagonalize a matrix and you can always pick $S_1\neq S_2$. $\endgroup$
    – Git Gud
    Nov 14, 2014 at 20:02
  • 1
    $\begingroup$ For example, if $S_1$ works, then so does $-S_1$. Where exactly in Strang did you read that? $\endgroup$
    – copper.hat
    Nov 14, 2014 at 20:02
  • $\begingroup$ The book states "diagonalizable matrices shares the same eigenvector matrix $S$ if and only if $AB=BA$" in a quite informal language. I think I misunderstood and I'm going to edit. Thank you for the comments! $\endgroup$ Nov 14, 2014 at 20:04
  • 3
    $\begingroup$ An accurate statement is "two matrices commute if and only if they are simultaneously diagonalizable" that is, if and only if you can write $A = S\Lambda_1S^{-1}, B = S\Lambda_2S^{-1}$ (but there's nothing requiring this). $\endgroup$
    – BaronVT
    Nov 14, 2014 at 20:08

2 Answers 2

9
$\begingroup$

The idea is to show that you can find a basis consisting of vectors that are eigenvectors of both $A$ and $B$. Then a proof goes by induction on the dimension of the space (or the size of the matrices, if you prefer that). The key observation is the following.

Let $V$ be the whole space ($\Bbb{C}^n$ or $\Bbb{R}^n$, depending). Let $\lambda$ be an eigenvalue of $A$. Consider the corresponding eigenspace $V_\lambda$. Then it follows that $B(V_\lambda)\subseteq V_\lambda$. This is because for all $x\in V_\lambda$ we have $$ A(Bx)=(AB)x=(BA)x=B(Ax)=B(\lambda x)=\lambda (Bx) $$ proving that $Bx\in V_\lambda$.

This holds for all eigenvalues of $A$. If there is more than one eigenspace, then they all have dimensions $<\dim V$, and induction hypothesis kicks in: by the above observation it is enough to settle the question for all those smaller spaces as by diagnoalizablity of $A$ the whole space is a direct sum of $V_\lambda$:s.

OTOH, if one of the $V_\lambda$:s is the whole space, then $A$ is a scalar matrix, and thus diagonalized by any matrix $S$. In that case it suffices to simply diagonalize $B$.

The base case of $1\times 1$ matrices is trivial.


[Edit]

What seems to be missing from the above is that the subspace $V_\lambda$ also has a basis consisting of eigenvectors of $B$. This can be shown as follows. Diagonalizability of $A$ means that $$ V=V_\lambda\oplus\left(\bigoplus_{\mu\neq\lambda}V_\mu\right) $$ is a sum of eigenspaces of $A$. Call that other summand $V_{\neq\lambda}$. Both $V_\lambda$ and $V_{\neq\lambda}$ are stable under $B$, because the above argument also shows that $B(V_\mu)\subseteq V_\mu$ for all $\mu$. If $\beta$ is any eigenvalue of $B$, and $U_\beta$ is the corresponding eigenspace, then any vector $y\in U_\beta$ can be uniquely written in the form $y=y_1+y_2$ with $y_1\in V_\lambda$, $y_2\in V_{\neq\lambda}$. Here $By=\beta y=(\beta y_1)+(\beta y_2)$. But as $By_1\in V_\lambda$ and $By_2\in V_{\neq\lambda}$ we must have $By= By_1+By_2$. By the direct sum property we can conclude that $By_1=\beta y_1$ and $By_2=\beta y_2$. Therefore $$ U_\beta=(U_\beta\cap V_\lambda)\oplus (U_\beta\cap V_{\neq\lambda}). $$ The claim follows from this. [\Edit].

$\endgroup$
8
  • 1
    $\begingroup$ I'm fairly sure that this has been explained on the site already, but my first search didn't show anything explicit - the result of simultaneous diagonalizability was used/mentioned many times. So I decided to post a sketch. Will delete if a suitable duplicate shows up. $\endgroup$ Nov 14, 2014 at 21:11
  • $\begingroup$ Thank you so much for your answer! I hope I've understood: if $A:V\to V$, $V=\bigoplus_i E_i$ where each $E_i$ is a space generated by eigenvectos corresponding to a set of eigenvalues of $A$ all distinct from each other. Then the restriction of $A$ to each $E_i$ and the result explained by alexjo prove the lemma. Have I misunderstood anything? Why is $V$ such a direct sum? Is every $E_i$ orthogonal to the other ones? I heartily thank you again! $\endgroup$ Nov 15, 2014 at 10:36
  • 1
    $\begingroup$ @DavideZena: You don't get orthogonality unless $A$ is Hermitian (or real symmetric). But the some of eigenspaces is always direct. If $x_1,x_2,\ldots,x_m$ belong to different eigenvalues $\lambda_i$ of $A$, then from any linear dependency relation $\sum_ic_ix_i=0$ you get another one by applying $A$: $\sum_i \lambda_ic_ix_i=0$. Then working with these two relations you can eliminate one of the vectors, and proceed by induction. Diagonalizability implies that the sum of eigenspaces is the whole space. $\endgroup$ Nov 15, 2014 at 10:56
  • 1
    $\begingroup$ The idea is that $x\in V_\lambda\implies Bx\in V_\lambda$? So both $A$ and $B$ map $V_\lambda$ to itself. If $\dim V_\lambda<\dim V$, then we can apply induction to $V_\lambda$. $\endgroup$ Nov 15, 2014 at 19:35
  • 1
    $\begingroup$ @Davide: I added an explanation showing that the restriction of $B$ to $V_\lambda$ is also diagonalizable. This is needed for the induction to work. Sorry about not explaining that earlier. $\endgroup$ Nov 16, 2014 at 21:48
7
$\begingroup$

Proposition. Diagonalizable matrices share the same eigenvector matrix $S$ if and only if $AB = BA$.

Proof. If the same $S$ diagonalizes both $A = S\Lambda_1S^{-1}$ and $B = S\Lambda_2S^{-1}$, we can multiply in either order: $$AB = S\Lambda_1S^{-1}S\Lambda_2S^{-1}= S\Lambda_1\Lambda_2S^{-1} \;\text{and}\; BA = S\Lambda_2S^{-1}S\Lambda_1S^{-1}= S\Lambda_2\Lambda_1S^{-1}.$$ Since $\Lambda_1\Lambda_2 = \Lambda_2\Lambda_1$ (diagonal matrices always commute) we have $AB = BA$.

In the opposite direction, suppose $AB = BA$. Starting from $Ax =\lambda x$, we have $$ABx = BAx = B\lambda x =\lambda Bx.$$

Thus $x$ and $Bx$ are both eigenvectors of $A$, sharing the same $\lambda$ (or else $Bx = 0$). If we assume for convenience that the eigenvalues of $A$ are distinct (the eigenspaces are all one-dimensional), then $Bx$ must be a multiple of $x$. In other words $x$ is an eigenvector of $B$ as well as $A$. The proof with repeated eigenvalues is a little longer.

$\endgroup$
4
  • 1
    $\begingroup$ This is the idea (+1). Repeated eigenvalues can be handled for example by induction. $\endgroup$ Nov 14, 2014 at 21:12
  • 1
    $\begingroup$ Thank you very much! In the case where eigenvectors of $A$ arn't distinct and $\dim V_\lambda>1$ how can the proof be generalised? I heartily thank you!!! $\endgroup$ Nov 16, 2014 at 14:40
  • $\begingroup$ Proof is just a copy-paste of the proof given in G. Strang' s book. $\endgroup$ Oct 5, 2023 at 14:15
  • $\begingroup$ how should we prove that $Bx$ is a multiple of $x$ in the case of repeated eigenvalues ? $\endgroup$
    – PermQi
    Nov 9, 2023 at 7:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .