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Suppose that I have two matrices $A$ and $B$, and I want them to share a common eigenvector $x$. For simplicity let's just assume that the eigenvalue associated with $x$ is $1$ for both matrices, so $Ax=x$ and $Bx=x$. Is there a simple condition on $A$ and $B$ which is both necessary and sufficient for this to occur?

Edit: loup blanc's answer covers the case where the eigenvalues are not known, which is generally much more interesting than the case I was asking about, which is when both eigenvalues are 1. The solution to my case is just that $\ker(A-I) \cap \ker(B-I) \ne 0$. I would still be interested if someone found an even simpler condition which is equivalent to this, though.

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  • $\begingroup$ I seems to me that it is clear that $x$ should be an eigenvector of $AB$. As you can easily compute the eigenvectors of the product, you can then try to check which are eigenvectors of both $A$ and $B$. $\endgroup$ – Fernando Muro Oct 9 '13 at 15:19
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Let $A,B$ be two $n\times n$ matrices with entries in a field $K$. Then $A,B$ have a common eigenvector iff $\cap_{k,l=1}^{n-1}\ker([A^k,B^l])\not=\{0\}$.

This result is due to D. Shemesh. Common eigenvectors of $2$ matrices. Linear algebra and appl., 62, 11-18, 1984.

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    $\begingroup$ For the few of us who do not have access to the above paper, a sketch of proof might be helpful: for the converse implication, one shows that the intersection $V$ of kernels under consideration is stable under both $A$ and $B$, and that $A$ and $B$ induce commuting endomorphisms of $V$. $\endgroup$ – Clément de Seguins Pazzis Oct 10 '13 at 12:56
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    $\begingroup$ I just had a chance to look at the paper for five minutes or so, and I guess that I've been kind of stupid the whole time. The main reason he ends up with such a complicated condition is that he assumes that the associated eigenvalues are not known. In my problem, I know the eigenvalues, so I could just write $\ker(A-I) \cap \ker(B-I) \ne 0$. This isn't as constructive as I was hoping it would be (it basically seems like restating the problem, whereas I was hoping for a more straightforward condition to enforce), but it is correct. $\endgroup$ – sasquires Oct 10 '13 at 15:42
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This may be a partial solution to your problem:

I claim that if there exists a shared eigenvector, $x$ of $A$ and $B$ with common eigenvalue of $1$ then $\det(AB - BA) = \det[A,B] = 0$.

Proof:

Suppose that there exists a shared eigenvector $x$ such that $Ax=x$ and $Bx=x$. Then, as Muro suggested, $ABx=x=BAx$. Hence, $(AB - BA)x = 0$ for some $x \neq 0$. This implies that the matrix $AB - BA = [A, B]$ is not invertible. So $\det([A,B]) = 0$.

I don't know if the condition is both necessary and sufficient. In particular, I do not know if converse to the statement above is true.

(Edit) I forgot to mention that if $A$ and $B$ commute, then they share a common eigenvector. This is a standard exercise in linear algebra.

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  • $\begingroup$ If Ax=ax and Bx=bx, in'nit ABx= A(bx) = b Ax = ba x = ab x = a Bx = B(ax) = BAx, for every matrices A, B, and eigenvector x with eigenvalues a, b with respect to these matrices? Then, a, b, should not be necessarily 1. $\endgroup$ – SashaKolpakov Oct 9 '13 at 15:52
  • $\begingroup$ @SashaKolpkov: I just put the restriction (that the shared eigenvector has a shared eigenvalue of 1) that sasquires suggests in his problem. You are right, however, that the converse should be written up differently. $\endgroup$ – Mustafa Said Oct 9 '13 at 16:05
  • $\begingroup$ Thanks, I think that this is both intuitive and computationally straightforward, although the question of the converse is still open. I'm kicking myself for not thinking of it, because it's the natural generalization of the standard theorem that $A$ and $B$ can be simultaneously diagonalized if $[A,B]=0$ (although I suppose that this corresponds to the converse). $\endgroup$ – sasquires Oct 9 '13 at 19:19
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I'm not sure if this is what you're looking for, but Donu Arapura's website contains notes on algebraic geometry. Starting on page 24 of the notes, he proves that the set of pairs of matrices $(A,B)\in M_n(k)\times M_n(k)$ (with $k$ algebraically closed) having a common eigenvector is Zariski closed in $M_n(k)\times M_n(k)$.

In particular, for each $n$, there is a system of polynomials in the entries of two arbitrary $n\times n$ matrices with the property that all of the polynomials vanish iff the matrices share a common eigenvector. So, once you know these polynomials, you have an "simple, easily computable" way to check whether or not a pair of matrices shares a common eigenvector.

Unfortunately, he mentions that starting with $n=3$, the computation of what these polynomials is "painfully slow" on his computer. (When $n=2$, is turns out the polynomial is $\det(AB-BA)$). He also mentions, and later proves, that for $n > 2$, the system of polynomials must consist of more than one polynomial. Thus, $\det(AB-BA)=0$ is necessary for sharing a common eigenvalue, but not sufficient as other answers have shown.

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This has to be a comment, but it's too long, I'm afraid. Definitely, if $A$ and $B$ have a common eigenvector, then $\det (AB-BA) = 0$. The converse is not true, since $A\text{:=}\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$ and $B\text{:=}\left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \\ \end{array} \right)$ produce the following Mathematica output:

In[156]:= Det[A.B - B.A]

Out[156]= 0

In[154]:= N[Eigenvectors[A]]

Out[154]= {{1., 0., 0.}, {-1., 0., 1.}, {-1., 1., 0.}}

In[155]:= N[Eigenvectors[B]]

Out[155]= {{2.24698, 1.80194, 1.}, {-0.801938, 0.445042, 1.}, {0.554958, -1.24698, 1.}}

Here, I do not find that $A$ and $B$ share an eigenvector, although $\det (AB-BA) = 0$.

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    $\begingroup$ The only thing which struck me about this example is that $A$ has the eigenvalue $0$ with multiplicity 2, and I was hoping that the "converse" discussed above might be true when all eigenvalues have multiplicity 1 (the typical case I'm interested in). But I've just constructed a simple example where it doesn't: $A=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}$, $B=A^T$. In this case, $\textrm{det}([A,B])=0$ but none of the eigenvectors are equal. $\endgroup$ – sasquires Oct 9 '13 at 21:17
  • $\begingroup$ @sasquires: all right, I see what you mean. I just picked up the above two matrices virtually at random (I mean I did several tries, but without too much consideration). Your example is much more clever in this regard and finally dots the "i". $\endgroup$ – SashaKolpakov Oct 9 '13 at 22:02

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