# Do commuting matrices share the same eigenvectors?

In one of my exams I'm asked to prove the following

Suppose $$A,B\in \mathbb R^{n\times n}$$, and $$AB=BA$$, then $$A,B$$ share the same eigenvectors.

My attempt is let $$\xi$$ be an eigenvector corresponding to $$\lambda$$ of $$A$$, then $$A\xi=\lambda\xi$$, then I want to show $$\xi$$ is also some eigenvector of $$B$$ but I get stuck.

• It's enough to make both matrices diagonalizable. – Omnomnomnom Apr 9 '15 at 14:40
• Take $A=I$ and $B=2I$, then they have different eigenvalues but they commute. What you can say, though, is that if $AB=BA$ then $A$ is diagonalizable iff $B$ is and then they can be diagonalized simultaneously – Shai Deshe Apr 9 '15 at 14:43
• @Vim: read math.stackexchange.com/questions/6258/…. – Dietrich Burde Apr 9 '15 at 14:43

(Modified) Answer of Qiaochu Yuan here: This is false in a sort of trivial way. The identity matrix $I$ commutes with every matrix and has eigenvector set all of the underlying vector space $V$, but no other matrix except a multiple of the identity matrix has this property.

Commuting matrices do not necessarily share all eigenvector, but generally do share a common eigenvector.

Let $$A,B\in\mathbb{C}^{n\times n}$$ such that $$AB=BA$$. There is always a nonzero subspace of $$\mathbb{C}^n$$ which is both $$A$$-invariant and $$B$$-invariant (namely $$\mathbb{C}^n$$ itself). Among all these subspaces, there exists hence an invariant subspace $$\mathcal{S}$$ of the minimal (nonzero) dimension.

We show that $$\mathcal{S}$$ is spanned by some common eigenvectors of $$A$$ and $$B$$. Assume that, say, for $$A$$, there is a nonzero $$y\in \mathcal{S}$$ such that $$y$$ is not an eigenvector of $$A$$. Since $$\mathcal{S}$$ is $$A$$-invariant, it contains some eigenvector $$x$$ of $$A$$; say, $$Ax=\lambda x$$ for some $$\lambda\in\mathbb{C}$$. Let $$\mathcal{S}_{A,\lambda}:=\{z\in \mathcal{S}:Az=\lambda z\}$$. By the assumption, $$\mathcal{S}_{A,\lambda}$$ is a proper (but nonzero) subspace of $$\mathcal{S}$$ (since $$y\not\in\mathcal{S}_{A,\lambda}$$).

We know that for any $$z\in \mathcal{S}_{A,\lambda}$$, $$Bz\in \mathcal{S}$$ since $$\mathcal{S}_{A,\lambda}\subset\mathcal{S}$$ and $$\mathcal{S}$$ is $$B$$-invariant. However, $$A$$ and $$B$$ commute so $$ABz=BAz=\lambda Bz \quad \Rightarrow\quad Bz\in \mathcal{S}_{A,\lambda}.$$ This means that $$\mathcal{S}_{A,\lambda}$$ is $$B$$-invariant. Since $$\mathcal{S}_{A,\lambda}$$ is both $$A$$- and $$B$$-invariant and is a proper (nonzero) subspace of $$\mathcal{S}$$, we have a contradiction. Hence every nonzero vector in $$\mathcal{S}$$ is an eigenvector of both $$A$$ and $$B$$.

EDIT: A nonzero $$A$$-invariant subspace $$\mathcal{S}$$ of $$\mathbb{C}^n$$ contains an eigenvector of $$A$$.

Let $$S=[s_1,\ldots,s_k]\in\mathbb{C}^{n\times k}$$ be such that $$s_1,\ldots,s_k$$ form a basis of $$\mathcal{S}$$. Since $$A\mathcal{S}\subset\mathcal{S}$$, we have $$AS=SG$$ for some $$G\in\mathbb{C}^{k\times k}$$. Since $$k\geq 1$$, $$G$$ has at least one eigenpair $$(\lambda,x)$$. From $$Gx=\lambda x$$, we get $$A(Sx)=SGx=\lambda(Sx)$$ ($$Sx\neq 0$$ because $$x\neq 0$$ and $$S$$ has full column rank). The vector $$Sx\in\mathcal{S}$$ is an eigenvector of $$A$$ and, consequently, $$\mathcal{S}$$ contains at least one eigenvector of $$A$$.

EDIT: There is a nonzero $$A$$- and $$B$$-invariant subspace of $$\mathbb{C}^n$$ of the least dimension.

Let $$\mathcal{I}$$ be the set of all nonzero $$A$$- and $$B$$-invariant subspaces of $$\mathbb{C}^n$$. The set is nonempty since $$\mathbb{C}^n$$ is its own (nonzero) subspace which is both $$A$$- and $$B$$-invariant ($$A\mathbb{C}^n\subset\mathbb{C}^n$$ and $$B\mathbb{C}^n\subset\mathbb{C}^n$$). Hence the set $$\mathcal{D}:=\{\dim \mathcal{S}:\mathcal{S}\in\mathcal I\}$$ is a nonempty subset of $$\{1,\ldots,n\}$$. By the well-ordering principle, $$\mathcal{D}$$ has the least element and hence there is a nonzero $$\mathcal{S}\in\mathcal{I}$$ of the least dimension.

• It is also probably worth adding that the S as above will be spanned by some common eigenvectors of A and B which all have identical eigenvalues, since you showed that each vector in S is also an eigenvector of both A and B. – ashu May 18 at 0:03

The answer is in the book Linear Algebra and its Application by Gilbert Strang. I'll just write down what he said in the book.

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 one dimensional – then $$Bx$$ must be a multiple of $$x$$. In other words $$x$$ is an eigenvector of $$B$$ as well as $$A$$.

There's another proof using diagonalization in the book.

• So the assertion holds if $A$ and $B$ only have eigenvalues of multiplicity 1. This is probably the best we can hope for. – Klaus Draeger Apr 9 '15 at 16:23

As noted in another answer, the statement is not true as stated, just take $$A = \begin{bmatrix}1&0\\0&1\\\end{bmatrix}, \qquad B = \begin{bmatrix}1&0\\0&2\\\end{bmatrix}.$$

What is true is that, if $A$ and $B$ are diagonalizable, then $A$ and $B$ can be simultaneously diagonalized. Thanks to Thomas Andrews for pointing out an oversight.

Applying $B$ to both sides of $\lambda \xi = A \xi$ you get $\lambda (B \xi) = B A \xi = A (B \xi)$, so either $B \xi = 0$, or $B \xi$ is an eigenvector for $A$ with respect to the eigenvalue $\lambda$.

In any case $B$ maps the eigenspace $W$ of $A$ relative to the eigenvalue $\lambda$ into itself. On $W$, $A$ acts like the scalar $\lambda$. Now one can put $B$ in diagonal form on $W$ without changing the scalar shape of $A$ on $W$.

Matrices $$A = \pmatrix{0&1\\0&0}, B = \pmatrix{1&0\\0&1}$$ commute, but they don't share the eigenvector $$\pmatrix{0\\1}$$ of $$B.$$

• (0,1) is an eigenvector of A with eigenvalue zero. More precisely, the eigenbasis of both of these matrices is the same in C^{2}- – James Smithson Jul 18 '17 at 18:11

Let $$S$$ be a set of commuting matrices over an algebraically closed field $$F$$. As Algebraic Pavel said above, there may not be a common basis of eigenvectors (since any of them may not be diagonalizable!) but there must be at least a common eigenvector. Let us prove that this can also be seen as an easy consequence of Burnside's theorem on matrix algebras:

Burnside's theorem on matrix algebras states that if $$F$$ is algebraically closed, $$V$$ is a finite-dimensional $$F$$-vector space and $$S$$ is a proper subalgebra of $$\text{End}(V)$$ then there exists a nontrivial $$S$$-invariant subspace, i.e, there exists $$W\leq V$$ with $$0\neq W\neq V$$ such that $$s(W)\subseteq W$$ for every $$s\in S$$.

Suppose $$S\subseteq M_n(F)$$ with $$n>1$$ is commuting. Observe that a subspace of $$F^n$$ is $$S$$-invariant if and only if it is invariant for $$\langle S\rangle$$, the subalgebra of $$M_n(F)$$ generated by $$S$$. Since $$S$$ is commuting, $$\langle S\rangle$$ is also commuting and therefore $$\langle S\rangle\neq M_n(F)$$. Burnside's theorem applies, and so there exists a proper and nontrivial subspace $$V\leq F^n$$ which is invariant for all $$S$$. If $$V$$ has dimension more than $$1$$ then $$\langle S\rangle\neq\text{End}(V)$$, since $$\langle S\rangle$$ is commuting, and we can apply Burnside's theorem again. By induction there exists an $$S$$-invariant subspace of dimension $$1$$, and so a common eigenvector for the matrices in $$S$$.

The other counterexample can be generated with a real skew-symmetric matrix, denote it $$K$$. Consider matrix dimension $$4 \times 4$$. Take for example $$K=\begin{bmatrix} 0 & 3 & 0 & 0 \\ -3 & 0 & 0 & 0\\ 0 & 0 & 0 & 4\\ 0 & 0 & -4 & 0 \end{bmatrix}\,.$$ Now if $$K$$ is skew-symmetric then $$K^2$$ is symmetric. $$K$$ and $$K^2$$ commute as polynomials.

Now for $$K$$ we have no real eigenvectors at all, but for $$K^2$$ (as it is symmetric) there are eigenvectors which can be presented as only real. The matrices do not share any eigenvectors.

• From the above Klaus Draeger's comment "So the assertion holds if $A$ and $B$ only have eigenvalues of multiplicity $1$. This is probably the best we can hope for". Here in the case $K^2$ we have eigenvalues $-9$ and $-16$, both of multiplicity 2. Interesting that $K$ has eigenvalues only of multiplicity 1 (although conjugate in pairs). Taking this into account it is not surprise that for example $K$ and $K^3$ share the same ( complex) eigenvectors - here we have condition of multiplicity 1 fulfilled. – Widawensen Jan 25 '18 at 10:04