If $ A $ and $ A^T $ commute, do they share the same eigenvectors when they are not diagonalizable?

Question

I saw this on Strong's book Introduction to Linear Algebra, but it didn't prove it:

If $ A $ and $ A^T $ commute, then they share the same eigenvectors.

And now I am wondering whether it is still true even if they may not be diagonalizable.

There are many similar questions like "$A$ and $B$ are both diagonalizable and commute if and only if they share the same eigenvectors", but I am still wondering whether it is still true when $B=A^T$ and $A$ is not diagonalizable.

Answer 1

I assume that you are working over $\mathbb{R}$.

Let $v$ be an eigenvector of $A$. We prove that $v$ is also an eigenvector of $A^T$.

Since $v$ is an eigenvector of $A$, there exists $\lambda \in \mathbb{R}$ such that $Av=\lambda v$. Then \begin{align*} 0 = ||Av-\lambda v||_2^2 & = (Av-\lambda v)^T(Av-\lambda v) \\ &= v^TA^TAv - \lambda v^TA^Tv - \lambda v^TAv + \lambda^2 v^Tv \\&= v^TA A^Tv - \lambda v^TA^Tv - \lambda v^TAv + \lambda^2 v^Tv \\ & = (A^Tv-\lambda v)^T(A^Tv-\lambda v)\\ & = ||A^Tv-\lambda v||_2^2 \end{align*}

and hence $A^Tv = \lambda v$, which means that $v$ is an eigenvector of $A^T$.

Answer 2

UPDATE: To complement the answer above, for matrices over a field $\mathsf k$ such as $\mathbb R$, which have nontrivial finite extension, then, as the comments below point out in somewhat different language, there are two distinct notions of "diagonalizability", or to put in another way, two distinct ways a matrix can fail to be diagonalizable. The first is when the characteristic polynomial $\chi_A(t)$ does not split into a product of linear factors over $\mathsf k$. By considering the inclusion $\mathrm{Mat}_n(\mathsf k) \subseteq \mathrm{Mat}_n(\bar{\mathsf k})$, i.e. viewing $\mathrm{Mat}_n(\mathsf k)$ as a subset of the space of matrices $\mathrm{Mat}_n(\bar{\mathsf k})$ over an algebraic closure $\bar{\mathsf k}$ of $\mathsf k$, we can remove this obstruction to diagonalizing $A$, and obviously for a given matrix, it suffices to extend $\mathsf k$ to a splitting field $K$ of $\chi_A(t) \in \mathsf k[t]$. A matrix which becomes diagonalizable over the algebraic closure of $\mathsf k$ is called semisimple.

However, it may be the case that even after extension to $\bar{\mathsf k}$ a given matrix remains impossible to diagonalize. In this case, $\bar{\mathsf k}^n$ is not equal to the direct sum of the eigenspaces of $A$, only its generalised eigenspaces, that is, if $\sigma(A)$ denotes the set of eigenvalues of $A$ over $\bar{\mathsf k}$, then $$ \bar{\mathsf k}^n = \bigoplus_{\lambda \in \sigma(A)} G_{\lambda}, \quad G_{\lambda} = \{ v \in \overline{\mathsf k}^n: (A-\lambda.I_n)^N(v)=0, \quad \text{for all } N\gg 0\}. $$ If, for $\lambda \in \sigma(A)$, we let $n_\lambda$ be the minimal integer such that $(A-\lambda.I_n)^{n_{\lambda}}(v)=0$ if $v \in G_{\lambda}$, then if, for some $\lambda \in \sigma(A)$ we have $n_\lambda>1$, $A$ is not diagonalizable.

Now set $\mathsf k = \mathbb R$ and take $\bar{\mathsf k}=\mathbb C$. Let $\langle -,-\rangle$ be the standard positive definite Hermitian form given by $\langle \varepsilon_i, \varepsilon_j\rangle = \delta_{ij}$ where $\{\varepsilon_i: 1 \leq i \leq n\}$ is the standard basis of $\mathbb C^n$ and $$ \langle \lambda u +v,w\rangle = \bar{\lambda}\langle u,w\rangle+\langle v,w\rangle, \quad \overline{\langle u,v\rangle} = \langle v,u\rangle \quad (u,v,w \in \mathbb C^n, \lambda \in \mathbb C) $$ Note that we have a natural inclusion map from $\mathbb R^n \hookrightarrow \mathbb C^n$, and that the restriction of $\langle -,-\rangle$ to $\mathbb R^n$ is the standard positive definite symmetric bilinear form on $\mathbb R^n$. Recall that the adjoint of a linear map $\alpha\in \mathrm{End}(\mathbb C^n)$ is the linear map $\alpha^*\colon \mathbb C^n \to \mathbb C^n$ characterized by the property that $$ \langle A^*v_1,v_2 \rangle = \langle v_1, Av_2\rangle, \quad \forall v_1,v_2 \in V. $$ If $\mathcal B= \{\varepsilon_1, \ldots,\varepsilon_n\}$ is an orthonormal basis of $\mathbb C^n$, then the matrix of a linear map with respect to it is given by $\langle \alpha(\varepsilon_i),\varepsilon_j\rangle$ from which it follows that if $A$ is the matrix of $\alpha$ with respect to $\mathcal B$, then the matrix of $\alpha^*$ is given by $\overline{A}^\intercal$.

Question: Which linear maps $\alpha \colon \mathbb C^n\to \mathbb C^n$ have the property that there is an orthonormal basis $\mathcal B$ of $\mathbb C^n$ with respect to which $\alpha$ is given by a diagonal matrix?

Now if we suppose that $\alpha \in\mathrm{End}(\mathbb C^n)$ is a linear map for which there is a orthonormal basis of $\mathbb C^n$ with respect to which the matrix $D$ of $\alpha$ is diagonal, then the matrix of $\alpha^*$ with respect to the same basis is $\bar{D}$ (the matrix obtained by taking the complex conjugates of its entries) and since all diagonal matrices commute, it follows that a necessary condition for such a basis to exist is that $\alpha\alpha^* = \alpha^*\alpha$.

In fact this condition is also sufficient:

Claim: Let $\alpha \in \mathrm{End}(\mathbb C^n)$ and suppose that $\alpha \alpha^* = \alpha^*\alpha$. Then there is an orthonormal basis $\mathcal B$ of $\mathbb C^n$ such that the matrices of $\alpha$ and $\alpha^*$ with respect to $\mathcal B$ are diagonal.

We claim that $\alpha$ preserves the orthogonal complement of $E_{\lambda}$. Indeed if $w \in E_{\lambda}^\perp$ then for any $v \in E_{\lambda}$ we have $$ \langle \alpha(w),v\rangle = \langle w, \alpha^*(v)\rangle $$ hence the claim follows is we can show that $\alpha^*(E_\lambda)\subseteq E_{\lambda}$. But this is easy: if $\alpha(v)=\lambda v$ then $$ \alpha(\alpha^*(v)) = \alpha^*(\alpha(v)) = \alpha^*(\lambda v) = \lambda.\alpha^*(v) $$

Since $\mathbb C^n = E_{\lambda} \oplus E_{\lambda}^{\perp}$ is an orthogonal direct sum, the result then follows by induction on dimension (the case of dimension 0 being trivial).

Now if $A\in \mathrm{Mat}_n(\mathbb R)\subseteq \mathrm{Mat}_n(\mathbb C)$ satisfies $A^\intercal A = A A^{\intercal}$ then taking $\alpha$ to be the linear map given by multiplication by the matrix $A$, $\alpha^*$ is the linear map given by multiplication by $A^\intercal$ and thus applying the above we find that $A$ is diagonalizable.