# When are the Eigenvectors of a A and $A^T$ the same?

I have a matrix $$A$$ which has distinct, real eigenvalues. Will $$A$$ and $$A^T$$ have the same eigenvectors. I thought this result was false, but I am reading a paper which uses this as a fact when $$A$$ is a $$2\times 2$$ real matrix.

• It is indeed false. Check $A = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$ for example. Sep 9, 2020 at 21:29
• Is the paper taking the transpose of the eigenvectors as well as the transpose of the matrix? Sep 9, 2020 at 21:31
• Oops, I misread the question. Deleting my previous comments because they are misleading. Sep 9, 2020 at 21:31
• This seems strange, but eigenvalues are indeed equal, maybe authors just mixed it somehow? For eigenvectors, right eigenvectors of $A^t$ are left eigenvectors of $A$. Sep 9, 2020 at 21:54
• @PeterFranek it turns out the author stated that they were left eigenvectors and I just read that they were eigenvectors Sep 10, 2020 at 18:22

With the hypothesis that the eigenvalues of $$A$$ are distinct, $$A$$ and $$A^T$$ have the same eigenvectors iff $$A$$ is normal ($$A^T A = A A^T$$), and with the further hypothesis that the eigenvalues of $$A$$ are real, $$A$$ and $$A^T$$ have the same eigenvalues iff $$A$$ is symmetric ($$A^T = A$$). So any non-symmetric matrix with real distinct eigenvalues is a counterexample; in the $$2 \times 2$$ case we can take, for example,

$$A = \left[ \begin{array}{cc} 0 & -2 \\ 1 & 3 \end{array} \right].$$

$$A$$ has eigenvectors $$v_1 = \left[ \begin{array}{c} -1 \\ 1 \end{array} \right]$$ and $$v_2 = \left[ \begin{array}{c} -2 \\ 1 \end{array} \right]$$ with eigenvalues $$\lambda_1 = 2$$ and $$\lambda_2 = 1$$. Its transpose

$$A^T = \left[ \begin{array}{cc} 0 & 1 \\ -2 & 3 \end{array} \right]$$

has the same eigenvalues but eigenvectors $$w_1 = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right], w_2 = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$$.

You'll notice in the above example that $$v_1$$ is orthogonal to $$w_2$$ and vice versa. This generalizes: if $$v_i$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda_i$$ and $$w_j$$ is an eigenvector of $$A^T$$ with eigenvalue $$\lambda_j \neq \lambda_i$$, then

$$w_j^T A v_i = w_j^T (\lambda v_i) = \lambda_i w_j^T v_i$$

but we also have

$$w_j^T A v_i = (A w_j)^T v_i = (\lambda_j w_j)^T v_i = \lambda_j w_j^T v_i$$

which gives $$w_j^T v_i = 0$$. This says that if $$A$$ has distinct real eigenvalues then the eigenvectors of $$A$$ and $$A^T$$ are dual bases with respect to the inner product (up to scale), which among other things is one way to prove the above result: if the eigenvectors of $$A$$ aren't orthogonal then they can't be their own dual basis.

• Is it true that: "$A$ and $B$ (with distinct real eigenvalues same as $A$) have the same eigenvectors iff $B A = A B$"? I think the the last equality holds iff $B=A^k$ for some $k$. (so in case $B=A^T$ it should be $B=A^k$. isn't?) Sep 10, 2020 at 4:57
• The first statement is true but the second isn't. Any polynomial in $A$ commutes with $A$. Sep 10, 2020 at 5:20
• I actually meant that. $P(A)=I+aA+...$. So $A^T$ is a polynomial of $A$. Sep 10, 2020 at 5:23
• Yes, for a matrix with distinct eigenvalues that's equivalent, but normality is much easier to check! Sep 10, 2020 at 5:36