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24 (A paradox for instructors) If $A^TA = AA^T$, then $A$ and $A^T$ share the same eigenvectors (true). $A$ and $A^T$ always share the same eigenvalues. Find the flaw in this conclusion: $A$ and $A^T$ must have the same X and same A. Therefore $A$ aequals $A^T$.

How does one prove the first sentence above, that $A$ and $A^T$ have the same eigenvectors, using the material found in an introductory Linear Algebra course? I mistakenly marked this as "trivial" in my solution document, only to be rightfully called on it by a student! I have tried using direct calculation as well as the Spectral Theorem (the one the class has is that $A$ is symmetric iff it is orthogonally diagonalizable). The students do not know any complex numbers, so I would prefer solutions that avoid them if possible, but I'll take such a solution if that's the best we can do. They also have no knowledge of ground fields; everything is real in this class.

Johnny Apple
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3 Answers3

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As written (with other notations, in the hermitian context and with non necessarily finite dimension) in the question of the second proposed duplicate, if $A$ (hence also $A-\lambda I$) is normal and $Av=\lambda v$ then $$\begin{align}\|A^Tv-\lambda v\|^2&= \langle(A-\lambda I)^Tv,(A-\lambda I)^Tv \rangle\\&=\langle(A-\lambda I)(A-\lambda I)^Tv,v\rangle\\&=\langle(A-\lambda I)^T(A-\lambda I)v,v \rangle\\&=\langle(A-\lambda I)^T0,v \rangle\\&=0 \end{align}$$ hence every eigenvector for $A$ is an eigenvector for $A^T.$

Similarly (or by symmetry) every eigenvector for $A^T$ is an eigenvector for $A.$

Anne Bauval
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Note that it suffices to prove the statement when $A$ is nonsingular, since otherwise we may replace $A$ with $A - \nu I$ for some $\nu$ not an eigenvalue of $A$ without changing the hypotheses or conclusion of the problem.

Let's start by assuming that $A$ (and therefore also $A^T$) is diagonalizable. If $A$ and $A^T$ are two commuting diagonalizable operators, then they may be simultaneously diagonalized, so let $v_1, \dots, v_n$ be a basis of simultaneous eigenvectors for $A$ and $A^T$. Recall that the spectrum of $A$ is the same as the spectrum of $A^T$, and moreover that the multiplicities are the same, since the two matrices share characteristic polynomials.

Suppose $\lambda$ is an eigenvalue of multiplicity $k$ for $A$ and $A^T$, and that WLOG $v_1, \dots, v_k$ is a basis for the $\lambda$-eigenspace of $A$. If not all of $v_1, \dots, v_k$ are $\lambda$-eigenvectors for $A^T$, then $A^Tv_i = \lambda v_i$ for some $i > k$. Letting $\lambda_i$ be the eigenvalue of $A$ associated to $v_i$, we have $$\lambda_i^2 \|v_i\|^2 = \langle Av_i, Av_i\rangle = \langle A^TA v_i, v_i\rangle = \lambda_i \lambda \|v_i\|^2.$$

Since $\lambda \neq \lambda_i$ and $\|v_i\| \neq 0$, the only way this can happen is if $\lambda_i = 0$. But this contradicts our assumption that $A$ is nonsingular. We conclude that the $\lambda$-eigenspace of $A$ is the same as the $\lambda$-eigenspace of $A^T$, which implies that all eigenvectors are shared with the same eigenvalues. This concludes the proof for the diagonalizable case.

If $A$ is not diagonalizable, then let $V$ be the maximal diagonalizable subspace for $A$ (i.e. spanned by all eigenvectors). We claim that this is invariant under $A^T$. Indeed, if $Av = \lambda v$, then $$AA^Tv = A^TAv = \lambda A^T v,$$ so we conclude that $A^T$ preserves the $\lambda$-eigenspace of $A$, and hence $A^T$ preserves $V$. (This is not very different than the usual argument for showing that two commuting diagonalizable matrices may be simultaneously diagonalized.) Therefore, restricting $A$ and $A^T$ to $V$ and applying the diagonalizable case, we conclude that the eigenvectors of $A$ are a subset of the eigenvectors of $A^T$. However, the same argument works symmetrically to show that the eigenvectors of $A^T$ are a subset of the eigenvectors of $A$, so these two sets are equal.

CJ Dowd
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Idea for an elementary proof (Linear Algebra I):

First of all, we observe that if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ then $A^Tv$ is another eigenvector with eigenvalue $\lambda$ under the condition $A^TA=AA^T$: $$Av=\lambda v\implies A^TAv=A^T\lambda v\implies A(A^Tv)=\lambda(A^Tv).$$ It follows that if $\dim(Ker(A-\lambda I))=1$ then $A^Tv=kv$ for some constant $k$ so $v$ is an eigenvector of $A^T$ too. Furthermore, $A^Tv=kv\implies (Av)^Tv=kv^Tv\implies \lambda ||v||^2=k||v||^2$ so that $k=\lambda$.

I am thinking about the case $\dim(Ker(A-\lambda I))>1$.

Bob Dobbs
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    Actually $k=\lambda$ and the result holds whatever the dimensions (see the community wiki answer or the proposed duplicates). – Anne Bauval Apr 29 '23 at 12:46