Let $A$ be an $n\times n$ matrix such that $A^2=A^t$. Then prove that possible real eigenvalues of $A$ are $0,1$.

Let $\lambda$ be an eigenvalue of $A$ then $\lambda^2$ is eigenvalue of $A^2$.

As $A^2=A^t$, $\lambda^2$ is eigenvalue of $A^t$..

Eigenvalue of $A$ are same as eigenvalue of $A^t$..

So, real eigenvalues of $A^t$ are $\{\lambda_1,\cdots,\lambda_r,\lambda_1^2,\cdots,\lambda_r^2\}$.

As number of real eigenvalues are fixed we must have $\lambda_i^2=\lambda_i$ or $\lambda_j$ or $\lambda_j^2$..

$\lambda_i^2=\lambda_j^2$ and $\lambda_i\neq \lambda_j$ implies $\lambda_i=-\lambda_j$ i do not see any contradiction here..

$\lambda_i^2=\lambda_j$ i do not know what to conclude...

$\lambda_i^2=\lambda_i$ then $\lambda_i=0$ or $1$.. which is what i want..

Help me to clear this...

  • $\begingroup$ @amd I don't agree to close this question because the answers in the previous "similar questions" are far from satisfying. $\endgroup$ – Jean Marie Mar 7 '16 at 9:21
  • $\begingroup$ @JeanMarie : Please vote to reopen $\endgroup$ – user312648 Mar 7 '16 at 9:22
  • $\begingroup$ Where must I vote ? $\endgroup$ – Jean Marie Mar 7 '16 at 9:23
  • $\begingroup$ Just below the content of the question there are options.. share, cite, edit, reopen.. Please click reopen @JeanMarie $\endgroup$ – user312648 Mar 7 '16 at 9:24
  • $\begingroup$ I have not the "reopen" option ; not enough ranking may be... $\endgroup$ – Jean Marie Mar 7 '16 at 9:26

I think you should consider the fact that corresponding eigenvectors of $A$ and $A^2$ are the same. So these $\lambda$ and $\lambda^2$ are both eigenvalues of the same eigenvectors. Since eigenvalues must be unique, then $\lambda = \lambda^2$, which leads $\lambda = 0$ or $\lambda = 1$.


Suppose $\lambda$ is a real eigenvalue of $A$, with eigenvector $v$; then $Av=\lambda v$ and, easily, also $A^2v=\lambda^2v$. Therefore $$ A^tv=A^2v=\lambda^2v $$ and so $v^tA=\lambda^2v^t$. Hence $$ v^tAv=\lambda^2(v^tv) $$ but, on the other hand, $$ v^tAv=v^t(\lambda v)=\lambda(v^tv) $$ Since $v^tv\ne0$, we get $\lambda^2=\lambda$.

  • $\begingroup$ I missed this point it seems.. "$\lambda$ and $\lambda^2$ are eigenvalues of same eigenvector" $\endgroup$ – user312648 Mar 7 '16 at 8:23
  • $\begingroup$ $Av=\lambda v$ implies $A^2v=\lambda^2 v$ i.e., $A^t v=\lambda^2 v$.. Now i have to see if $A v=\lambda v$ implies $A^t v=\lambda v$... $\endgroup$ – user312648 Mar 7 '16 at 8:26
  • $\begingroup$ @cello see the edit. $\endgroup$ – crbah Mar 7 '16 at 8:31
  • $\begingroup$ Ok Ok... So, we are not saying $Av=\lambda v$ implies $A^t v=\lambda v$.. we are only saying that $\lambda=\lambda^2$. fine.. thanks. $\endgroup$ – user312648 Mar 7 '16 at 8:34

If $\lambda$ is an eigenvalue, so is $\lambda^2$, and so is, for the same reason, $\lambda^{4}$, $\lambda^{8}$... or any $\lambda^{2^k}$.

Thus, in the case $\lambda$ is not in $\{-1,0,1\}$, it would generate an infinite spectrum, which cannot be.

It remains the case $\lambda=-1$ that has to be eliminated, because all eigenvalues of $A$ are eigenvalues of $A^2$, and these are $\geq 0$.

  • $\begingroup$ I did not understand last line... Suppose $-1$ is an eigenvalue of $A$ then $-1$ is also an eigenvalue of $A^2$ then what? $\endgroup$ – user312648 Mar 7 '16 at 8:39
  • $\begingroup$ All eigenvalues of $A$ are eigenvalues of $A^2$. But all eigenvalues of $A^2$ are positive. Thus, it is impossible for an eigenvalue of $A$ to be negative. $\endgroup$ – Jean Marie Mar 7 '16 at 8:53
  • $\begingroup$ @JeanMarie How do you prove that the real eigenvalues of $A^2$ are $\geq 0$ ? I need to triangularize $A$ to see that, do you have a different proof ? $\endgroup$ – Gabriel Romon Mar 7 '16 at 10:40
  • 2
    $\begingroup$ @JeanMarie (continued) after some lengthy manipulations, I found an elegant proof that $-1$ cannot be an eigenvalue. Suppose that $Av=-v$. Then $A^Tv=A^2v=A(Av)=-Av=v$. Since $A^2=A^T$, $A$ commutes with $A^T$ ($AA^T=A^TA$), which allows me to "break some symmetry". Indeed, $0\leq ||Av||^2 = v^TA^TAv=v^TAA^Tv=v^TAv=-v^Tv=-||v||^2$. Hence $v=0$ ! $\endgroup$ – Gabriel Romon Mar 7 '16 at 14:31
  • 1
    $\begingroup$ @LeGrandDODOM splendid proof. I felt intuitively that one had to deal with quadratic forms to get out of this marsh: you found the path... Thank you very much for having pointed out the deficiency of my arguments. $\endgroup$ – Jean Marie Mar 7 '16 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy