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Are the eigenvalues of the matrix $ AB $ equal to the eigenvalues of the matrix $ BA $ . Where the matrices A And B of sizes $ {3}\mathrm{\times}{5} $ and $ {5}\mathrm{\times}{3} $ Respectively .then how can we find the Jordan form of the matrix $ BA $ if we have the matrix:

$ {AB}\mathrm{{=}}\left[{\begin{array}{l} {{1}\hspace{0.33em}{1}\hspace{0.33em}{0}}\\ {{0}\hspace{0.33em}{1}\hspace{0.33em}{0}}\\ {{0}\hspace{0.33em}{0}\hspace{0.33em}\mathrm{{-}}{1}} \end{array}}\right] $

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    $\begingroup$ If $A B x = \lambda x$, then $(B A) B x = \lambda B x$. When $x \neq 0$, and $\lambda \neq 0$, this proves that $B x$ is an eigenvector of $B A$ for the same eigenvalue. For $\lambda = 0$, it is an eigenvalue if and only if $A$ or $B$ is singular, then it is also an eigenvalue of $B A$ $\endgroup$ – Gribouillis Jan 10 '17 at 17:41
  • $\begingroup$ Ok I will try it thanks $\endgroup$ – user401187 Jan 10 '17 at 17:42
  • $\begingroup$ That means that the eigenvalues of both matrices are equal .Is it right ? $\endgroup$ – user401187 Jan 10 '17 at 17:43
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    $\begingroup$ Indeed :) but AB and BA can be square when A and B are only rectangular! Also note the result that $B P(AB) = P(BA) B$ when $P$ is a polynomial. This can be useful for Jordan with $P(X) = (X-\lambda)^k$. $\endgroup$ – Gribouillis Jan 10 '17 at 17:58
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    $\begingroup$ The more natural question, if $A,B$ are not square and $AB$ is smaller than $BA$ (which we may assume WLOG), is whether the eigenvalues of $AB$ are contained in those of $BA$. The spectra can't possibly be the same, taking multiplicities into account. So even if you do have this containment, there is a followup question of "what extra eigenvalues does $BA$ have?" $\endgroup$ – Ian Jan 10 '17 at 18:15
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Denote $$C=\begin{bmatrix}{\lambda I}&{A}\\{B}&{I}\end{bmatrix},\quad D=\begin{bmatrix}{-I}&{0}\\{B}&{-\lambda I}\end{bmatrix}.$$ Then, $$CD=\begin{bmatrix}{\lambda I}&{A}\\{B}&{I}\end{bmatrix}\begin{bmatrix}{-I}&{0}\\{B}&{-\lambda I}\end{bmatrix}=\begin{bmatrix}{-\lambda I+AB}&{-\lambda A}\\{0}&{-\lambda I}\end{bmatrix},$$ $$DC=\begin{bmatrix}{-I}&{0}\\{B}&{-\lambda I}\end{bmatrix}\begin{bmatrix}{\lambda I}&{A}\\{B}&{I}\end{bmatrix}=\begin{bmatrix}{-\lambda I}&{- A}\\{0}&{BA-\lambda I}\end{bmatrix}.$$ Using that $\det(CD)=\det(DC)$ we get: $$\det (AB-\lambda I)(-\lambda)^n=(-\lambda)^n\det(BA-\lambda I)$$ So $\det(AB-\lambda I)=\det(BA-\lambda I)$ that is, $AB$ y $BA$ have the same characteristic polynomial as a consequence the same eigenvalues and besides (this is important), with the same algebraic multiplicity.

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  • $\begingroup$ I think it's like a special case .we can't depend on it $\endgroup$ – user401187 Jan 10 '17 at 18:15
  • $\begingroup$ It seems that your problem has two parts. If $A$ and $B$ are square matrices with the same order, the previous proof is general. $\endgroup$ – Fernando Revilla Jan 10 '17 at 18:21
  • $\begingroup$ Yes. I think that $\endgroup$ – user401187 Jan 10 '17 at 18:23
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Assuming $A$ and $B$ are square: $BA$ is invertible, so $B$ is invertible. Note that $$ B^{-1}(BA)B= AB $$ so $AB$ is similar to $BA$.


Per the clarification: it is well known that $A$ and $B$ will have the same non-zero eigenvalues, as explained in the other answer. What's more: $AB$ and $BA$ have the same rank, which means that $BA$ (a $5 \times 5$ matrix) has eigenvalue $0$ with algebraic and geometric multiplicity $2$.

We must exclude, however, the possibility that $BA$ is diagonalizable. Equivalently, we want to show that since $(AB - I)^2$ has a lower rank than $(AB - I)$, $(BA - I)^2$ has a lower rank than $(BA - I)$.


$AB$ has an eigenvector $x$ assoicated with $1$, and a generalized eigenvector $y$ satisfying $ABy = y + x$. Thus, we see that $BA$ has eigenvector $Bx$, since $$ (BA)(Bx) = B(AB)x = Bx $$ moreover, $BA$ has generalized eigenvector $By$ satisfying $$ (BA)(By) = B(AB)y = B(y + x) = (By) + (Bx) $$ Thus, $BA$ indeed fails to be diagonalizable. Thus, we know its Jordan form.

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  • $\begingroup$ But the matrices aren't square .in the assumption $\endgroup$ – user401187 Jan 10 '17 at 18:04
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    $\begingroup$ Well, you added that after I wrote my answer. I'll add something $\endgroup$ – Omnomnomnom Jan 10 '17 at 18:18
  • $\begingroup$ Ok I am waiting $\endgroup$ – user401187 Jan 10 '17 at 18:39
  • $\begingroup$ Are there any ideas? $\endgroup$ – user401187 Jan 10 '17 at 19:31
  • $\begingroup$ Thanks really a lot but I didn't get it what should I try to do firstly? $\endgroup$ – user401187 Jan 10 '17 at 20:06

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