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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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    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 $B A$ – 2017-01-10
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    Ok I will try it thanks – 2017-01-10
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    That means that the eigenvalues of both matrices are equal .Is it right ? – 2017-01-10
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    I think so yes, provided $A$ and $B$ are square matrices. – 2017-01-10
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    They are not in the assumption. – 2017-01-10
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    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 $P(X) = (X-\lambda)^k$. – 2017-01-10
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    @Math1000 No, $A$ could be $p\times q$ and $B$ $q\times p$. – 2017-01-10
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    @Gribouillis Indeed. I will retract my previous false statements :) – 2017-01-10
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    So now should I find the eigenvalues of AB and their corresponding eigenvectors that are the same for both of matrices AB and BA. – 2017-01-10
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    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?" – 2017-01-10
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    The Jordan form of $AB$ and $BA$ are not necessarily equal, even when $A$ and $B$ are square. So in the general case it's not true that one is determined by the other, though perhaps in this case it is. – 2017-01-10

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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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    But the matrices aren't square .in the assumption – 2017-01-10
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    Well, you added that after I wrote my answer. I'll add something – 2017-01-10
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    Ok I am waiting – 2017-01-10
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    Are there any ideas? – 2017-01-10
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    Thanks really a lot but I didn't get it what should I try to do firstly? – 2017-01-10
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    See my latest edit. – 2017-01-10
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    İs there any difference that the matrices are complex matrices – 2017-01-10
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    @user401187 there is no difference – 2017-01-10
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    So what is the relationship between the Jordan form and the matrix is not diagonalizable ? – 2017-01-11
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    @user401187 a matrix is diagonalizable if and only if its Jordan form is diagonal, which is to say that all Jordan blocks have size $1$. – 2017-01-11
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    In our example we obtained that AB fails to be diagonalizable soo – 2017-01-11
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    @user401187 we know the eigenvalues of $BA$, we know that it has two Jordan blocks for $0$, and we know that it has one Jordan block for $1$. So, we know the Jordan form of $BA$. – 2017-01-11
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    Thanks really ..but still don't know how we proved that the eigenvalues are the same ????? – 2017-01-11
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    Well, the **non-zero** eigenvalues are the same. The other answer provides a valid proof. See also [this post](http://math.stackexchange.com/questions/124888/are-the-eigenvalues-of-ab-equal-to-the-eigenvalues-of-ba-citation-needed) – 2017-01-11
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    I have just focused on why the matrix failed to be diagonalisable – 2017-01-11
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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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    I think it's like a special case .we can't depend on it – 2017-01-10
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    It seems that your problem has two parts. If $A$ and $B$ are square matrices with the same order, the previous proof is general. – 2017-01-10
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    Yes. I think that – 2017-01-10