Geometric multiplicity for non zero eigen values of matrices $AB$ and $BA$. As lot of information is given in this site about eigen values of $AB$ and $BA$ for square matrices $A$ and $B$. As characteristics polynomial of $AB$  and $BA$ are same so both have same set of eigen values with multiplicity. Now I want to know about geoemetric multiplicity and as one of $AB$ and $BA$ may become zero and other not even diagonalizable so i can conclude that geometric multiplicity of eigen value $0$ may not equal. Now what about geometric multiplicity of common non zero eigen values? Are they will be same? i.e. if $a\neq 0$ then can we say $$Geo.Mult_a(AB) =Geo.Mult_a(BA)? $$ please explain or give counter example. Thanks.
 A: Here is a somewhat different explanation for the equality of dimensions of the eigenspaces of $AB$ and $BA$ for nonzero eigenvalues than in the other answers (so far); it gives rise to the somewhat stronger result that the Jordan types (lists of sizes of Jordan blocks) are also the same for nonzero eigenvalues. For any linear operator $T$ there is a unique $T$-stable complementary subspace$~W$ to the generalised eigenspace for the eigenvalue$~0$. There are several ways to describe it: over an algebraically closed field, $W$ is the (direct) sum of all other generalised eigenspaces; it is the image of $T^k$ for sufficiently large$~k$ ($k=n$, the dimension of the space, is certainly sufficient); if $Q$ is the quotient of the characteristic polynomial by any factors$~X$ it contains, then $W=\ker(Q[T])$.
Now let $T$ be the linear operator given by $AB$ and let $W_0$ be this subspace$~W$ for it. By construction the restriction of $T$ to $W_0$ is invertible (does not have $0$ as eigenvalue). If $W_1$ is the image of $W_0$ under multiplication by $B$, we have linear maps $b:W_0\to W_1$ (given by multiplication by $B$) and $a:W_1\to W_0$ (given by multiplication by $A$) whose composition $a\circ b$ is that invertible restriction of $T$ to $W_0$, so $a$ and $b$ must each be invertible. Starting with $T'$ given by $BA$ instead of $AB$, one sees that its subspace $W$ is in fact $W_1$. Now the restriction $a\circ b$ of $T$ to $W_0$ is conjugate to the restriction $b\circ a$ of $T'$ to$~W_1$, since $ab=a(ba)a^{-1}$. Since all (generalized) eigenspaces for nonzero eigenvalues of $AB$ respectively of $BA$ are contained in $W_0$ respectively $W_1$, one gets the desired result.
A: It's true. Let $x_1,x_2,\ldots,x_k$ be a basis of the eigenspace of $AB$ corresponding to a nonzero eigenvalue $\lambda$. Then $Bx_1,Bx_2,\ldots,Bx_k$ are linearly independent, for, if $\sum_ic_iBx_i=0$, then $\lambda\sum_ic_ix_i=A(\sum_ic_iBx_i)=0$ and hence all $c_i$s are zero. However, as $BA(Bx_i)=B(ABx_i)=\lambda Bx_i$, each $Bx_i$ is an eigenvector of $AB$ corresponding to the eigenvalue $\lambda$. Therefore the geometric multiplicity of $\lambda$ in $BA$ is greater than or equal to the geometric multiplicity of $\lambda$ in $AB$. The reverse inequality is also true if we interchange the roles of $A$ and $B$ in the above. Therefore the geometric multiplicities of $\lambda$ in $AB$ and $BA$ are the same.
A: Hint:
If $\lambda \ne 0$ is an eigenvalue of $AB$ and $BA$, check that the linear maps $$\ker(AB-\lambda I) \to \ker (BA - \lambda I), \quad x \mapsto Bx$$
$$\ker(BA-\lambda I) \to \ker (AB - \lambda I), \quad x \mapsto Ax$$
are injective. It follows $\dim \ker (AB - \lambda I) = \dim \ker (BA - \lambda I)$.
