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.
Geometric multiplicity for non zero eigen values of matrices $AB$ and $BA$.
eigenvalues-eigenvectorslinear algebra
Related Solutions
If an $N\times N$ matrix $A$ has an eigenvalue $\lambda$, there are infinitely many vectors $v$ satisfying $Av = \lambda v$; that is, $A$ has infinitely many eigenvectors for the eigenvector $\lambda$. In fact,
$$E_{\lambda} := \{v \in \mathbb{R}^N \mid Av = \lambda v\}$$
is a subspace of $\mathbb{R}^N$.
This answers your first question, but let me point out an error in your thinking.
When we say $A$ has $N$ eigenvalues, we mean that the characteristic equation for $A$, $|\lambda I - A| = 0$, has $n$ zeroes; this is always true by the fundamental theorem of algebra, but some of the eigenvalues may be complex (e.g. a $2\times 2$ rotation matrix). Note, we count eigenvalues with multiplicity, so $\lambda$ could be repeated multiple times. We call the order of the zero $\lambda$ the algebraic multiplicity of $\lambda$.
For any eigenvalue $\lambda$, some say that $\lambda$ has $k$ corresponding eigenvectors if $\dim E_{\lambda} = k$ (this terminology is not often defined but is instead used in verbal communication). If $A$ has distinct eigenvalues $\lambda_1, \dots, \lambda_M$, one might say that $A$ has $\dim E_{\lambda_1} + \dots + \dim E_{\lambda_M}$ corresponding eigenvectors. The dimension of $E_{\lambda}$ is called the geometric multiplicity of $\lambda$.
We have the following result relating the two notions of multiplicity:
The geometric multiplicity is less than or equal to the algebraic multiplicity.
There are cases where the geometric multipicity of $\lambda$ is strictly less than the algebraic multiplicity, and therefore $A$ has less than $N$ corresponding eigenvectors. For example,
$$A = \left[\begin{matrix}1 & 1\\ 0 & 1\end{matrix}\right]$$
has a repeated eigenvalue of $1$ but only has a one-dimensional eigenspace.
To reconcile the difference between geometric multiplicity and algebraic multiplicity, one can consider generalised eigenvectors.
Every symmetric matrix is diagonalizable, hence the algebraic and geometric multiplicities of eigenvalues coincide.
Best Answer
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.