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If A has eigenvector $\mathbf{v}_1$ so that $A\mathbf{v}_1=\lambda_1\mathbf{v}_1$and B has eignenvector $\mathbf{v}_2$ so that $B\mathbf{v}_2=\lambda_2\mathbf{v}_2$, then what can you say about AB? can you say $AB\mathbf{v_3}=\lambda_3\mathbf{v}_3$? and what would be the relationship between $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$ and what would be the relationship between $\lambda_1,\lambda_2,\lambda_3$?

Edit $A,B$ are 3 by 3 matrices and $\lambda_1,\lambda_2,\lambda_3$ can be real or complex numbers and $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$ is a triple.

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Any (real) $3\times 3$ matrix will have eigenvectors, so $AB$ certainly has eigenvectors. But there need not be any relation between $\mathbf{v}_2$ and $\mathbf{v}_3$, or between $\mathbf{v}_1$ and $\mathbf{v}_3$ (although there can be relations between them, depending on the specific $A$ and $B$, or on the choice of $\mathbf{v}_2$). I find this question somewhat confusing. – Arturo Magidin Mar 17 '11 at 13:44

It is possible to have $2 \times 2$ diagonal matrices $A$ and $B$ so that $A$ has eigenvalues $0$ and $\lambda_1$ and $B$ has eigenvalues $0$ and $\lambda_2$, but $AB$ is the $0$ matrix. This can be extended to $3\times 3$ matrices.

It can also be the case for $2 \times 2$ matrices or larger that the only eigenvalues of $A$ and $B$ are $0$, but $AB$ has an arbitrary nonzero eigenvalue.

If either of $A$ or $B$ is not invertible, then $AB$ can't be, so it must have $0$ as an eigenvalue.

If the eigenvectors $v_1$ and $v_2$ are parallel, then $v_1$ must be an eigenvector of $AB$ of eigenvalue $\lambda_1 \lambda_2$.

However, my guess is that if $v_1$ and $v_2$ are not parallel, and $\lambda_1$ and $\lambda_2$ are not $0$, then there is no restriction on the set of eigenvalues of $AB$.

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