# Invariant Subspace Problem

• Aug 9th 2010, 09:30 PM
akolman
Invariant Subspace Problem
Hello, I need help with this problem

Let $\displaystyle A$ be a $\displaystyle n \times n$ matrix, $\displaystyle B$ a $\displaystyle k \times k$ and $\displaystyle X$ a $\displaystyle n \times k$ such that $\displaystyle AX=XB$.
Show:

1. $\displaystyle X$ is an invariant subspace relative to $\displaystyle A$.
2. If $\displaystyle X$ is of column rank $\displaystyle k$, the every eigenvalue of $\displaystyle B$ is an eigenvalue of $\displaystyle A$.

• Aug 20th 2010, 04:03 AM
halbard
Invariant Subspace Problem
Quote:

1. $\displaystyle X$ is an invariant subspace relative to $\displaystyle A$.
I suppose by this wording you mean that the subspace $\displaystyle \mathop{\textrm{Ran}}X$ is invariant under $\displaystyle A$?

Let $\displaystyle u\in\mathop{\textrm{Ran}}X$. Then $\displaystyle u=Xc$ for some $\displaystyle c\in\mathbb R^k$.

Hence $\displaystyle Au=AXc=XBc$ and so $\displaystyle Au\in\mathop{\textrm{Ran}}X$; $\displaystyle \mathop{\textrm{Ran}}X$ is therefore invariant under $\displaystyle A$.

Quote:

2. If $\displaystyle X$ is of column rank $\displaystyle k$, the every eigenvalue of $\displaystyle B$ is an eigenvalue of $\displaystyle A$.
The condition on the column rank implies that if $\displaystyle v\in\mathbb R^k$ is non-zero then $\displaystyle Xv$ is also non-zero.

Now let $\displaystyle \lambda$ be an eigenvalue of $\displaystyle B$ with non-zero eigenvector $\displaystyle v\in\mathbb R^k$.

Let $\displaystyle u=Xv$ so that, by the remark above, $\displaystyle u$ is non-zero. Then

$\displaystyle Au=AXv=XBv=X(\lambda v)=\lambda Xv=\lambda u$.

Since $\displaystyle u$ is non-zero, clearly $\displaystyle \lambda$ is an eigenvalue of $\displaystyle A$.