# Invariant Subspace Problem

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

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

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

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

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

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

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

Quote:

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

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

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

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

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