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Thread: Invariant Subspace Problem

  1. #1
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    Acolman, Mexico
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    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 $\displaystyle A$.


    Thanks in advance.
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  2. #2
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    Mar 2009
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    Invariant Subspace Problem

    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 $\displaystyle A$.

    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 $\displaystyle A$.
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