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Thread: Invariant Subspace Representation

  1. #1
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    Invariant Subspace Representation

    If S1, S2 are T-invariant subspaces of V, with the intersection of S1 and S2 being the zero vector, then find the form of the representation of T in a basis {alpha(1),...,alpha(n1), beta(1),...,beta(n2), v(1),...,v(n3)} and alpha vectors represent S1, beta represent S2, and v vectors represent V.

    I really don't know what "the representation of T" means. Is it talking about the fact that T can be written in block diagonal form?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by letitbemww View Post
    If S1, S2 are T-invariant subspaces of V, with the intersection of S1 and S2 being the zero vector, then find the form of the representation of T in a basis {alpha(1),...,alpha(n1), beta(1),...,beta(n2), v(1),...,v(n3)} and alpha vectors represent S1, beta represent S2, and v vectors represent V.

    I really don't know what "the representation of T" means. Is it talking about the fact that T can be written in block diagonal form?
    I think what they're saying is that if $\displaystyle S_1\cap S_2=\{\bold{0}\}$ and $\displaystyle \{\alpha_1,\cdots,\alpha_{n_1}$ is a basis for $\displaystyle S_1$, $\displaystyle \{\beta_1,\cdots,\beta_{n_2}\}$ is a basis for $\displaystyle S_2$, and $\displaystyle \{\gamma_1,\cdots,\gamma_{n_3}\}$ are vectors obtained by extending $\displaystyle \{\alpha_1,\cdots,\alpha_{n_1},\beta_1,\cdots,\bet a_{n_2}\}$ to a basis for $\displaystyle V$ then if $\displaystyle \mathcal{B}$ is the ordered basis $\displaystyle (\alpha_1,\cdots,\alpha_{n_1},\beta_1,\cdots,\beta _{n_2},\gamma_1,\cdots,\gamma_{n_3})$ then $\displaystyle [T]_\mathcal{B}$ can be written as $\displaystyle D_1\oplus D_2\oplus M$ where $\displaystyle D_1,D_2$ are diagonal.
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    Oh I think I get it up to where you have M. What is M? Just any nxn matrix?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by letitbemww View Post
    Oh I think I get it up to where you have M. What is M? Just any nxn matrix?
    No, I must have been half-asleep when I wrote this--what I said was wrong assuming I'm interpreting you correctly. Let $\displaystyle D$ and $\displaystyle D'$ be the matrices one gets when computing $\displaystyle \left[T_{\mid S_1}\right]_{(\alpha_1,\cdots,\alpha_n)}$ and similarly for $\displaystyle D'$. Then, to get your matrix imagine taking the block diagonal matrix $\displaystyle D\oplus D'$ then padding "zero rows" below them to get a $\displaystyle (n_1+n_2+n_3)\times(n_1+n_2)$ matrix and then add on $\displaystyle n_3$ columns which can be anything to get a $\displaystyle (n_1+n_2+n_3)\times(n_1+n_2+n_3)$ matrix. Does that make sense
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    Oh that makes a little more sense. I'm just wondering why it can be anything in the n3 columns. I thought it would be the direct sum of D, D', and [T|V]_(v_1,...,v_n3). How can it be anything in the n3 columns?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by letitbemww View Post
    Oh that makes a little more sense. I'm just wondering why it can be anything in the n3 columns. I thought it would be the direct sum of D, D', and [T|V]_(v_1,...,v_n3). How can it be anything in the n3 columns?
    For the first two collections of columns (those corresponding to the bases of $\displaystyle S_1$ and $\displaystyle S_2$) we know that $\displaystyle T$ is invariant under those 'blocks' and so the first two collections of columns will look like a block diagonal matrix. We don't know anything about the third collection of columns (in terms of $\displaystyle T$) so they can be anything, yeah?
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    For the first two collections of columns (those corresponding to the bases of $\displaystyle S_1$ and $\displaystyle S_2$) we know that $\displaystyle T$ is invariant under those 'blocks' and so the first two collections of columns will look like a block diagonal matrix. We don't know anything about the third collection of columns (in terms of $\displaystyle T$) so they can be anything, yeah?
    Oh that's right. I didn't think about that. Thank you!
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