Invariant Subspace Representation

**letitbemww** · April 3rd 2011, 01:33 PM

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?

**Drexel28** · April 3rd 2011, 04:28 PM

Originally Posted by letitbemww

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 $S_1\cap S_2=\{\bold{0}\}$ and $\{\alpha_1,\cdots,\alpha_{n_1}$ is a basis for $S_1$ , $\{\beta_1,\cdots,\beta_{n_2}\}$ is a basis for $S_2$ , and $\{\gamma_1,\cdots,\gamma_{n_3}\}$ are vectors obtained by extending $\{\alpha_1,\cdots,\alpha_{n_1},\beta_1,\cdots,\bet a_{n_2}\}$ to a basis for $V$ then if $\mathcal{B}$ is the ordered basis $(\alpha_1,\cdots,\alpha_{n_1},\beta_1,\cdots,\beta _{n_2},\gamma_1,\cdots,\gamma_{n_3})$ then $[T]_\mathcal{B}$ can be written as $D_1\oplus D_2\oplus M$ where $D_1,D_2$ are diagonal.

**letitbemww** · April 3rd 2011, 10:06 PM

Oh I think I get it up to where you have M. What is M? Just any nxn matrix?

**Drexel28** · April 3rd 2011, 10:22 PM

Originally Posted by letitbemww

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 $D$ and $D'$ be the matrices one gets when computing $\left[T_{\mid S_1}\right]_{(\alpha_1,\cdots,\alpha_n)}$ and similarly for $D'$ . Then, to get your matrix imagine taking the block diagonal matrix $D\oplus D'$ then padding "zero rows" below them to get a $(n_1+n_2+n_3)\times(n_1+n_2)$ matrix and then add on $n_3$ columns which can be anything to get a $(n_1+n_2+n_3)\times(n_1+n_2+n_3)$ matrix. Does that make sense

**letitbemww** · April 3rd 2011, 10:44 PM

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?

**Drexel28** · April 3rd 2011, 11:01 PM

Originally Posted by letitbemww

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 $S_1$ and $S_2$ ) we know that $T$ 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 $T$ ) so they can be anything, yeah?

**letitbemww** · April 3rd 2011, 11:05 PM

Originally Posted by Drexel28

For the first two collections of columns (those corresponding to the bases of $S_1$ and $S_2$ ) we know that $T$ 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 $T$ ) so they can be anything, yeah?

Oh that's right. I didn't think about that. Thank you!

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