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How to find matrix form of an operator on a vector space V which is direct sum of its two subspaces?

So Let's say I have a vector space $V$ over a field $F$ which is n dimensional, and additionally that I have $V=B\oplus C$ with $B$ and $C$ subspaces. If I have a linear transformation $T:V\rightarrow V$ such that restricted to either of these subspaces, $T$ is invariant, then obviously we have a matrix in $F^{n\times n}$ corresponding to $T$. I've seen the claim that if we form the matrix $Q$ whose left-most columns are a basis for $B$ and the rest a basis for $C$ that $Q^{-1}TQ$ is block diagonal. I've never quite understood why this was true, nor have I been able to prove it...Certainly this is important for proving the existence of Jordan Normal Forms etc, so I'd like to be able to justify the result.

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marked as duplicate by fpqc, Arkamis, rschwieb, Jason DeVito, Noah Snyder Oct 24 '12 at 16:11

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up vote 1 down vote accepted

Let $D=\{v_1,\dots,v_k,v_{k+1},\dots,v_n\}$ be an ordered basis for $V$ where $\{v_0,\dots,v_k\}$ is a basis for $B$ and $\{v_{k+1},\dots,v_n\}$ is a basis for $C$. Ignoring the fact that $T$ is not actually a matrix, what you mean by the matrix "$Q^{-1}TQ$" is the matrix for $T$ with respect to the ordered basis $D$. Write $$[T]_D=\begin{bmatrix}T_{1,1} & T_{1,2} \\ T_{2,1} & T_{2,2}\end{bmatrix}$$ where $T_{1,1}$ is a $k \times k$ matrix and $T_{2,2}$ is a $(n-k) \times (n-k)$ matrix. You say that $B$ is $T$-invariant, so $Tv \in B$ for all $v \in B$. In other words, $$[T]_D \begin{bmatrix} v_1 \\ 0 \end{bmatrix} = \begin{bmatrix} w \\ 0 \end{bmatrix}$$ for all $k \times 1$ vectors $v_1$. Now you can deduce that $T_{2,1}=0$. Using the fact that $C$ is $T$-invariant, you can deduce that $T_{1,2}=0$.

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Maybe the link you're missing is that, if a matrix $Q$ contains a basis $Q=[b_1|...|b_n]$ as its column vectors, then for any given matrix $T$, the matrix of lin.transformation $v\mapsto Tv$ in the basis $(b_1,...,n_n)$ is just $Q^{-1}TQ$.

And this is because,

  1. $\ TQ=T[b_1|..|b_n] = [Tb_1|..|Tb_n]$
  2. $Q\cdot\begin{pmatrix} \alpha_1\\ \vdots \\ \alpha_n \end{pmatrix} = \sum_i\alpha_ib_i$
  3. $Q^{-1}(\sum_i \alpha_ib_i) = \sum_i\alpha_i\cdot Q^{-1}b_i = \sum_i\alpha_ie_i$ where $e_i$ is the $i$th member of the standard basis in $F^n$.
  4. So, altogether $Q^{-1}TQ$ is just the matrix of $T$ written in the new basis.
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