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Suppose $\alpha$ is a linear operator on a finite dimensional vector space V. I understand that theory says there is a Jordan basis for V in which the matrix representation of $\alpha$ with respect to that basis is in Jordan form. But, what if one knows that V is a direct sum of some $\alpha$-invariant subspaces: $W_1, W_2, ..., W_k?$

Does this extra assumption give for a shorter proof that there exists a basis, B, in which the matrix representation of $\alpha$ is the diagonal of block matrices (ie $[\alpha]_B = diag(A_1, ..., A_k)$ ? I cannot seem to tell how the problem this came from wants us to use the assumption. Does anyone have any hints at what they might be after? All I see to do is avoid the assumption all together...

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  • $\begingroup$ Could you clarify what your question is? There seem to be more than one in your post, but I haven't quite understood which are contextual and which you want answered. $\endgroup$ – Guido A. Sep 27 '18 at 21:53
  • $\begingroup$ Sure. The problem is to show that if V is finite dimensional, alpha is a linear operator on V, and V is a direct sum of alpha invariant subspaces W1 through Wk, that there exists a basis B in which the representation of alpha with respect to B is the diag of block matrices. The question was: how might they want us to use the assumption that V is a direct sum of those Wi? $\endgroup$ – Corey Prachniak Sep 27 '18 at 21:59
  • $\begingroup$ Okay, great, I think I got it. Thanks. $\endgroup$ – Guido A. Sep 27 '18 at 22:00
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Here's an idea: let $B_i = \{v^i_1 , \dots, v^i_{m_i}\}$ be a basis for $W_i$ for each $i \in [k]$. Now, since $V = \bigoplus_{i=1}^k W_i$, the set $\mathcal{B} = \cup_{i=1}^nB_i$ is a basis for $V$. Consider now the matrix of $\alpha$ in this basis, $A :=[\alpha]_{\mathcal{B}}$. Now, the $i$-th colum of $A$ consists of the coordinates of the $i$-th vector of $\mathcal{B}$. Since

$$ \mathcal{B} = \{v^1_1, \dots, v^1_{m_1}, v^2_1, \dots v^k_1, \dots , v^k_{m_k}\}, $$

The columns of $A$ that correspond to the vectors $v^j_1, \dots, v^j_{m_j}$ are the coordinates of the images of these, and since $\alpha$ is $W_j$ invariant, they will correspond to again, the vectors $v^j_1, \dots, v^j_{m_j}$. This tells us that in these columns, the nonzero entries of $A$ will be in the rows that correspond to $\{v^j_l\}_{l=1}^{m_j}$, which is precisely that $A$ is block diagonal.

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  • $\begingroup$ Thank you for your help $\endgroup$ – Corey Prachniak Sep 27 '18 at 22:13
  • $\begingroup$ No problem! You can be a bit more formal if you want, writing explicitly $A_{ij}$ and all, but this is the gist of it. $\endgroup$ – Guido A. Sep 27 '18 at 22:16

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