# Jordan Canonical Form of Linear Operator

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...

• 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. – Guido A. Sep 27 '18 at 21:53
• 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? – Corey Prachniak Sep 27 '18 at 21:59
• Okay, great, I think I got it. Thanks. – Guido A. Sep 27 '18 at 22:00

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.
• 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. – Guido A. Sep 27 '18 at 22:16