Help in understanding matrices acting on spaces I am reading a paper and I need help in understanding a proof. Here, $A, B, C$ are just some matrices, $K$ and $L$ are spaces, and $0 < a < 1$. I understood almost everything except the underlined parts, but I think that stemmed from the fact that I do not really understand the concept of $R$ acting on $H= K \oplus L$. Can someone help me with this? Thank you very much.

 A: For the first underlined bit, as I said in the comments, if you have a direct sum $H = K\oplus L$, you can say that an element of the form $[0, v]$ belongs to $L$ because it lies in $0\oplus L\cong L$, which is canonically included in $K\oplus L$.
For th underlined equation there are several things to consider. First, if $R$ is acting on $H=K\oplus L$, you can block-decompose $R$ in such a way that the  left blocks act on $K$ and the right blocks act on $L$. For that, your blocks need to have the correct dimensions, in particular the dimension of the identity matrix must be the same as the dimension of $K$ (think about it in coordinates if it helps).  That solves $t_1=\dim K$.
Next, we have proved that $(R-aI)$ is one-to-one on $K$, so the image of $K$ under $(R-aI)$ has the same dimension as $K$. Therefore, we have $\dim K = \dim (R-aI)K$.
Now, recall that we have decomposed $R$ in such a way that half of it acts on $K$ and the other half on $L$. You can do the same for $(R-aI)$. Then, $\dim(R-aI)K$ is precisely the rank of the half that acts on $K$ (because the map is one-to-one, and therefore the rank equals the number of columns, which also equals the dimension of the colummn space). You can now think of $(R-aI)$ acting on $H$ as a bigger matrix than acting on $K$ because we would be using the other half of the matrix as well. Thus, the rank of $(R-aI)$ must be at least the rank of the left half. This finally gives us $\dim(R-aI)K\leq \mathrm{rank}(R-aI)$.
