The matrix of an endomorphism

Question

Let $V$ be a vector space and $f: V\to V$ is an endomorphism. $V = V_1 \oplus V_2$ and $V$ has a finite dimension. For $i=1, 2$ is $N_i$ a basis of $V_i$ and we have that $f(V_i) \subset V_i$. Show that matrix $(f)_{NN}$ (matrix of an endomorphism), where $N = N_1 \cup N_2$, is a diagonal block matrix.

My solution:

If we put the vectors from $N_i$ into a matrix, let say $M_i$, the direct sum of those two matrices would exists, because $V = V_1 \oplus V_2$. The final matrix $M$ would have $M_1$ and $M_2$ on its diagonal, so it would be the block matrix:

$$ M =\begin{pmatrix} M_1 & 0 \\ 0 & M_2 \\ \end{pmatrix}=(1_V)_{NK} $$

The matrix of endomorphism can be written as:

$f_{NN} = (1_V)_{NK}*(f)_{KK}*(1_V)^{-1}_{NK}$

and I am thinking what about that $(f)_{KK}$.

Any suggestion how to solve this please?

Answer 1

This is simpler than you are making it.

If $N_1 = \{e_1, ..., e_k\}$ and $N_2 = \{e_{k+1}, ..., e_n\}$ then denoting the $ij$ element of $(f)_{NN}$ by $f_{ij}$, we have that $$f(e_j) = \sum_{i=1}^n f_{ij}e_i$$ by definition of that matrix.

However, we are given that $f(V_1) \subset V_1$. Now if $j \le k$ then $e_j \in N_1 \subset V_1$, and so $f(e_j) \in V_1$, and so can be written as a sum of vectors in $N_1$ only. That is, $$f(e_j) = \sum_{i=1}^k f_{ij}e_i$$ from which it follows that $f_{ij} = 0$ for $i > k$.

Conversely, since $f(V_2) \subset V_2$, if $j > k$, then $e_j \in N_2 \subset V_2$, so $f(e_j) \in V_2$ as well, and so can be expressed as a sum of vectors in $N_2$ only. Therefore if $i \le k$, then $f_{ij} = 0$.

That is, if $j \le k$ and $i > k$, or if $j > k$ and $i \le k$, then $f_{ij} = 0$, which is what it means for $(f)_{NN}$ to be block-diagonal.