Linear operator with invariant subspace - Representation wrt to a basis [Kolman, 6.3.12] 
A nonempty subspace $S$ of $V$ is called invariant under $L$ means : $L(S) \subseteq S$.
  Let $L: V \rightarrow V$ be a linear operator with invariant subspace $S.$
  Show that if $\dim S = m$ and $\dim V = n$, then $L$ has a representation with respect to a basis $\beta$ for $V$ of the form 
  $\begin{bmatrix}
        \color{#009900}{A_{m \times m}} & B_{m \times (n - m)} \\
     \color{#009900}{O_{(n - m) \times m}} & C_{(n - m) \times (n - m)} \\
       \end{bmatrix}$.
Answer: Since $\dim S = m$ is given, let $\beta = \{\mathbf{v_i}\}_{1 \le i \le m}$ be an ordered basis for $S$.
  Since any lin-ind subset of a vector space can be extended to a basis for this same vector space proper (Basis Extension Theorem),
  thus $T = \{\mathbf{v_i, v_{m + 1}, \cdots, v_n}\}_{1 \le i \le m}$ is an ordered basis for $V$.
  Since  $L(S) \subseteq S$, thus $\forall \; 1 \le j \le m, \; L(\mathbf{v_j}) \in S$.
  $\Longrightarrow L(\mathbf{v_j}) = \sum\limits_{1 \le j \le m}a_j\mathbf{v_j} + \sum\limits_{(m+1) \le k \le n}0\mathbf{v_k} $.
  $\Longrightarrow [L(\mathbf{v_j})]_T = \begin{bmatrix}
    a_1 & a_2 & \ldots & a_m & 0 & \ldots & 0 \\
    \end{bmatrix}^T$.

The above $[L(\mathbf{v_j})]_T$ holds for $\forall \; 1 \le j \le m$, so it describes only $\color{#009900}{\text{the first $m$ columns of the matrix representation}}$? What about submatrices $B, C$?
Source: P398, 6.3.12, Elementary Linear Algebra by Kolman. WARNING: Book's odiously vile. 

Supplementary dated Sep 12 2013 :
Doesn't the solution prove that the matrix representation of $L$ is only  $\begin{bmatrix}
        \color{#009900}{A_{m \times m}} \\ \color{#009900}{O_{(On - m) \times m}}  \\
       \end{bmatrix}$? Despite our apathy towards $B,C$, mustn't we conclude with $\begin{bmatrix}
        \color{#009900}{A_{m \times m}} & B_{m \times (n - m)} \\ \color{#009900}{O_{(n - m) \times m}} & C_{(n - m) \times (n - m)} \\
       \end{bmatrix}$ ?
 A: You are correct: the proof describes only the first $m$ columns. This is all that is necessary. The goal is to find a matrix representation where the lower $(n-m)\times m$ corner vanishes. To verify this property, we only need to consider the first $m$ columns. We don't care what $A,B,C$ are. 
Regarding your edit, we know from the problem statement $L$ is an operator from an $n$-dimensional vector space to itself, so it must be an $n\times n$ matrix (after we fix a basis). We can partition any $n\times n$ matrix into the following form (I have violated alphabetical order to conform with the problem statement)
$$\begin{bmatrix}
        {A_{m \times m}} & B_{m \times (n - m)} \\ {D_{(n - m) \times m}} & C_{(n - m) \times (n - m)} \\
       \end{bmatrix}.$$ 
Think of taking a $n\times n$ matrix and drawing a vertical line after the $m$th column and a horizontal line after the $m$th row. This gives you $4$ block matrices (here $A,B,C,D$).
Your problem asks us to show that if we fix a basis for $V$ as stipulated in the problem statement, then $D$ is the zero matrix $O$. To show this, we must only examine the first $m$ columns, since $D$ is contained in the first $m$ columns.
