Long comment:
$[T]_{\mathcal B}$ means both the elements of $V$ and $T(V)$ are expressed as linear combinations of the vectors from the basis $\mathcal B$. This is always possible because $T:V\to V\implies T(V)\subseteq V$.
Let $B_1=\left\{b_1,\ldots,b_k\right\}$ and let $X$ be some direct complement of $W$ in $V$ with a basis $B_2=\left\{b_{k+1},\ldots,b_n\right\}$.
Now, the columns of $[T]_{\mathcal B}$ are exactly the vectors $T\left(b_1\right), T\left(b_2\right),\ldots,T\left(b_k\right)$ written as linear combinations of $b_1,b_2,\ldots,b_k,b_{k+1},\ldots,b_n$ (in the same order).
@user7530 has already noted $X$ doesn't have to be invariant under $T$.
That means, for some $v_1\in X$, it might be possible that $T(v_1)\in X$.
So, for some $j\in\{k+1,\ldots,n\},\quad T\left(b_j\right)=\sum_{i=1}^n\alpha_ib_i$, where $\sum_{i=1}^k\alpha_ib_i$ isn't necessarily $0$, however,
we cannot express $T\left(b_1\right),\ldots,T\left(b_k\right)$ over $b_{k+1},\ldots,b_n$.
Whether or not $X$ is invariant under $T$ will only affect the block $B$.