# Thread: Show that there is a basis for beta

1. ## Show that there is a basis for beta

Let V be an n-dimensional vector space, and let T:V->V be a linear transformation. Suppose that W is a T-invariant subspace of V having dimension k. Show that there is a basis $\beta$ for V such that [T]_ $\beta$ has the form : (this is supposed to be a matrix im still working on my latex)

(A B
O C) where A is a kxk matrix and O is the (n-k)xk zero matrix.

2. You can start by choosing a basis $\{w_1,...,w_k\}$ for $W$, and then complete it to a basis $\{w_1,w_2,...,w_k,v_{k+1},...,v_n\}$ of $V$.

The first column of the matrix of T with respect to this basis is the coordinates of T(w_1) expressed in this basis: Writing $T(w_1) = a_1w_1+a_2w_2+...+a_kw_k+d_{k+1}v_{k+1}+...+d_nv_n ,$
the first column of the matrix consists of the numbers $a_1,a_2,...,a_k,d_{k+1},...,d_n.$

But $W$ is an invariant subspace of $T$, so since $w_1$ is a member of $W$, then $T(w_1)$ will also be a member of $W$. But this means that $T(w_1)$ can be expressed in the basis $\{w_1,w_2,...,w_k\}$ of $W$, so that all the scalars $d_{k+1},\ldots,d_n$ are $0$. In conclusion, the last $n-k$ entries in the first column of the matrix are 0.

Repeat this procedure for all the basis vectors $w_i$ to see that for each of the first $k$ columns, it is the case that the last $n-k$ entries are 0.