You can choose a base for and complete it to get a base of .
The main interest is to write in a simpler way.
Let V be a n-dimensional vector space, and let be a linear operator on
Suppose that is a -invariant subspace of ,
Show that there exists a basis for such that has the form
where is a matrix and is the zero matrix
Request
The question itself confuses me. I am not sure what the question wants
If is a basis for , then has n vectors and
is a matrix
So what is this question asking us to reduce the matrix to a matrix with the entries themselves being matrices?
That's one way of looking at it but it is really just saying that the matrix has the left k columns of the bottom n-k rows all 0s. Use girdav's suggestion- select an ordered basis for V such that the first k vectors form a basis for W. Remember that the matrix representation of a linear transformation, in a given basis, has columns that are the coefficients of the vector you get by applying the linear transformation to the basis vectors. Since the first k basis vectors are in W and W is T-invariant, what will those first k columns look like?