Question

Let T be a linear operator over the finite dimensional vectorspace V.
Show that T is diagonalizable if and only if T is a product of one dimensional T-invariant subspaces.

This is fairly trivial to show. I've provided a sketch of a proof below. Its fairly rough and I need to make it a bit more rigorous, but I'd like some comments.

Answer 1

1) Assume the operator is diagonalizable

Then there is a basis B of eigenvectors of T for the vectorspace V.
I claim that each eigenvector spans a one dimensional T-invariant subspace and the direct sum of these is V

Proof:
\[\text{Let $v \in B$, and let $W$ be the subspace generated by $v$.}\]\[W = \left\{av : a \in \mathrm{F}\right\}\]
\[\text{Let $u \in W$ and $b \in \mathrm{F}$:}\]\[T(u) = T(bv) = bT(v) = (b\lambda)v\]So W is T-invariant

The basis vectors span one dimensional subspaces, they are disjoint from each since the vectors are linearly independent. Since the number of vectors in a basis is equal to the dimension of the vector space the direct sum of these subspaces is the vector space.

2) Assume there are one dimensional T-invariant subspaces whose direct sum is V

Since each subspace is one dimensional we can form a basis for each subspace by taking a single vector from each subspace. Since the subspaces are disjoint from each other (direct sum) the set B consisting of the basis vectors taken from each subspace is linearly independent.

Since it is a direct sum of one dimensional subspaces, for the direct sum to be V there must be as many subspaces as the dimension of V. That means |B| = dim V and since B is also linearly independent then B must be a basis for V.

So it suffices to show that each vector in the basis for V is an eigenvector.
Since each basis vector belongs to a 1 dimensional T-invariant subspace the image of each such vector can be expressed as a linear combination of a single basis vector for the subspace.

\[\text{Let $v \in B$ be the basis vector for T-invariant subspace $W$}\]\[\text{Since W is T-invariant $T(v) \in W$}\]
Therefore the image of v can be written as a linear combination of a set of basis vectors for W, but v is also a basis for W.

\[T(v) = \lambda v\]
Therefore we have a basis for V of eigenvectors so T is diagonalizable.

Answer 2

I think it's fine. I don't think you have to be more rigorous (the parts you explained in words are well-founded and correct). I think more rigor would be redundant, it's fine as it is.