If $T(W)⊆W$ show $W$ spaned by eigenvectors.

Question

Let $T$ be a linear transformation of a finite dimensional real vector space $V$ and assume that $V$ is spanned by eigenvectors of $T$. If $T(W)⊆W$ for some subspace $W⊆V$, show that $W$ spaned by eigenvectors. Any suggestion? Thanks.

Answer 1

Note that the following are equivalent for $T\in L(V)$ on a finite-dimensional $K$-vector space $V$:

1- $T$ is diagonalizable.

2- there exists $p(X)\in K[X]$ which splits over $K$ with simple roots such that $p(T)=0$.

3- $V$ is spanned by eigenvectors of $T$.

Proof: the only non-completely standard implications are $3\Rightarrow 1$ or $2$. But if there is a spanning set of eigenvectors, one can extract a basis of eigenvectors from that. So $T$ is diagonalizable. $\Box$.

Now assuming $V$ is spanned by eigenvectors of $T$, we know that there exists a polynomial $p(X)$ splitting over $K$ with simple roots, such that $p(T)=0$. E.g. the minimal polynomial of $T$.

Then if $W$ is invariant under $T$, i.e. $T(W)\subseteq W$, we can consider $T_W$ the restriction of $T$ to $W$. Note that every power of $T$ leaves $W$ invariant as well whence $p(T_W)=p(T)_W=0$. Note also that every eigenvector of $T_W$ is an eigenvector of $T$. It follows that $W$ is spanned by eigenvectors of $T$.

Note: in short, if $T$ is diagonalizable, so is every restriction to an invariant subspace.