Invariant subspaces

Question

Let T be a linear operator on a finite dimensional vector space V over a field F such that every subspace of V is invariant under T then how to prove T is digonalizable ? Is the converse true ?

Answer 1

If $v_1,\ldots,v_n $ is a basis, then for each $j $ we have $Tv_j=\lambda _jv_j $ for some $\lambda \in F $, since the subspace $Fv_j $ is invariant. So $T $ is diagonalizable.

This property means that $T $ is diagonalizable in every basis, so it is very strong; so the converse does not hold. Let $V=\mathbb R^2$ and $$T=\begin {bmatrix}1&0\\0&2\end {bmatrix}. $$ Consider the subspace $$V_0=\left\{(a,a)^T: a\in\mathbb R\right\}. $$ Then $T $ is diagonalizable (in the canonical basis) but $V_0$ is not invariant.

Answer 2

The property implies that $T=cI$ for some scalar $c$.

First, since every subspace is invariant, every (non-zero) vector is an eigenvector. Now suppose $u$ and $v$ are independent, $Tu=au$, $Tv=bv$. Then $$T(u+v)=au+bv.$$But we also have $$T(u+v)=c(u+v)$$for some scalar $c$. Independence shows that $a=c=b$. So $T=c I$.