Let $T$ be a linear operator on $V$.If every subspace of $V$ is invariant under $T$,then $T$ is a scalar multiple of the identity operator.

Question

Let $T$ be a linear operator on $V$.If every subspace of $V$ is invariant under $T$,then $T$ is a scalar multiple of the identity operator.

This problem is from Hoffman Kunze form chapter Invariant Subspace.

Let $\alpha_1$ be a non zero vector then $T(\alpha_1)$ is in the subspace generated by $\alpha_1$ since every subspace is invariant under $T$.

hence $T(\alpha_1)$ = $\lambda_1\alpha_1$

Now considering another element $\alpha_2$ from the complement of the subspace generated by $\alpha_1$.

Similarly, $T(\alpha_2)$ = $\lambda_2\alpha_2$

Now $T(\alpha_1+\alpha_2)$ $=$ $T(\alpha_1)$ $+$ $T(\alpha_2)$ $=$ $\lambda_1\alpha_1$ + $\lambda_2\alpha_2$

But I cannot proceed further.

What about this theorem when $V$ is inifinte dimensional?

Answer 1

Let $V$ be a infinite dimensional vector space and $\{v,w\}$ be a linearly independent subset of $V$. Let $T(v)=\lambda v$ and $T(w)=\mu w$ for two scalars $\lambda,\mu$ using hypothesis. Now, $T\big(v+w)=\lambda v+\mu w$. Since $T(v+w)=c(v+w)$ for some scalar $c($by hypothesis$)$ we have, $c(v+w)=\lambda v+\mu v$, so that $(c-\lambda)v=(\mu-c)w$. Hence, $\mu=c=\lambda$ by linear independent assumption.

Since, $\{v,w\}$ is an arbitrary linearly independent subset we have $T$ is a scalar multiple of the identity operator.