# 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.

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?

• Also $T(\alpha_1+\alpha_2)=\lambda_3(\alpha_1+\alpha_2)$ Commented Sep 5, 2020 at 13:27
• can you please explain. Commented Sep 5, 2020 at 13:29
• It applies for every $\alpha_1, \alpha_2$ so $\lambda_1=\lambda_2=\lambda_3$. Commented Sep 5, 2020 at 13:32

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.