Some insight regarding a difficult problem on Linear Operators. I am required to prove the following Theorem howver despite thinking about the problem for some time i have not been able to come up with a dignified solution. Could you please provide some hints to get me going, **please do not present the complete solution.** > **Theorem.** Given that $V$ is a finite dimensional vector space and $T$ is a linear operator on $V$ such that given any subspace of $V$ say $U$ with $\dim U = \dim V-1$ is invariant under $T$ then $T$ is a scalar multiple of the identity operator.

The case $\dim V\le 1$ is trivial. Suppose $\dim V\ge2$ and consider a basis $\\{v_1,v_2,\dots,v_n\\}$ of $V$.

Consider $T(v_1)=\alpha_1v_1+\alpha_2v_2+\alpha_3v_3+\dots+\alpha_nv_n$; by assumption, $\langle v_1,v_3,\dots,v_n\rangle$ is $T$-invariant, so $\alpha_2=0$.

Similarly, $\langle v_1,v_2,v_4,\dots,v_n\rangle$ is $T$-invariant, so $\alpha_3=0$ and, doing the same for the other vectors, $T(v_1)=\alpha_1v_1$.

On the other hand, $v_1$ can be _any_ nonzero vector of $V$. Thus every nonzero vector of $V$ is an eigenvector.

The conclusion should now be easy.