Original question: Suppose $V$ is finite dimensional with $dim\ V\geq 3$ and $T\in\mathcal{L}(V)$ is such that every 2-dimensional subspace of $V$ is invariant under $T$. Prove that $T$ is a scalar multiple of the identity operator.
Reason asking: My proof is different from the proof of this solution guide Question 28. http://linearalgebras.com/5a.html
My proof: We argue by contradiction. Suppose $T$ is not a scalar multiple of the identity operator. Then there exists $u\in V$ such that $u$ and $Tu$ are linearly independent. Then they can be extended to basis of $V$ which are $\{u,Tu,v_1,\cdots,v_n\}$ with $n\geq 1$ since $dim\ V\geq 3$. Let $$U=span\{u,v_j\}$$where $1\leq j\leq n$. Clearly, $u\in U$ but $Tu\notin U$, a contradiction as we have found a 2-dimensional subspace that is not invariant under $T$. Therefore, $T$ is a scalar multiple of the identity operator.
Is this valid?
Additional request: Is there anyone who have alternative solution guide for Linear Algebra Done Right by Sheldon Axler other than the one I posted? I feel embarassing if I keep ask for verification of proof. It will be better if I have solution guide with me.