Proposition: If $v_1,v_2,..,v_n$ are eigenvectors of $T:V\to V$ corresponding to distinct eigenvalues and $W$ is a T-invariant subspace of V such that $\sum_{i=1}^nv_i\in W$ then $v_i\in W \space \forall i\in\{1,2,..,n\} $.
Proof:Let $\lambda_i$ be the eigenvalue corresponding $v_i$ and I be the identity transformation on $V$.
$$\sum_{i=1}^{m}v_i\in W\Rightarrow (T-\lambda_{m}I)\sum_{i=1}^{m}v_i\in W\text{ (as W is T-invariant)}\Rightarrow \sum_{i=1}^{m-1}(\lambda_{i}-\lambda_{m})v_{i}\in W\\
\Rightarrow (T-\lambda_{m-1}I)\sum_{i=1}^{m-1}(\lambda_{i}-\lambda_{m})v_i\in W\Rightarrow \sum_{i=1}^{m-2}(\lambda_{i}-\lambda_{m})(\lambda_i-\lambda_{m-1})v_{i}\in W\Rightarrow ...\\
\Rightarrow v_1\prod_{i=2}^{m}(\lambda_1-\lambda_i)\in W\Rightarrow v_1\in W\text{ (as $\lambda_i$ are distinct, $\prod_{i=2}^{m}(\lambda_1-\lambda_i)$ is non-zero)}\\
\text{and } \sum_{i=2}^nv_i\in W\;[\;\because\text{ W is a linear subspace}\;]$$ Repeating the same process we see that the proposition is true.
As $v_1,v_2,v_3$ correspond to distinct eigenvalues, they are linearly independent and consequently form a basis for $\mathbb{R}^3$.$v$ is an eigenvector of $T\Rightarrow cv$ is an eigenvector of T corresponding to the same eigenvalue if $c$ is non-zero. So if W is a $2$-dimensional T-invariant subspace and $a,b\neq 0$, $av_i+bv_j\in W\Rightarrow av_i,bv_j\in W\Rightarrow v_i,v_j\in W\Rightarrow W=span\{v_i,v_j\}$.
Best Answer
Since we are working over a complex inner product space it suffices to show that $V$ has an orthonormal basis consisting of eigenvectors of $T$. Again, since we are working over $\mathbb{C}$, we know that the characteristic polynomial of $T$ splits. Hence by Schur's theorem there exits an orthonormal basis $\{v_1,\dots,v_n\}$ such that the matrix representation of $T$, say $A$, is upper triangular. By definition, we have that $v_1$ is an eigenvector of $T$. We prove that each $v_i$ is also an eigenvector by induction. Suppose that $v_1,\dots,v_{k-1}$ are eigenvectors for some $k$. For $j<k$, let $\lambda_j$ be the eigenvector corresponding to $v_j$. Since $A$ is upper triangular $$T(v_k)=A_{1k}v_1+A_{2k}v_2+\dots+A_{kk}v_k.$$ Since our basis is orthonormal we have that $$A_{jk}=\langle T(v_k),v_j\rangle=\langle v_k, T^\ast(v_j)\rangle=\langle v_k,\eta_j v_j\rangle=\overline{\eta_j}\langle v_k,v_j\rangle=0$$where $\eta_j$ is the eigenvalue of $T^\ast$ corresponding to $v_j$. Since this holds for all $j<k$ it must be that $T(v_k)=A_{kk}v_k$.