All Invariant Subspaces of a Linear Transformation

Question

I got this problem:

Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation such that all it's eigenvalues are 1, 2 and 3 and the corresponding eigenvectors are $v_1, v_2$ and $v_3$ respectively, Find all the T invariant subspaces of $\mathbb{R}^3$

I found that:
All the T invariant subspace of dimension 0 are: $\{0\}$
All the T invariant subspace of dimension 1 are: $span\{v_1\}$ $span\{v_2\}$ and $span\{v_3\}$
All the T invariant subspace of dimension 3 are: $\mathbb{R}^3$

How do I show that all the T invariant subspace of dimension 2 are: $span\{v_1,v_2\}$ $span\{v_1,v_3\}$ and $span\{v_2,v_3\}$
In other words how can I show that this are ALL the 2 dimensional T invariant subspaces and that there are no other T invariant 2 dimensional subspaces?

Answer 1

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 $a,b\neq 0$, $av_i+bv_j\in W\Rightarrow av_i,bv_j\in W\Rightarrow v_i,v_j\in W\Rightarrow W=span\{v_i,v_j\}$.

Answer 2

Let $M$ be an invariant subspace with dimension 2. Since $T$ is invariant on the space, as a consequence of the fundamental theorem of algebra, $T|_M$ has an eigenvalue, which is one of $\lambda_1,\lambda_2,\lambda_3$. So $M$ will include one of $v_1, v_2, v_3$, say $v_1$. Adding a linear independent element to get a basis $\{v_1, w\}$ of $M$, one can show $sp\{w\}$ is an invariant space of $T$. By the similar argument, one can see $w$ is another eigenvector, or one of $v_2, v_3$.