What's the point of invariant subspaces I know what the definition of an invariant subspace is but I still don't really understand why it's important and/or what makes it useful. For example, part (b) of the following problem uses the fact that the eigenspace is $g$-invariant to solve the problem, and although I know it means that if $v \in V_\lambda$, then $g(v)\in V_\lambda$, I don't really understand what it means for this problem and how/why it's useful. I would really appreciate it if someone could explain to me, someone who isn't the best at this stuff, what it all means.

Let $f,g:V\to V$ be commuting linear transformations $fg=gf$.
(a) Show that any eigenspace of $f$ is $g$-invariant.
(b) If $K$ is algebraically closed and $V$ is finite-dimensional, prove that $f$ and $g$ have a common eigenvector.

(b) solution:

Let $a\in k$ be an eigenvalue of $f$. We know that $V_a$ is $g$-invariant, so we can consider the restriction $g|V_a$. It is a linear transformation on a finite-dimensional vector space $V_a$ over the algebraically closed field $K$, so it has at least one eigenvalue $b$. In other words, there exists $w\in V_a,w\neq0$ such that $$g(w)=bw.$$ But since $w\in V_a$, we also have $f(w)=aw$, so $w$ is an eigenvector for both $f$ and $g$.

 A: Let $T \colon V \rightarrow V$ be a linear operator. A $T$-invariant subspace $W \subseteq V$ is a subspace such that $T(W) \subseteq W$. This is a very restrictive condition on $W$. For example, if $W = \left< v \right>$ is a one-dimensional subspace spanned by $v$, then $W$ will be $T$-invariant if and only if $Tv \in W$ which means that there exists $\lambda \in \mathbb{F}$ such that $Tv = \lambda v$. That is, $v$ must be an eigenvector of $T$!
What's the point of invariant subspaces? If $W$ is a subspace of $V$, you can always restrict the linear map to $W$ and obtain a linear map $T \colon W \rightarrow V$ but the domain and codomain of $T$ are different and so this isn't an operator! If $W$ is $T$-invariant, you can restrict both the domain and codomain of $T$ and obtain an operator $T|_{W} \colon W \rightarrow W$. This is an extremely useful notion because it allows you to potentially understand a complicated operator $T$ defined on some big space $V$ by decomposing the space into a direct sum $V = W_1 \oplus \dots \oplus W_k$ of $T$-invariant subspaces $W_i$ and then understand the operators $T|_{W_i}$ (which are defined on smaller spaces and often much less complicated). If the $W_i$ weren't $T$-invariant, then $T$ would "mix $W_i$ with $W_j$" (for $j \neq i$) and you wouldn't gain much from such a decomposition but if the $W_i$ are $T$-invariant, $T$ doesn't "mix up" different parts and so you reduce your problem significantly. 
For example, each eigenspace $V_{\lambda} = \ker(T - \lambda I)$ is a $T$-invariant subspace and on each eigenspace, $T$ acts in the simplest possible way - it multiplies vectors by the scalar $\lambda$. Thus, if we take a complicated operator and manage to decompose our space into $V = \bigoplus_{\lambda} V_{\lambda}$, we understand exactly how $T$ acts on $V$. This is possible if $T$ is diagonalizable. If $T$ is not diagonalizable, there are other decompositions that you can get that tells you much about the operator (generalized eigenspaces, cyclic subspaces, etc).
