Let $V$ be a vector space over a field $F$ and let $\alpha \in End(V)$. IF $W$ and $Y$ are subspaces of $V$ which are invariant under $\alpha$, show that both $W+Y$ and $W\cap Y$ are invariant under $\alpha$.
I am very rusty on my math, trying to freshen up my skills for grad school but I am not sure where to start with this problem.
On
Since $\alpha$ is an element of $End(V)$, we know $\alpha$ is a linear operator (which means it satisfies the homogeneity and additivity properties). As suggested, you should refresh your memory on the properties of linear transformations. From there, the solution should be simple to find based on the suggestions given by gniourf_gniourf.
You need to show that $\alpha(W+Y)\subset W+Y$ and $\alpha(W\cap Y)\subset W\cap Y$.
Your hypotheses are: $\alpha(W)\subset W$, $\alpha(Y)\subset Y$ (and other stuff like $\alpha$ is linear from $V$ to $V$).
I'll show you the strategy for $\alpha(W+Y)\subset W+Y$: let $v\in W+Y$ and let's show that $\alpha(v)\in W+Y$ (I hope you agree this is how we should proceed). We need to translate the fact that $v\in W+Y$ into something useful: namely, this means that there exists $w\in W$ and $y\in Y$ such that $v=w+y$. Now, since $\alpha$ is linear, $\alpha(v)=\alpha(w+y)=\alpha(w)+\alpha(y)$. From our hypotheses we know that $\alpha(w)\in W$ and $\alpha(y)\in Y$, hence $\alpha(w)+\alpha(y)\in W+Y$ and we're done.
Try the intersection on your own, using the same kind of reasoning. Hint: $v\in W\cap Y$ means that $v$ is both in $W$ and $Y$. Good luck.