Exercise Let $E$ be a projection on $V$ and let $T$ be a linear operator on $V$. 1) Prove that the range of $E$ is invariant under $T$ if and only if $ETE=TE$. 2) Prove that both the range and null space of $E$ are invariant under $T$ if and only if $ET=TE$.
1)
Note that any projection $E$ is represented by a matrix that is "part of" an identity matrix. For simplicity's sake, let us consider the basis $B=\{\alpha_1,…,\alpha_r,…,\alpha_n\}$ such that $E_{ii}=1$ for $i\leq r$ and $0$ elsewhere.
Let $\alpha=(a_1,…,a_n)$, then $T(E(\alpha))=T(a_1,…,a_r,…,0)=\beta$. If we assume $T$ to be invariant over the range $W$ of $E$, then $\beta\in W$. $$\beta=(\beta_1,…,\beta_r,…,0),\qquad E\beta=\beta$$
Therefore $ETE=TE$. If we assume to the contrary that $ETE\neq TE$, then $TE\notin W$ so $T$ is not invariant over $W$.
After some struggling this seemed to go pretty well. Though the last argument makes me think that there may be a much faster way to prove this. Part 2 is where I've been bumping my head against a wall though.
2) Consider the same basis $B$ and let $TE\neq ET$. Then there exists some $\alpha\in V$ such that $T(a_1,…,a_r,…,0)\neq(Ta_1,…,Ta_r,…,0)$, but if this is the case, $T$ is not invariant to $W$. Assuming that $T$ is invariant to $W, \quad T(a_1, …, a_r,…,0)\in W$ for all $\alpha\in W$. Therefore $$T(a_1, …, a_r,…,0)=(Ta_1, …, Ta_r,…,0)$$
And $TE=ET$.
The way I've been treating the transformation (just pulling it inside the vector), is that something that is allowed? I've been able to convince myself but wouldn't be able to defend it to someone else.
Best Answer
If $ETE=TE\ $ then $$T(E(V))=E(T(E(V)))\leqslant E(V)$$ and so $E(V)$ is $T$-invariant.
Conversely, if $E(V)$ is $T$-invariant then any $TE(x)$ is some $E(y)$, and then $$ETE(x)=EE(y)=E(y)=TE(x)\ \text{and so}\ ETE=TE.$$
For the rest, write $F=I-E$ and check in the usual way that $F$ is also a projection, and that moreover $\text{im} F=\ker E$ and $\text{im} E=\ker F$.
We then have (by the first part) that $\ker E=\text{im} F$ is $T$-invariant if and only if $FTF=TF$. Putting $F=1-E$ and simplifying we get that $\ker E$ is $T$-invariant if and only if $ETE=ET$.
Hence both image and kernel are $T$-invariant if and only if $TE=ET$.