# When is $PQ=QP$? $Q$ and $P$ are projections

I have a question: If $P$ and $Q$ are projections. Does it follow that $P$ and $Q$ commute? If yes, how can we prove it? If not, what is the relationship between $PQ$ and $QP$?

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In $\Bbb R^2$ let $P$ be the projection to the $x$-axis and $Q$ the projection to the subspace generated by $\langle 1,1\rangle$. Let $v=\langle 1,1\rangle$. What are $P(Q(v))$ and $Q(P(v))$? Are they the same?

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following your example, they are not the same – M.Krov Apr 12 '12 at 20:25
@M.Krov: That’s correct. – Brian M. Scott Apr 12 '12 at 20:31
@M.Krov: Let $V=\operatorname{Im}(P),W=\operatorname{Im}(Q)$. Let $U$ be the orthogonal complement of $V\cap W$. You want $V\cap U$ and $W\cap U$ to be orthogonal. – Brian M. Scott Apr 12 '12 at 23:30

Hint. The image of $PQ$ lies in $\mathrm{Im}(P)$, and the image of $QP$ lies in $\mathrm{Im}(Q)$. So we need $P(\mathrm{Im}(Q))\subseteq \mathrm{Im}(Q)\cap\mathrm{Im}(P)$, and $Q(\mathrm{Im}(P)) \subseteq \mathrm{Im}(P)\cap\mathrm{Im}(Q)$. So we need $\mathrm{Im}(Q)$ to be $P$-invariant, and $\mathrm{Im}(P)$ to be $Q$-invariant. When does that occur?

To answer the question you had asked in comments and since deleted, under what conditions do we have $QP=PQ$?

As noted in the hint, $\mathrm{Im}(P)$ must be $Q$-invariant, and $\mathrm{Im}(Q)$ must be $P$-invariant. A similar computation shows that $\mathrm{ker}(P)$ must be $Q$-invariant, and that $\mathrm{ker}(Q)$ must be $P$-invariant. The key observation to make is:

Proposition. Let $V$ be a vector space, and let $T$ be a projection on $V$ (that is, $T$ is a linear transformation $V\to V$ and $T^2=T$). If $W$ is a subspace of $V$ such that $T(W)\subseteq W$ (that is, if $W$ is $T$-invariant), then $W=(W\cap\mathrm{Im}(T)) + (W\cap \mathrm{ker}(T))$.

Proof. Let $w\in W$. Then $T(w)\in W$, hence $w-T(w)\in W$. Since $w-T(w)\in W\cap\mathrm{ker}(T)$ and $T(w)\in W\cap \mathrm{Im}(T)$, and $w=T(w)+(w-T(w))$, we conclude that $W\subseteq (W\cap\mathrm{Im}(T))+(W\cap\mathrm{ker}(T)) \subseteq W$, giving equality. $\Box$

In fact, the converse also holds (prove it).

Use the above to conclude that $PQ=QP$ if and only if $$V=(\mathrm{Im}(P)\cap\mathrm{Im}(Q)) +(\mathrm{Im}(P)\cap\mathrm{ker}(Q))+(\mathrm{ker}(P)\cap\mathrm{Im}(Q)) + (\mathrm{ker}(P)\cap\mathrm{ker}(Q))$$ in which case, the sum is a direct sum.

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Isn't the last sum always a direct sum? – user26770 Apr 13 '12 at 0:40
@Boston: Yes, because $\mathrm{Im}(P)\cap\mathrm{ker}(P) = \mathrm{Im}(Q)\cap\mathrm{Ker}(Q) = \{0\}$. – Arturo Magidin Apr 13 '12 at 2:01