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Let $H$ be a Hilbert space, and $V_{1}, V_{2}$ are two finite dimensional subspaces of $H$. If $P_1,P_2$ are two orthogonal projections, $P_1:H\to V_1$ and $P_2:H\to V_2$, and $P_2\circ P_1=P_2\circ P_1\circ P_2$. How to show that $\dim\; range(P_2\circ P_1)\leq \dim(V_1)$?, where "$\circ$" means composition of the two projections.

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What have you done so far? –  Arkamis Nov 9 '12 at 4:21

2 Answers 2

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Suppose $A: U \to V$, and suppose $u_1,...,u_k$ is a basis for $U$. Then $Au_1,...,A u_k$ must span ${\cal R}(A)$ (the range of $A$). Hence it is always the case that $\dim {\cal R}(A) \leq \dim U$ (where $U$ is the domain of $A$, of course).

Also, if $S,T$ are subspaces, with $S \subset T$, it follows that $\dim S \leq \dim T$.

We have ${\cal R}(P_1) \subset V_1$, hence $\dim {\cal R}(P_1) \leq \dim V_1$

Now consider the restriction of $P_2$ to ${\cal R}(P_1)$, ie, $\left.P_2\right|_{{\cal R}(P_1)}: {\cal R}(P_1) \to V_2$, then from above we have $\dim {\cal R}(\left.P_2\right|_{{\cal R}(P_1)}) \leq \dim {\cal R}(P_1) \leq \dim V_1$. However, we have ${\cal R}(P_2 \circ P_1) = {\cal R}(\left.P_2\right|_{{\cal R}(P_1)})$, hence the result follows

Note: The only relevant restriction on the operators $P_1,P_2$ is that ${\cal R}(P_1) \subset V_1$. The orthogonality or fact that they are projections are irrelevant.

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$P_2:P_1(H)\to V_2$ has range at most as big as dim$P_1(H)$ which is dim$V_1$. This restriction is exactly the composition you ask. The condition on $P_1$ and $P_2$ that you have is irrelevant. In fact, Range$(P_2\circ P_1)\leq$min(dim $V_1$, dim $V_2$).

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