Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
What have you done so far? – Emily Nov 9 '12 at 4:21
up vote 0 down vote accepted

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.

share|cite|improve this answer

$P_2:P_1(H)\to V_2$ 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 $V_2$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.