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Let $P_1,P_2$ be two $n\times n$ projection-matrices such that the column space of $P_2$ is contained in the column space of $P_1$. Then we have that $P_1-P_2$ is also a projection matrix with the rank of $P_1-P_2$ being $\operatorname{rank}(P_1) - \operatorname{rank}(P_2)$.

I do not see why have have $\operatorname{rank}(P_1-P_2) = \operatorname{rank}(P_1) - \operatorname{rank}(P_2)$ ?

Do you see why this is?, can you please explain it?

EDIT: A "projection-matrix" here means a matrix of an orthogonal projection. In other words, a symmetric idempotent matrix.

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    $\begingroup$ This only works if you restrict to orthogonal projection matrix. $\endgroup$ Commented Jun 25, 2019 at 0:38
  • $\begingroup$ Certainly if we're doing orthogonal projections, $P_1-P_2$ gives the projection onto the orthogonal complement of the image of $P_2$ in the image of $P_1$. $\endgroup$ Commented Jun 25, 2019 at 0:42
  • $\begingroup$ @user10354138 Yes you are right, we assume it is symmetric and idempotent? I thought an orthogonal projection matrix was one with orthogonal columns, but that is wrong? $\endgroup$ Commented Jun 25, 2019 at 0:42
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    $\begingroup$ You’re confusing “orthogonal matrix” with “matrix of an orthogonal projection.” $\endgroup$ Commented Jun 25, 2019 at 5:15
  • $\begingroup$ @amd I think it is cleard up now, so we can get back to the main question. $\endgroup$ Commented Jun 25, 2019 at 5:33

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Note: I'm assuming orthogonal projections.

It is not hard to check that the two projections commute (that is, the order of application does not matter). That means they can be diagonalized in a common basis. Indeed, it is not hard to write it down in that basis (sort the basis so that the zero diagonal elements occur last). If you do so, you can immediately see why that claim is true.

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  • $\begingroup$ Indeed, it's easy to show that $P_1P_2=P_2P_1=P_2$ (if they are othoprojectors). But after that you don't have to diagonalize the matrices: $(P_1-P_2)^2=P_1^2-P_1P_2-P_2P_1+P_2^2=P_1-P_2$. $\endgroup$ Commented Jun 26, 2019 at 12:55
  • $\begingroup$ @SergeiGolovan: That equation shows that the difference is a projector. But how does it show the rank? $\endgroup$ Commented Jun 26, 2019 at 17:42
  • $\begingroup$ You're right, it doesn't. Though if sum of two projectors ($P_2$ and $P_1-P_2$) is a projector then it's rank equals the sum of ranks. This has to be proved separately though. So maybe indeed diagonalizing makes the proof shorter. $\endgroup$ Commented Jun 26, 2019 at 17:47
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Hint: Use the fact that if $A$ is idempotent, then $rankA=traceA$. Note that $P_1$, $P_2$ and $P_1-P_2$ are all projections and hence idempotent.

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