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By theorem of orthogonal projections we have $P^T=P=P^2$ and $Q^T=Q=Q^2$. By definition we need to show that $(P-Q)^T(P-Q) = I = (P-Q)(P-Q)^T$. Note that $P+Q = I \iff P = I - Q$. Observe \begin{equation} \begin{split} (P-Q)^T(P-Q) &= (P^T-Q^T)(P-Q) \\ &= P^TP - P^TQ - Q^TP + Q^2 \\ &= P - PQ - QP + Q \\ &= (I-Q) - (Q-Q^2) - (Q-Q^2) + Q\\ &= I. \end{split} \end{equation} And \begin{equation} \begin{split} (P-Q)(P-Q)^T &= (P-Q)(P^T-Q^T) \\ &= PP^T -PQ^T -QP^T + QQ^T \\ &= P^2 - PQ - QP + Q^2 \\ &= P - PQ-QP - Q \\ &= (I-Q) - (I-Q)Q-Q(I-Q)+Q \\ &= I-Q +Q \\ &= I. \end{split} \end{equation} Hence $P-Q$ is an orthogonal matrix.

Is this a valid prove? Have I made correct use of the carrying out of the transposes?

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    $\begingroup$ This looks right to me. Note however that you only need to go one way, since for square matrices if $AB = I$ then $BA = I$ automatically. Also, I think the "key insight" here is that if $P, Q$ are projections with $P + Q = I$, then $PQ = 0 = QP$. We then get $(P - Q)^2 = P^2 - QP - PQ + Q^2 = P + Q = I$, which is more or less what this boils down to checking once you throw the transposes away. $\endgroup$
    – Joppy
    Oct 28, 2021 at 13:00
  • $\begingroup$ Oh yes. I see now. Easily showing that $(P-Q)^T = P-Q$ then I can do the easy $(P-Q)^2$ calculation to make it much shorter. $\endgroup$
    – Seong
    Oct 28, 2021 at 13:16

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