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Let A,B be orthogonal matrices of order $n \geq 2 $. $\det A = 1, \det B = -1$.

There exist $a \in [0,1]$ such that $aA + (1-a)B$ is projection.

I know that the claim above is false. I fail to come up with a counterexample, so I decided to use $PP = P$ property of projection with hope to run into some contradiction.

$$a^2A^2 + a(1-a)(AB + BA) + (1-a)^2B^2 = aA + (1-a)B$$

And here I stuck. Could you help me: how can I utilize the fact that A,B are orthogonal.

Ava Skovko
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1 Answers1

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The claim is not necessarily false if $P$ was allowed to be an orthogonal projection. For a trivial example, take $A=I$ and $a=1$. The identity matrix is certainly a projection.

The claim is, however, false if $P$ is assumed to be a "strictly" nonorthogonal projection. We know that a projection $P$ is nonorthogonal if and only if $\|P\|_2>1$, see, e.g., here. However, for any $x$, $$ \|[aA+(1-a)]x\|_2\leq a\|Ax\|_2+(1-a)\|Bx\|_2=a\|x\|_2+(1-a)\|x\|_2=\|x\|_2, $$ which implies that $\|aA+(1-a)B\|_2\leq 1$.

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Algebraic Pavel
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  • On a second thought, I agree with you that the claim is correct. It must be a misprint in my answers. Sorry for bothering you. – Ava Skovko May 06 '15 at 14:25
  • @AvaSkovko What do you mean? $aA+(1-a)B$ cannot be a nonorthogonal projection since otherwise we might get a contradiction. It might, however, be an orthogonal projection unless there would be some additional hypotheses on $A$ and $B$ or possibly $a$. – Algebraic Pavel May 06 '15 at 14:37