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Searching for some help with this exam review proof. If we let $x$ be a eigenvector for $Q$, that is, a non zero vector satisfying $Qx=cx$ for some scalar $c$. How do I show that $c= \pm1.$

EDIT: $Q$ is a matrix with orthonormal columns

Gerry Myerson
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tshelton
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2 Answers2

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We can prove this easily in terms of vector norms. $$\left|\left|{Qx}\right|\right| = \sqrt{(Qx)^\top(Qx)} = \sqrt{x^\top (Q^\top Q) x} = \sqrt{x^\top I x} = \sqrt{x^\top x} = \sqrt{\left|\left| x\right|\right|^2} = \left|\left|x\right|\right|$$.

Then, since $Qx = cx$, $$\left|\left|x\right|\right|=\left|\left| Qx\right| \right| = \left|\left|cx\right|\right| = \left|c\right|\left|\left|x\right|\right|$$

This implies $|c| = 1$. If we assume that $Q$ has only real eigenvalues, $c = \pm 1$.

learnmore provides a counterexample that shows $c$ does not have to equal $\pm 1$ if we consider complex eigenvalues.

Bryan Kaperick
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  • How does $|c|=1\implies c=\pm 1$ – Learnmore Nov 25 '16 at 03:53
  • What is $|i|$or $|-i|$? – Learnmore Nov 25 '16 at 03:55
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    @learnmore You are correct that this is not true in general since $c$ can be complex, but given the context of the question, it seems like they are discussing only real eigenvalues. If we consider that case, the question itself is incorrect. – Bryan Kaperick Nov 25 '16 at 03:56
  • OP did not ever mention that the eigen values are real – Learnmore Nov 25 '16 at 03:57
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    Based on their phrasing of the question, it seems that they have been given this result from their instructor, and since this question is explicitly for the purpose of studying for a test given by that same instructor, it seems that this is the desired scope of the question. – Bryan Kaperick Nov 25 '16 at 03:58
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Your question is false ;

Take $Q=$\begin{bmatrix}0 &-1\\1&0\end{bmatrix}

Learnmore
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    The OP asked to prove that every orthonormal matrix has eigen values $1$ or $-1$ which is not the case here@GerryMyerson – Learnmore Nov 25 '16 at 03:22
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    The eigenvalues of this matrix are indeed $\pm 1$, as the characteristic polynomial is $p (t)=t^2-1$. – Dave Nov 25 '16 at 03:23
  • Why the downvotes ?Anyone please.. – Learnmore Nov 25 '16 at 03:56
  • @learnmore Because your original matrix wasn't a counterexample. Now that you've fixed it, I've turned the down into an up. –  Nov 25 '16 at 04:02
  • Yeah,I missed that minus sign@Bye_World – Learnmore Nov 25 '16 at 04:04
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    BTW...a *question* is neither true nor false. An assumption can be false, and an answer can be false. But also an answer can fail to "answer" or address a question, and your "answer" is such a case. – amWhy Nov 26 '16 at 00:36
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    And an answer that fails, and upon being called out, the answerer deletes the failure, to replace it after the error is pointed out, without acknowledging the original error, isn't worth an upvote, in the first place. – amWhy Nov 26 '16 at 00:40