# let $x$ be a eigenvector for $Q$ [duplicate]

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

• What is Q? Is it a projection? – ET93 Nov 25 '16 at 3:06
• @ET93 sorry Q is a matrix with orthonormal columns – tshelton Nov 25 '16 at 3:06
• math.stackexchange.com/questions/653133/… – ET93 Nov 25 '16 at 3:11
• @ET93 does this solve my proof $|\lambda|^2x^tx=(Ax)^{t}Ax={x}^{t}A^{t}Ax=x^tx.$ So $|\lambda|=1$. Then $\lambda=-1$ or $1$. – tshelton Nov 25 '16 at 3:15
• $\lambda$ can be complex. – ET93 Nov 25 '16 at 3:52

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.

• How does $|c|=1\implies c=\pm 1$ – Learnmore Nov 25 '16 at 3:53
• What is $|i|$or $|-i|$? – Learnmore Nov 25 '16 at 3:55
• @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 3:56
• OP did not ever mention that the eigen values are real – Learnmore Nov 25 '16 at 3:57
• 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 3:58

Take $Q=$\begin{bmatrix}0 &-1\\1&0\end{bmatrix}
• 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 3:22
• The eigenvalues of this matrix are indeed $\pm 1$, as the characteristic polynomial is $p (t)=t^2-1$. – Dave Nov 25 '16 at 3:23