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Show that the eigenvalues of a unitary matrix have modulus $1$.

I know that a unitary matrix can be defined as a square complex matrix $A$, such that

$$AA^*=A^*A=I$$

where $A^*$ is the conjugate transpose of $A$, and $I$ is the identity matrix. Furthermore, for a square matrix $A$, the eigenvalue equation is expressed by $$Av=\lambda v$$

If I use the relationship $u v=v^*u$ and take the conjugate transpose of this equation then $$v^*A^*=\lambda^*v^*$$

But now I got stuck. Can someone help?

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  • $\begingroup$ i think you can find your answer in part 4 of this post math.stackexchange.com/questions/40246/… $\endgroup$ – K.K.McDonald Mar 28 '16 at 20:46
  • $\begingroup$ I actually looked at that post but it was difficult to understand. I think they do it in another way. $\endgroup$ – Alim Teacher Mar 28 '16 at 20:51
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You multiply your two relations to obtain

\begin{align} v^*A^*Av &=\lambda^* v^*\lambda v \\ v^*Iv &=\left(\lambda^*\lambda\right) v^*v \\ v^*v &=\left(\lambda^*\lambda\right) v^*v \\ ||v||^2 &= |\lambda|^2 ||v||^2 \\ \sqrt{1} &=|\lambda| \\ 1 &=|\lambda| \end{align}


Recall that the modulus of a complex number $\lambda = a + bi$, also called the "complex norm", is denoted $|\lambda|$ and defined by $|\lambda| = |a + bi| = \sqrt{a^2 + b^2}$ and $\lambda^*\lambda = (a -bi)(a + bi) = a^2 + b^2$. Hence $\lambda^*\lambda = |\lambda|^2.$

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$\Delta$ as $\lambda$

$Av=\Delta v$

$(Av)^*=(\Delta v)^*$

$v^*A^*=\Delta^*v^*$

$v^*A^*Av=\Delta^*v^*\Delta v$

As $A^*A=I$

$v^*Iv=\Delta^*\Delta v^*v$

$v^*v=\Delta^*\Delta v^*v$

$(1-\Delta^*\Delta)v^*v=0$

Since $v$ is not equal to zero.

Hence

$1-\Delta\Delta^*=0\implies \Delta^*\Delta=1$

$|\Delta|^2=1\implies |\Delta|=1$.

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  • $\begingroup$ What was the point changing $\lambda$ to $\Delta$? See meta.math.stackexchange.com/questions/5020/… for help formatting in MathJax. $\endgroup$ – Daniel Buck Aug 15 '18 at 17:31
  • $\begingroup$ I've added formatting to your answer, hope that's ok. Note that two * format as italic in HTML which was why some stuff rendered funny.. $\endgroup$ – Daniel Buck Aug 15 '18 at 17:39
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The result that you seek follows from the following.

Lemma. If $A$ is unitary and $\vert \vert x \vert \vert_2 = 1$, then $\vert\vert Ax \vert\vert_2 =1$.

Proof. By definition, $$ \vert\vert Ax \vert\vert_2^2 =\langle Ax, Ax \rangle = (Ax)^*(Ax) = x^*A^*A x = x^*x = \vert\vert x \vert\vert_2^2 = 1.$$


If $(\lambda,v)$ is an eigenpair of $A$, then, without loss of generality, it may be assumed that $\vert\vert v \vert\vert_2 = 1$. Following the lemma above and the absolute homogeneity property of vector norms, notice that $$ |\lambda| = |\lambda|\vert\vert v \vert\vert_2 = \vert\vert \lambda v \vert\vert_2 = \vert\vert Av \vert\vert_2 = 1,$$ as desired.

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A unitary matrix $U$ preserves the inner product: $\langle Ux, Ux\rangle =\langle x,U^*Ux\rangle =\langle x,x\rangle $.

Thus if $\lambda $ is an eigenvalue, $Ux=\lambda x$, we get $\vert\lambda \vert^2\langle x,x\rangle =\langle \lambda x,\lambda x\rangle =\langle Ux, Ux\rangle =\langle x,x\rangle $.

So $\vert \lambda\vert^2=1\implies \vert \lambda\vert=1$.

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