Rayleigh quotient of matrix $A$ is diagonal entry of $Q^*AQ$ Let $A\in \mathbb{C}^{m\times m}$ be a matrix. Show that if $z\in\mathbb{C}$ is a Rayleigh quotient of $A$ then $z$ is a diagonal entry of $Q^*AQ$, with $Q$ a unitary $m\times m$ matrix.
My approach:
We have that given a non-zero vector $x$, $z=\frac{x^TAx}{x^Tx}$. Let $x:=Qy$ for some unitary $Q$. Then
$$z=\frac{(Qy)^*AQy}{\|y\|\|y\|}$$
This shows that $z$ is a Rayleigh quotient for $Q^*AQ$.
WLOG, let $(Q^*AQ)_{ii}$ be the $ii$th diagonal entry of $Q^*AQ$. Then $(Q^*AQ)_{ii}= q_i^*Aq_i$. Set $q_i=\frac{Qy}{\|y\|}$. Construct $Q$ by a Gram-Schmidt process. This proves that there exists unitary $Q$ such that $z$ is a diagonal entry of $Q^*AQ$ for $x$.
Do you think my proof is consistent?
Would appreciate some feedback.
 A: You have the right idea, but a few aspects of your proof are problematic:

We have that given a non-zero vector $x$, $z=\frac{x^TAx}{x^Tx}$. 

Your phrasing is a bit off here.  You're implying that for any choice of non-zero vector $x$, we have $z=\frac{x^TAx}{x^Tx}$, which is not what you mean.  I would say
By the definition of a Rayleigh quotient, there exists a non-zero vector $x$ such that $z = \frac{x^*Ax}{x^*x}$
On to the next part:

Let $x:=Qy$ for some unitary $Q$. Then
  $$z=\frac{(Qy)^*AQy}{\|y\|\|y\|}$$
  This shows that $z$ is a Rayleigh quotient for $Q^*AQ$.

I agree with your conclusion that $z$ is a Rayleigh quotient of $Q^*AQ$.  The rest is confusing.  Writing $x:=Qy$ makes no sense in this context: $x$ was defined earlier, so we're not defining it now.  Also, you never said what $y$ is.  

WLOG, let $(Q^*AQ)_{ii}$ be the $ii$th diagonal entry of $Q^*AQ$. 

WLOG means "without loss of generality".  I have no idea how that is supposed to apply here.

Then $(Q^*AQ)_{ii}= q_i^*Aq_i$. Set $q_i=\frac{Qy}{\|y\|}$. Construct $Q$ by a Gram-Schmidt process. 

This is confusing.  To begin, you're defining $q_i$ to be the $i$th column of $Q$.  Then, you're saying (or trying to say) that we should construct a $Q$ that has $q_i$ as its $i$th column.  Why are we constructing $Q$ if we already have $Q$?

This proves that there exists unitary $Q$ such that $z$ is a diagonal entry of $Q^*AQ$ for $x$.

The "for $x$" at the end here makes no sense; perhaps it's there by mistake.

Here's a more concise proof, which is readable to me:

Because $z$ is a Rayleigh quotient of $A$, there exists a unit vector $x$ such that $z = \frac{x^*Ax}{x^*x} = x^*Ax$. Let $Q$ denote a unitary matrix that has $x$ as its first column (such a matrix could be constructed by a Gram-Schmidt process).  We note that
  $$
(Q^*AQ)_{11} = q_1^*Aq_1 = x^*Ax = z
$$
  Thus, we see that $z$ is a diagonal entry of the matrix $Q^*AQ$.

