Dgdrxrt

I'm perusing this paper. In page 8, I came across this:

My question is about equation 45. Also in this question I don't care about neither $u$ nor $M$, I mentioned them just to give some context.

In short my problem is:

Let $y in mathbbR^n$, $x in mathbbR^m$, $A in mathbbR^n times m$.

$|.|$ is the euclidean vector norm, $|.|_op$ is a matrix norm defined as:
$$|A|_op = max_x frac$$

So I have: $|y^TAx| leq |y|space|A|_op|x|$. As long as I have an inequality with the absolute value of something, it is the same as $y^TAx leq |y|space|A|_op|x|$

Now I will multiply both sides by 2 and multiply the left side by:
$$fracsqrttausqrttaufrac(tausigma (tausigma $$

Then I get:
$$2y^TAx leq 2
fracA
(tausigma
fracA^TA
sqrttau$$

using $a^2+b^2 geq 2ab$:
$$2y^TAx leq
frac^2_op tau
(tausigma |y|^2 +
fracA^TA
tau |x|^2 $$

In equation 45, the term $|A|^2_op$ doesn't appeard, it is replaced by $|A^TA|_op$. I know that $|A|^2_op geq |A^TA|_op$, so I don't know why the authors make this replacement. In the paper, they say that $A$ is any matrix with no special properties.

The next line comes from doing:
$$(tausigma |A^TA|_op)^frac12z = fracA^TA
(tausigma $$
$$tausigma |A^TA|_op z = |A^TA|_op tau$$
$$z = frac1sigma$$

Why do authors replace $|A|_op^2$ with $|A^TA|_op$ in equation 45? With this replacement the inequalities don't necessarily hold. Is there a property I am missing?

I'll use $langle x,yrangle$ to denote the usual inner product $y^*x$.

It is always true that $$|A|^2_op=|A^*A|_op,$$ where $A^*$ denotes the adjoint (the conjugate transpose, so just the transpose if $A$ is real). This is very easy to prove. The reverse inequality from the one you mention is the easy one:
beginalign
|A|^2&=maxAx=maxx\
&=maxx\
&leq |A^*A|,
endalign
where the inequality is simply Cauchy-Schwarz and definition of the operator norm. Now, from the above inequality, you have
$$
|A|^2leq|A^*A|leq|A^*|,|A|.
$$
So, for nonzero $A$ (the zero case is trivial) you have $|A|leq|A^*|$. Apply the above to $A^*$, to get $|A^*|leq|A^**|=|A|$, so $|A^*|=|A|$. If you now go back to the inequalities we had,
$$
|A|^2leq|A^*A|leq|A^*|,|A|=|A|^2.
$$
So $|A|^2=|A^*A|$.