For a matrix $A$, is $\|A\| \leq {\lambda}^{1/2}$ true? In class I saw a proof that went something along these lines:
Define $\|A\| = \sup \dfrac{\|Av\|}{\|v\|}$ for  v in V, where the norm used is the standard (Does this even exist?) Euclidean norm in V. 
$\|Av\|^2 = <Av, Av> = <A^TAv,v>$ where $<,>$ denotes a dot product. 
Note that $A^TA$ is a non-negative matrix, so we can apply the Spectral theorem.  
$<A^TAv,v>$ $=$ $<\sum_{i=1}^r{\lambda}_iE_iv,v>$   $=$ $\sum_{i=1}^r{\lambda}_i<E_iv,v>$ $\leq$ $ \lambda_1 <v,v>$ 
where ${\lambda}_k$ is some eigenvalue and ${\lambda}_1$ is the max eigenvalue (they're all real and non-negative). 
So we have $\dfrac{\|Av\|}{\|v\|} \leq {\lambda}_1^{1/2}$. 
Is this always correct? What assumptions were made during this proof? I'm asking this because after Googling the problem a  little bit I couldn't find things that resembled this.
Thanks for your time. 
 A: The matrix $A^TA$ is symmetric and thus it is orthogonal diagonalizable:
$$
A^TA=U^TDU,
$$
where $U$ is an orthogonal matrix, i.e., $U^TU=I$, $\|Ux\|=\|x\|$, for all $x\in\mathbb R^n$, and $D$ diagonal:
$$
D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n),
$$
where the $\lambda_j$'s are non-negative, as $A^TA$ is clearly positive definite. We can assume that $\lambda_k=\max_{j=1}^n \lambda_j$.
Now
$$
\langle A^TAx,x\rangle=\langle U^TDUx,x\rangle=\langle DUx,Ux\rangle.
$$
So
$$
\sup\frac{\langle A^TAx,x\rangle}{\|x\|^2}=\sup\frac{\langle DUx,Ux\rangle}{\|x\|^2}
=\sup\frac{\langle DUx,Ux\rangle}{\|Ux\|^2}=\sup\frac{\langle Dx,x\rangle}{\|x\|^2},
$$
as $f(x)=Ux$ is one-to-one and onto. But if $x=(x_1,\ldots,x_n)$, then
$$
\langle Dx,x\rangle=\lambda_1x_1^2+\cdots+\lambda_nx_n^2\le \lambda^2_k \|x\|^2,
$$
and the left hand-side inequality becomes equality for $x=e_k$.
Thus
$$
\sup\frac{\langle Dx,x\rangle}{\|x\|^2}=\lambda_k^2.
$$
But
$$
\sup\frac{\langle Dx,x\rangle}{\|x\|^2}=\sup\frac{\langle A^TAx,x\rangle}{\|x\|^2}=\sup\frac{\langle Ax,Ax\rangle}{\|x\|^2}=\sup\frac{\|Ax\|^2}{\|x\|^2},
$$
and hence
$$
\sup\frac{\|Ax\|}{\|x\|}=\lambda_k.
$$
