let $A$ be an n by n matrix, show that $||A||_{OP} \leq ||A||_{HS} \leq \sqrt{n} ||A||_{OP}$ We are given $A \in M_{n}(\mathbb R)$ and the following norms:
$||.||_{e}$ is the standard euclidean norm of $\mathbb R^n$.
$||A||_{OP}$ is the operator norm of $A$, meaning $||A||_{OP} = sup_{||v||_e=1} ||Av||_e$
$||A||_{HS}$ is the Hilbert-Schmidt norm. Meaning $||A||_{HS} = trace(A^TA)$
Show that $||A||_{OP} \leq ||A||_{HS} \leq \sqrt{n} ||A||_{OP}$
Firstly I found out what the hilbert-schmidt norm is, and $trace(A^TA) = \sum a^2, a\in A$
How can I show it is bigger than the operator norm?
 A: An equivalent definition for the Hilbert-Schmidt norm is $||A||_{HS} = (\sum_j \sigma_j(A)^2)^{1/2}$ where $\sigma_k(A)$ is the $k$-th singular value of the matrix $A$. These are just the eigenvalues of the matrix $\sqrt{A^*A}$.  Then you can show that the largest singular value of $A$ equals the operator norm of the matrix $A$.
Now clearly, $||A|| = \sigma_1(A) \leq (\sum_j \sigma_j(A)^2)^{1/2} = ||A||_{HS}$.
A: First, note that $||A||_{op} \leq {\lambda_1}^\frac{1}{2}$ where ${\lambda}_1 $ is $A^TA$'s max eigenvalue. The proof is as follows:
$||Av||^2 = <Av, Av> = <A^TAv,v>$ 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 <Iv,v> ={\lambda}_1 ||v||^2$ 
where ${\lambda}_k$ is some eigenvalue and ${\lambda}_1$ is the max eigenvalue (they're all real and non-negative). 
So we have $\frac{||Av||}{||v||} \leq {\lambda}_1^{1/2}$. 
Regarding $||A||_{HS}$, note that $A^TA$ has a diagonal form, for the reasons mentioned above. Therefore, 
$$||A||_{HS}^2 = \sum_{i=1}^r \lambda_i$$
It's trivial that ${\lambda_1}^\frac{1}{2} \leq {\lambda_1}^\frac{1}{2} +\sum_{i=2}^r {\lambda_i}^\frac{1}{2}  $ and thus: $||A||_{Op} \leq ||A||_{HS} $.
Note that for the second inequality, note that $||A||_{HS}$ is maximised $\iff$ all of $A^TA$'s eigenvalues are equal. What value does $||A||_{HS}$ get in that case?
Hope this helps.
