How does one prove that the spectral norm is less than or equal to the Frobenius norm?
The given definition for the spectral norm of $A$ is the square root of the largest eigenvalue of $A*A$. I don't know how to use that. Is there another definition I could use? We also have $\displaystyle{\max\frac{\|Ax\|_p}{\|x\|_p}}$, but if $p=2$ it's the Frobenius norm, right?
Best Answer
The Frobenius norm of $A$ is the squareroot of the sum of all of the eigenvalues (the trace) of $A^*A$, and since all eigenvalues of $A^*A$ are nonnegative, it follows that the largest eigenvalue is less than or equal to the sum of all of the eigenvalues.
No, if $p=2$ that is another characterization of the spectral norm. A proof of this is sketched in Michalis's answer.