Given the matrix $A = \begin{bmatrix}
a & b\\
c & d\\
\end{bmatrix}$ and the norms
$$\lVert A\rVert_{E} = (a^{2}+b^{2}+c^{2}+d^{2})^{\frac{1}{2}}$$
$$\lVert A\rVert_{op} = \sup_{\lVert\mathbf{v}\rVert_{2}=1}\lVert A\mathbf{v}\rVert_{2}$$
I want to find a positive constant $C$ so that
$$\lVert A\rVert_{op} \geq C\lVert A\rVert_{E}$$
What I have tried:. Letting $\mathbf{v}=(v_{1},v_{2})^{T}$ gives me
$$\lVert A\rVert_{op} = \sup_{\lVert\mathbf{v}\rVert_{2}=1} \sqrt{v_{1}^{2}(a^{2}+c^{2}) + v_{2}^{2}(b^{2}+d^{2}) + 2v_{1}v_{2}(ab+cd)}$$
For the upper bound I was able to use AM-GM. However, I'm not sure what inequalities I need to get the lower bound. In particular, what inequality do I need to get rid of the $(ab+cd)$ term? I am not sure whether taking negatives of the AM-GM works because I can't be sure whether the overall expression under the square root is still positive. I have also tried using the equality of $p-norms$ and using the infinity norm gave me
$$\lVert A\rVert_{op} \geq \sup_{\lVert\mathbf{v}\rVert_{\infty}=1} \max\{|av_{1}+bv_{2}|, |cv_{1}+dv_{2}|\}$$
but again, not sure how to relate this to $(a^{2}+b^{2}+c^{2}+d^{2})^{\frac{1}{2}}$. (I tried triangle inequality but it gave me inequality in the wrong direction).
Bottom Line: How can I get rid of the $2v_{1}v_{2}(ab+cd)$ term to achieve the desired inequality and find the constant?
Note: I am not allowed to use the theorem that says any two norms on a finite dimensional space are equivalent.
Best Answer
You may take $C=\frac{1}{\sqrt{2}}$.
Your $\|\cdot\|_{op}$ is just the induced $2$-norm, i.e. the largest singular value of a matrix, while your $\|\cdot\|_E$ is just the Frobenius norm of a matrix. Both norms are unitarily invariant. Therefore, by singular value decomposition, if $\sigma_1\ge\sigma_2$ are the two singular values of a $2\times2$ matrix $A$, then $$ \|A\|_{op}=\sigma_1\ge\frac{1}{\sqrt{2}}\sqrt{\sigma_1^2+\sigma_2^2}=\frac{1}{\sqrt{2}}\|A\|_E. $$ In general, if the matrix is $n\times n$, you may take $C=1/\sqrt{n}$.