Why is that the matrix $1$-norm and $\infty$-norm are equal to the operator norm, while 2 norm is not? Given a matrix $A$
I had always assumed that the operator norm of a matrix:
$\|A\|_p = \sup\limits_{\|v\|_p \neq 0}\dfrac{\|Av\|_p}{\|v\|_p}$
was the same as the norm of a matrix:
${\displaystyle \|A\|_{p}= \left(\sum _{i=1}^{m}\sum _{j=1}^{n}|a_{ij}|^{p}\right)^{1/p}}$
However, this post seems to shatter my assumption: 2-norm vs operator norm
Upon further examination, it seems that the operator norm and matrix norm only coincide (=) for matrix $1$ or $\infty$-norm (and extremely limited cases for matrix 2-norm): https://en.wikipedia.org/wiki/Matrix_norm
Why is this so? Is there a theorem that relates the operator norm with matrix $1$ or $\infty$ norm and show their equality?
Note:

 A: No, the vector $1$-norm or $\infty$-norm are not the same as their induced counterparts. E.g.
\begin{align*}
\|(1,0,0,1)\|_1=2&\ne1=\max_{\|v\|_1\ne0}\frac{\|I_2v\|_1}{\|v\|_1},\\
\|(1,1,1,1)\|_\infty=1&\ne2=\max_{\|v\|_\infty\ne0}\frac{\left\|\pmatrix{1&1\\ 1&1}v\right\|_\infty}{\|v\|_\infty}.
\end{align*}
In fact, the induced $1$-norm of a matrix $A$ is given by the maximum $1$-norm of all columns. Unless $A$ has $n-1$ zero columns, its induced $1$-norm is always strictly smaller than its vector $1$-norm.
Also, the the induced $\infty$-norm of a matrix $A$ is given by the maximum $1$-norm of all rows. In most cases, it is different from the vector $\infty$-norm of $A$.
A: As mentioned by @user1551 operator norm induced by 1-norm and $\infty$-norm are not equal to matrix 1 and $\infty$ norm. Moreover Forbenius norm or matrix-2 norm is always greater than equal to operator norm induced by 2-norm because operator norm induced by 2-norm is maximum singular value(by definition) and matrix 2-norm is $(\sum_i\sigma_i^2)^\frac{1}{2}$. To see this note :$$\left(\sum _{i=1}^{m}\sum _{j=1}^{n}|a_{ij}|^{2}\right)=trace(A^TA)=\sum_i eigenvalue_i(A^TA)=\sum_i singularvalue_i(A)^2=\sum_i \sigma_i^2$$
