I am trying to prove or disprove the compatibility (over $\mathbf{R}^n$) of the Frobenius norm of square matrices and the $1$-norm for vectors. That is the norms
$$\|A\| := \sqrt{\sum_{j=1}^n\sum_{i=1}^n a_{j,i}^2}$$
and
$$\|\mathbf{x}\|_1 := \sum_{i=1}^n |x_i|$$
respectively. When I restrict my view to the case of $n=2$, if
$$A = \begin{bmatrix} a & b \\c& d \end{bmatrix}$$
and $\mathbf{x} = \begin{bmatrix} e \\ f \end{bmatrix}$ then I would essentially need to show that
$$|ea+fb|+|ec+fd|\leq \sqrt{a^2 +b^2 +c^2+d^2} (|e|+|f|).$$
But this seems true and increasing the dimension would just make it more true. Any guidance on proving or a counterexample for
$$\|A \mathbf{x}\|_1 \leq \|A\| \|\mathbf{x}\|_1$$
would be much appreciated.
Best Answer
Consider the case where $A=\begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix}$ and $\mathbf{x}=\begin{bmatrix} 1\\0\end{bmatrix}.$ Then $A\mathbf{x}=\begin{pmatrix} 1\\1\end{pmatrix}.$ If compatibility holds we have $$ \begin{align} 1+1 &=\left|\left|\begin{pmatrix} 1\\1\end{pmatrix}\right|\right| =\left|\left|A\mathbf{x}\right|\right|_1 \le ||A||\cdot ||\mathbf{x}||_1\\ &= \left|\left|\begin{pmatrix}1 & 0\\1&0\end{pmatrix}\right|\right|\cdot\left|\left|\begin{pmatrix}1\\ 0\end{pmatrix}\right|\right|_1=\sqrt{2}. \end{align} $$