does $\|A\| > \|B\|$, $A$ and $B$ matrices, imply that $\|Ax\| >\|Bx\|$ for all $x$ in some vector space? I'm wondering if I can prove the first inequality in a question that I am working on, does that make the second inequality automatic? 
Thanks,
 A: No, remind that all vector spaces include $x = 0$.
A: Strictly speaking, the answer to the question in your question title is "yes" if you replace the strict inequality $>$ by weak inequality $\ge$. There will always be at least a one-dimensional subspace of the whole vector space in which the inequality is true. But the answer for the question about whether it's true for the whole space is of course "no" if the space has dimension greater than 1. But the one-dimensional case is probably not what you had in mind.
A: This system produces a counterexample.
\begin{bmatrix} 
\ 1 & 0
\\0 & 1  
\end{bmatrix}
Determinant of this is 1
\begin{bmatrix} 
\ 10 & 0
\\0 & 0  
\end{bmatrix}
Determinant of this is zero, less than the first matrix
\begin{bmatrix} 
\ 1 & 0
\end{bmatrix}
Each of the multiplications of the matrices by this vector are 
\begin{bmatrix} 
\ 1 & 0
\end{bmatrix}
and
\begin{bmatrix} 
\ 10 & 0
\end{bmatrix}
Respectively.
The length of each of these vectors is 1 and 10 respectively. The second matrix multiplication being the larger of the two.
A: You probably mean $||A||$ to be the norm of the operator $A$. Also, you probably mean $\ge $ rather then $>$. The answer is still No. 
Consider the operators $A = \left( \begin{array}{cc} 5 & 0 \\ 0 & 1\end{array} \right )$ and $B = \left( \begin{array}{cc} 2 & 0 \\ 0 & 2\end{array} \right )$. 
