Spectral Radius and Norm of multiplied vector

Question

Let $\mathbf{A}$, $\mathbf{B}$ be square matrices of equal dimensions, $\mathbf{w}$ a vector of compatible dimensions and $\rho$ be the spectral radius operator.

Does the following hold?

If $\rho (A) < \rho(B)$ then:

$ || \mathbf{A} \mathbf{w} || < || \mathbf{B} \mathbf{w} || $.

If yes, why?

Answer 1

It's false:

$A=\left(\begin{array}{cc}1 & 0 \\0 & 2 \\\end{array}\right)$, $B=\left(\begin{array}{cc}3 & 0 \\0 & 1 \\\end{array}\right)$

$ρ(A)<ρ(B)$, but $\left\|A\left(\begin{array}{c}0\\1\\\end{array}\right)\right\|>\left\|B\left(\begin{array}{c}0\\1\\\end{array}\right)\right\|$

Answer 2

No. Example: $$A=I, \quad B=\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} $$ We have that $\rho(A)=1<2=\rho(B)$ but $\lVert A\begin{bmatrix} 0 \\ 1 \end{bmatrix} \rVert > \lVert B \begin{bmatrix} 0 \\ 1 \end{bmatrix}\rVert = 0,$ in any vector norm.