Showing a stochastic matrix transforms any vector into a vector with a smaller (or equal) sum of absolute value of components

Question

The problem is from David Lay's textbook in linear algebra:

If $A$ is an $m\times m$ stochastic matrix, and $A\vec{x} = \vec{y}$, show that $|y_1| + ... + |y_m| \leq |x_1| + ... + |x_m|$.

First, since A's columns are probability vectors, I re-wrote $\vec{y}$ in two ways:

$\vec{y} = x_1\vec{p_1} + ... + x_m\vec{p_m}$

$\vec{y} = y_1\vec{e_1} + ... + y_m\vec{e_m}$ (where $e_i$ are the standard basis vectors)

From this, I saw that the sum of the components in both representations must be the same:

$y_1 + ... + y_m = x_1 + ... x_m$

Taking the absolute value of both sides, and applying triangle inequality gives:

$|y_1 + ... + y_m| \leq |x_1| + ... + |x_m|$

This is closer, but obviously I'm still not there. Any help on how to meaningfully get $|y_1| + ...+ |y_m|$ into the picture?

Answer 1

$$y_j=\sum_kx_kp_{j,k},$$so $|y_j|\le\sum_k|x_k|p_{j,k}$, so $$\sum_j|y_j|\le\sum_j\sum_k|x_k|p_{j,k}=\sum_k\sum_j=\dots$$