$A$ is $M_{n\times n}$ matrix and $x,y$ are column vector. I want to show that $||A(x-y)||_2\leq ||A||_2||x-y||_2$ with the Euclidean norm.
I know that for the Cauchy-Schwarz inequality, both vectors should be of the same dimension. Is it right to argue the above statement using the Cauchy-Schwarz inequality as it contains one matrix and another column vector?
Any help is appreciated.
This is true for every linear operator $T: V\to W$ between two normed spaces $(V,\|\cdot \|_V)$ and $(W,\|\cdot \|_W)$ (we say that $T$ is linear if $T(x+y)=T(x)+T(y)$ and $T(\lambda x)=\lambda T(x)$ for all $x,y\in V$ and $\lambda\in \mathbb{K}$).
We say that a linear operator $T: V\to W$ is bounded if there exists $M>0$ such that $\|x\|_V\leq 1$ implies $\|T(x)\|_W\leq M$. In such cases, we define the operator norm by \begin{equation} \|T\|=\sup\{\|T(x)\|_W : x\in V, \;\|x\|_V\leq 1\} \end{equation} If $x\neq 0$ and $x\in V$, then $$\left\|T\left(\frac{x}{||x||_V}\right)\right\|_W\leq \sup\{\|T(x)\|_W : x\in V, \;\|x\|_V\leq 1\}=\|T\|,$$ since the point $x/\|x\|_V$ has norm $1$. Then, for $x\neq 0$ and $x\in V$ (the case $x=0$ is trivial), we have $$\|T(x)\|_W=\left\|T\left(x\right)\frac{\|x\|_V}{\|x\|_V}\right\|_W=\left\|\|x\|_VT\left(\frac{x}{\|x\|_V}\right)\right\|_W=\|x\|_V \left\|T\left(\frac{x}{\|x\|_V}\right)\right\|_W \leq \|T\|\|x\|_V$$
In particular, if $T:V\to W$ is a linear operator between normed spaces with finite dimension $n$ with the Euclidean norm whose associated matrix in some basis is $A$, then you have your result.