TITLE: Lipschitz Continuity of Linear Map Between Finite Dimensional Vector Spaces
QUESTION [10 upvotes]: I am trying to show from first principles that a linear map $T:X \rightarrow Y$ between finite dimensional vector spaces $X$ and $Y$ is Lipschitz. I'm rather stuck and would appreciate a hint how I might continue the argument I have constructed so far:
To show that $T$ is Lipschitz, it is necessary to show for $x,y \in U$ that $|T(x) - T(y)| \leq \lambda|x - y|$ for some $\lambda > 0$. If $u_1, \dots, u_n$ is a basis for $X$, there exists scalars $a^i, b^i$ such that $x = a^iu_i$ and $y=b^iu_i$. Therefore,
$$
|T(x - y)| \leq |T(a^iu_i)| + |T(b^iu_i)| = |a^i|\cdot|T(u_i)| + |b^i|\cdot|T(u_i)| \leq k \sum_{i=1}^n|T(u_i)|
$$
where 
$$k = \max_{j=1, \dots n}|a^j + b^j|$$
So, I've succeeded in finding a bounds of sorts for $T$ but I'm not sure how to translate the bound $|T(u_i)|$ to $u_i$ and other than showing something rather obvious so far, I'm not really sure that the above argument gets me a lot closer to my goal. How could I proceed with this?

REPLY [10 votes]: There are a few ways you could proceed. For example, it is equivalent to showing that the following quantity is bounded: $\sup_{z\in X, \|z\| = 1}\|Tz\|$. Indeed, $\|T(x) - T(y)\|\leq \lambda\|x - y\|$ if and only if 
$$
\frac{\|T(x) - T(y)\}}{\|x - y\|}\leq \lambda,
$$
assuming that $x\neq y$. Now by linearity of $T$, the above inequality holds if and only if
$$
\left\|T\left(\frac{x - y}{\|x - y\|}\right)\right\|\leq \lambda.
$$
Rewriting this with $z = x - y$, we get the claim.
To bound this, notice that $\|Tz\|\leq (|a_1| + \cdots + |a_n|)\max_{1\leq i\leq n}\|Te_i\|$, where $\{e_i\}$ is an orthonormal basis for $X$ and $z = \sum_{i=1}^n a_i e_i$. Since $e_1,\dots,e_n$ are orthonormal and $\|z\| = 1$, this puts a uniform (i.e. independent of $z$) bound on $|a_1|+\cdots+|a_n|$ (prove this).