Uniform continuity of a linear transformation Let $T:\mathbb R^3\rightarrow\mathbb R^2$ be a linear transformation. Then, is it right to say that $T$ is uniformly continous?

Yes:
$$ T(x+h)-T(x) = T(h) $$
since $T$ is linear. For any $h$, this is bounded by $\lVert T \rVert \lVert h \rVert$, where the second term is just the Euclidean norm of $h$, and the first term is the norm of $T$ defined as $\sup_{\lVert x \rVert =1} \lVert T(x) \rVert$. Hence $T$ is Lipschitz, and so uniformly continuous.
