TITLE: A linear operator on a finite dimensional Hilbert space is continuous
QUESTION [15 upvotes]: How do I show that a linear function from a Hilbert space $H$ to itself is continuous if $H$ is finite dimensional?
Also, what would be an example of a linear function from a Hilbert space to itself which is not continuous?

REPLY [6 votes]: Here's another similar solution:
Let $\beta = \{e_1, e_2, \dots, e_n \}$ be a basis for $X$. Then for all $x$ in $X$ we have : $$\|Tx\| = \| \sum_{k=1}^n{\alpha_k Te_k}\| \le \sum_{k=1}^n{\| \alpha_kTe_k\|}=\sum_{k=1}^n{\vert \alpha_k \vert \ \| Te_k\|} $$
Let $\displaystyle \max_{1 \leq k \leq n} \| T e_k \| = M.$ Then $\|Tx\| \le M \sum_{k=1}^n{\vert \alpha_k \vert}$.
But $\| x\|_1 = \sum_{k=1}^n{\vert \alpha_k \vert}$ is a norm as you may check. So $\|Tx\| \le M \|x\|_1$ and since all norms on a finite dimensional space are equivalent there exists $R >0$ such that $\|x\|_1 \le R\|x\|$ and thus $$\|Tx\| \le C \|x\|$$ where $C = M.R$ .