Let $H$ be a finite dimensional Hilbert space. Let $T:H\to H$ be a linear operator. Show that $T$ is bounded.
my work: Since $H$ is finite dimensional, there exists an orthonormal basis $\{e_1,…,e_n\}$. I want to show that if $v \in H$ and $\|v\|=1$, then $\langle Tv,v \rangle$ is bounded. Let $v=c_1e_1+…+c_ne_n$, then $$\langle Tv,v\rangle=\big\langle c_1T(e_1)+c_2T(e_2)+…+c_nT(e_n),c_1e_1+…+c_ne_n\big\rangle$$
Can anyone tell me how to proceed or is there any better way to prove this?
Best Answer
Consider a basis of $H$, say $\{e_i\}$. Then, note that for any $x = \sum_{i=1}^n x_ie_i$ (where $n = \dim H$) we have $$||Tx||^2 = \langle Tx,Tx\rangle = \left\langle \sum_{i=1}^n x_i T(e_i),\sum_{i=1}^n x_i T(e_i) \right\rangle = \sum_{i,j=1}^nx_i \overline{x_j}\left\langle T(e_i),T(e_j)\right\rangle$$
Now, let $M = \displaystyle\max_{1 \leq i,j \leq n} \langle T(e_i),T(e_j)\rangle$, existing because we are taking the maximum over a finite set. Then, we have: $$ ||Tx||^2 = \sum_{i,j=1}^n x_i \overline{x_j}\langle T(e_i)T(e_j)\rangle \leq M \sum_{i,j=1}^n x_i \overline{x_j} \leq M ||x||^2 $$
taking square roots on both sides tells us that the norm of $T$ exists, and is less than or equal to $\sqrt M$.