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?

  • $\begingroup$ One can argue as follows: A linear map of finite dimensional vsp is necessarily continuous (the topology is independent of the norm one chooses). Then the image of the compact (closed) unit ball is necessarily bounded, as it is compact (this is Heine-Borel) $\endgroup$
    – kesa
    Nov 1, 2017 at 12:17

2 Answers 2


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 $\sqrt M$.

  • $\begingroup$ In your proof @астонвіллаолофмэллбэрг, I have confusion about your last inequality. Can you explain your last inequality? $\endgroup$
    – Amin
    Nov 1, 2017 at 13:35
  • $\begingroup$ @Amin which one? The fact that $\sum_{i,j=1}^n x_ix_j = ||x||^2$? But is this not true by just expansion? $\endgroup$ Nov 2, 2017 at 9:48
  • $\begingroup$ I mean the last inequality would be equality, isn't it? $\endgroup$
    – Amin
    Nov 2, 2017 at 9:53
  • 1
    $\begingroup$ It is very late now, but it would be equality $\endgroup$ Aug 6, 2018 at 2:28
  • $\begingroup$ Thanks for your comment. $\endgroup$
    – Amin
    Aug 6, 2018 at 4:33

Notice that $$\|v\|^2 = \left\|\sum_{i=1}^n c_ie_i\right\|^2 = \sum_{i=1}^n |c_i|^2$$

Thus for $\|v\| = 1$ we have:

\begin{align}\left|\langle Tv, v\rangle\right| &= \left|\left\langle\sum_{i=1}^n c_iTe_i, \sum_{j=1}^n c_je_j \right\rangle\right| \\ &= \left|\left\langle\sum_{i=1}^n c_i\left(\sum_{j=1}^n \langle Te_i, e_j\rangle e_j\right), \sum_{i=1}^n c_ie_i \right\rangle\right| \\ &= \left|\left\langle\sum_{j=1}^n \left(\sum_{i=1}^n c_i\langle Te_i, e_j\rangle\right)e_j, \sum_{i=1}^n c_ie_i \right\rangle\right| \\ &= \left|\sum_{j=1}^n \overline{c_j}\left(\sum_{i=1}^n c_i\langle Te_i, e_j\rangle \right)\right| \\ &\le \sum_{j=1}^n |c_j|\left(\sum_{i=1}^n |c_i||\langle Te_i, e_j\rangle| \right) \\ &\le \max\limits_{1 \le i \le j \le n}|\langle Te_i, e_j\rangle| \cdot \sum_{j=1}^n |c_j|\left(\sum_{i=1}^n |c_i| \right) \\ &= \max\limits_{1 \le i \le j \le n}|\langle Te_i, e_j\rangle| \cdot \left(\sum_{j=1}^n |c_j|\right)^2 \\ &\stackrel{CSB}{\le} \max\limits_{1 \le i \le j \le n}|\langle Te_i, e_j\rangle| \cdot \sqrt{\sum_{i=1}^n |c_i|^2}\sqrt{\sum_{i=1}^n 1^2} \\ &\le \max\limits_{1 \le i \le j \le n}|\langle Te_i, e_j\rangle| \cdot \sqrt{n} \cdot \|v\|\\ &\le \max\limits_{1 \le i \le j \le n}|\langle Te_i, e_j\rangle| \cdot \sqrt{n} \end{align}

Therefore, the set $$\big\{\left|\langle Tv, v\rangle\right| : v \in H, \|v\| = 1\big\}$$ is bounded.


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