# Dual Vector Spaces with Orthonormal Basis

I'm really stuck on the following quesiton.

Let $U$ and $V$ be finite dimensional vector spaces over the complex numbers with bases $e_1,..,e_n$ of $U$ and $f_1,...,f_m$ of $V$. They also have dual spaces $U^*$ and $V^*$ with bases $e^i$ and $f^i$ respectively.

Then assume that both spaces are Hermitian.

Let $T_U:U\to U^*$ be defined by $T_U(w)(u) = \langle w,u\rangle \forall w,u\in U$.

Prove that if the basis $e_i$ is orthonormal then $T_U(e_i) = e^i$.

I have tried showing that

$T_U(e_i)(u) = \langle e_i, u\rangle$ $= \langle e_i,x_1e_1+...+x_ne_n\rangle$ $= \langle e_i, x_1e_1\rangle + \langle e_i, x_2e_2\rangle +\cdots +\langle e_i, x_ne_n\rangle$ $=x_1\langle e_i,e_1\rangle + x_2\langle e_i,e_2\rangle +\cdots +x_n\langle e_i,e_n\rangle$

So we are left with, $x_i\langle e_i,e_i \rangle$ since the others are mutually orthogonal.

I'm having trouble understanding why this implies the desired result.

• Please use \langle and \rangle to get $\langle\cdots\rangle$. (I fixed it for you this time.) Oh, and what role is $V$ playing here? – Harald Hanche-Olsen Nov 28 '13 at 18:09

To see that

$T_U(e_i) = e^i, \tag{1}$

first compute

$T_U(e_i)(e_j) = \langle e_i, e_j \rangle = \delta_{ij} \tag{2}$

by orthonormality of the $e_i$. Then note that

$e^i(e_j) = \delta_{ij} \tag{3}$

as well, this time by the duality if the bases $e_i$, $e^j$. Then since $T_U(e_i)$ and $e^i$ agree on the basis elements $e_j$ of $U$, it follows by linearity thet agree on every $u = \sum u_j e_j \in U$:

$T_U(e_i)(\sum u_j e_j) = \sum u_j T_U(e_i)(e_j) = \sum u_j e^i(e_j) = e^i(\sum u_j e_j); \tag{4}$

thus

$T_U(e_i) = e^i, \tag{5}$

and we are done!QED

Hope this helps. Holiday Cheers,

and as always,

Fiat Lux!!!

• Thanks, I almost had it. I think my troubles stemmed from not entirely understanding the definition of the map. I got a bit worried because I had $T_U$ defined as $T_U(w)(u)$ but I had to prove something about $T_U(w)$. I didn't realise that this meant I had to prove it for all $u$. Thanks everyone – benjiebob Nov 28 '13 at 18:30
• @ benjiebob: Yup, you sure did! Glad to help out on this bee-a-ooo-ti-ful Turkey Day here in Beserkeley, California; and thanks for the "acceptance"! – Robert Lewis Nov 28 '13 at 18:34

The property of the dual basis is that $e^{i}(e_{j}) = \delta_{i,j}$. So, why don't you consider how $T_{U}(e_{i})$ and $e^{i}$ act on the basis of $U$?

Hint: First of all you don't need $V$ for your question, right? So let's first talk about your Hermitian. If $U$ is a Hermitian space, then there is a Hermitian form $$\langle, \rangle: U \times U \longrightarrow \Bbb C$$

Now your map $T_U$ is not defined in a clear way. It should go from $U \to U^*$, but then, you have to define

$$T_U: U \to U^*, u \mapsto \langle -,u \rangle$$

and then $T_U(u) \in U^*$ for $u \in U$. So to prove that $T_U(e_i) = e^i$, you just have to know how the $e^i$ is defined, namely via the Kronecker Delta: $$e^i (e_j) = \delta_{ij}$$

So start with $u = \sum_{i=1}^{n} e_i$ and show what happens if you let $T_U$ act on $u$.