# Showing that $\langle u, v \rangle = T(u)^T \hspace{1mm} T(v)$ is an inner product on $V$ for $T: V → {\Bbb R^n}$

Let $$V$$ be a real vector space and let $$T$$ be one-to-one linear transformation with $$T: V → {\Bbb R^n}$$. Show that for u, v in $$V$$,
$$\langle u, v \rangle = T(u)^T \hspace{1mm} T(v)$$
defines an inner product on V

I am not sure how the inner product relates to linear transformation. Do we need to use the definition of inner product? $$\langle u, v \rangle = u^T v$$

Definition (Lang): Let $$V$$ be a vector space over a field $$K$$. A scalar product [a.k.a. an inner product] on $$V$$ is a rule which to any pair of elements $$v,w\in V$$ associates a scalar, denoted by $$\langle v , w \rangle$$, or also $$v \cdot w$$, satisfying the following properties:

$$\qquad$$ SP $$1$$. $$\space$$ We have $$\langle v , w \rangle$$ = $$\langle w , v \rangle$$ for all $$v,w\in V$$.

$$\qquad$$ SP $$2$$. $$\space$$ If $$u, v, w$$ are elements of $$V$$, then $$\langle u , v + w \rangle = \langle v , w \rangle + \langle v , w \rangle.$$

$$\qquad$$ SP $$3$$. $$\space$$ If $$x \in K$$, then $$\langle xu, v\rangle = x \langle u , v \rangle \qquad and \qquad \langle u, xv\rangle = x \langle u , v \rangle$$

As a general rule, when wanting to see if some $$X$$ is a $$Y$$ on some structure $$Z$$, you will want to go back to the original definition of $$Y$$ and check to see if your hypothesis satisfies all of the necessary conditions.

Lastly, the inner/scalar product is related to linear transformations due to SP $$2$$ and SP $$3$$ (which are the definition of a linear transformation).

As an exercise for when you finish your problem, prove the following (credit due to Lang, 1966):

$$1. \quad$$ Let $$V$$ be the space of continuous real-valued functions on the interval $$[0,1]$$ (i.e., $$V = \mathbb{C}[0,1]$$). If $$f, g \in V$$, show that $$\langle f, g \rangle = \int_{0}^{1}f(t)g(t) \space dt$$ is an inner product.

Hope this helps you figure this out!

We have to verify that the function $$\langle \cdot, \cdot \rangle$$ does define an inner product on $$V$$. Since $$V$$ is a real vector space, an inner product on $$V$$ is a function $$\langle \cdot, \cdot \rangle$$ such that $$\forall u, v, w \in V$$ and $$\forall a, b \in \mathbb{R}$$, \begin{align} & \langle u, v\rangle = \langle v, u \rangle, \\ & \langle au + bw, v \rangle = a \langle u, v \rangle + b \langle w, v \rangle, \\ & \langle u, u \rangle \geq 0 \quad\text{and}\quad \langle u, u \rangle = 0 \iff u = 0. \end{align}

To verify the first condition, we see that for all $$u, v \in V$$, $$\langle u, v \rangle = (Tu)^\top Tv = \sum_{j=1}^n (Tu)_j (Tv)_j = (Tv)^\top Tu = \langle v, u \rangle,$$ where $$(Tu)_j$$ is the $$j$$-th component of the vector $$Tu$$ in $$\mathbb{R}^n$$.

The second condition is quite straightforward from the linearity of $$T$$.

The third condition can be verified since $$T$$ is one-to-one. First, we have for all $$u \in V$$ that $$\langle u, u \rangle = (Tu)^\top Tu = \sum_{j=1}^n (Tu)_j (Tu)_j = \sum_{j=1}^d (Tu)_j^2 \geq 0.$$ For $$\langle u, u \rangle = 0$$, we should have that $$(Tu)_j = 0$$ for all $$j$$, so that $$Tu = 0$$. But since $$T$$ is one-to-one, we know that $$Tu = 0$$ if and only if $$u=0$$. Hence we find that $$\langle u, u\rangle = 0$$ if and only if $$u = 0$$, as it should be for $$\langle \cdot, \cdot \rangle$$ to be an inner product on $$V$$.