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

Question

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? $$\langle u, v \rangle = u^T v$$

Answer 1

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 $V$.

Answer 2

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!