Construction of Inner Product From a Linear Transformation

Given any linear transformation
$T$
from a vector space
$V$
to another vector space
$W$
, we can construct an inner product on
$W$
:
$p_T(\mathbf{v}_1 , \mathbf{v}_2) = \langle T \mathbf{v}_1 , T \mathbf{v}_2 \rangle$
for all
$\mathbf{v}_1 , \: \mathbf{v}_2 \in V$

We need to check that all the conditions of an inner product are satisfied.
The inner product is symmetric:
\begin{aligned} \langle T \mathbf{v}_1, T\mathbf{v}_2 \rangle &= \langle T \sum_iv_1^i \mathbf{e}_i, T \sum_j v^j_2 \mathbf{e}_j \rangle \\ &= \langle \sum_i v^i_1 T \mathbf{e}_i, \sum_j v_2^j T \mathbf{e}_j \rangle \\ &= \sum_{i \: j} v_1^i v_2^j \langle T \mathbf{e}_1, T\mathbf{e}_2 \rangle \\ &= \sum_{j \: i} v_2^j v_1^i \langle T \mathbf{e}_2, T\mathbf{e}_1 \rangle \\ &= \langle T \mathbf{v}_2, T\mathbf{v}_1 \rangle \end{aligned}

The inner product is positive definite:
Let
$T \mathbf{v} = \mathbf{w} = \sum_i w_i \mathbf{e}_i$

\begin{aligned} \langle T \mathbf{v}, T\mathbf{v} \rangle &= \langle \sum w^i \mathbf{e}_i , \sum_j w_j \mathbf{e}_j \rangle \\ &= \sum_{i, \: j} w^i w^j \\ &= \sum_i (w^i)^2 \end{aligned}

The inner product is linear in both argumnets:
\begin{aligned} \langle \alpha T \mathbf{v}_1, T\mathbf{v}_2 \rangle &= \langle \alpha T \sum_iv_1^i \mathbf{e}_i, T \sum_j v^j_2 \mathbf{e}_j \rangle \\ &= \langle \alpha \sum_i v^i_1 T \mathbf{e}_i, \sum_j v_2^j T \mathbf{e}_j \rangle \\ &= \alpha \langle \sum_i v^i_1 T \mathbf{e}_i, \sum_j v_2^j T \mathbf{e}_j \rangle \\ &= \sum_{i \: j} \alpha v_1^i v_2^j \langle T \mathbf{e}_1, T\mathbf{e}_2 \rangle \\ &= \sum_{i \: j} v_1^i \alpha v_2^j \langle T \mathbf{e}_2, T\mathbf{e}_1 \rangle \\ &= \langle T \mathbf{v}_1, \alpha T\mathbf{v}_1 \rangle \end{aligned}