Question

Let T be a one-to-one linear transformation from a vector space V into \({\mathbb{R}^n}\). Show that u, v in V, the formula \(\left\langle {{\bf{u}},{\bf{v}}} \right\rangle = T\left( {\bf{u}} \right) \cdot T\left( {\bf{v}} \right)\) defines an inner product on V.

Short answer

Formula \(\left\langle {{\bf{u}},{\bf{v}}} \right\rangle = T\left( {\bf{u}} \right) \cdot T\left( {\bf{v}} \right)\) is a inner product space.

Step-by-step solution

Check for the first Axiom

Consider the following expression:

\(\begin{aligned}\left\langle {{\bf{u}},{\bf{v}}} \right\rangle &= T\left( {\bf{u}} \right) \cdot T\left( {\bf{v}} \right)\\ &= T\left( {\bf{v}} \right) \cdot T\left( {\bf{u}} \right)\\ &= \left\langle {{\bf{v}},{\bf{u}}} \right\rangle \end{aligned}\)

Thus, Axiom (1) is proved.

Check for the second Axiom

Consider the following expression:

\(\begin{aligned}\left\langle {{\bf{u}} + {\bf{v}},{\bf{w}}} \right\rangle &= \left[ {T\left( {{\bf{u}} + {\bf{v}}} \right)} \right] \cdot T\left( {\bf{w}} \right)\\ &= \left( {T\left( {\bf{u}} \right) + T\left( {\bf{v}} \right)} \right) \cdot T\left( {\bf{w}} \right)\\ &= T\left( {\bf{u}} \right) \cdot T\left( {\bf{w}} \right) + T\left( {\bf{v}} \right) \cdot T\left( {\bf{w}} \right)\\ &= \left\langle {{\bf{u}},{\bf{w}}} \right\rangle + \left\langle {{\bf{v}},{\bf{w}}} \right\rangle \end{aligned}\)

Thus, Axiom (2) is proved.

Check for the third Axiom

Consider the following expression:

\(\begin{aligned}\left\langle {c{\bf{u}},{\bf{v}}} \right\rangle &= T\left( {c{\bf{u}}} \right) \cdot T\left( {\bf{v}} \right)\\ &= c\left[ {T\left( {\bf{u}} \right)} \right] \cdot T\left( {\bf{v}} \right)\\ &= c\left\langle {{\bf{u}},{\bf{v}}} \right\rangle \end{aligned}\)

Thus, Axiom (3) is proved.

Check for the fourth Axiom

Consider the following expression:

\(\begin{aligned}\left\langle {{\bf{u}},{\bf{u}}} \right\rangle &= T\left( {\bf{u}} \right) \cdot T\left( {\bf{u}} \right)\\ &= {\left\| {T\left( {\bf{u}} \right)} \right\|^2}\\ \ge 0\end{aligned}\)

Thus, Axiom (4) is proved.

The above equation implies that \({\bf{u}} = 0\), therefore, T is a one-to-one transformation.

The formula defined \(\left\langle {{\bf{u}},{\bf{v}}} \right\rangle = T\left( {\bf{u}} \right) \cdot T\left( {\bf{v}} \right)\) is an inner product space.