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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.

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.

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.

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.