Question

Use the inner product axioms and other results of this section to verify the statements in Exercises 15–18.

\(\left\langle {{\rm{u,v}}} \right\rangle = \frac{1}{4}{\left\| {{\rm{u}} + {\rm{v}}} \right\|^2} - \frac{1}{4}{\left\| {{\rm{u}} - {\rm{v}}} \right\|^2}\).

Short answer

The statement \(\left\langle {{\rm{u,v}}} \right\rangle = \frac{1}{4}{\left\| {{\rm{u}} + {\rm{v}}} \right\|^2} - \frac{1}{4}{\left\| {{\rm{u}} - {\rm{v}}} \right\|^2}\) is verified.

Step-by-step solution

Apply axiom 2

According to axiom 2, if \({\bf{u}}\) and \({\rm{v}}\) be pair of vectors in a vector space \(V\), then, the inner product on \(V\), relates a real number \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle \)and satisfies the axiom, \(\left\langle {{\bf{u}} + {\rm{v,}}\,{\rm{w}}} \right\rangle = \left\langle {{\rm{u,w}}} \right\rangle + \left\langle {{\rm{v,w}}} \right\rangle \) for all \({\bf{u}}\), \({\rm{v}}\), \({\rm{w}}\)and scalars \(c\).

Apply axiom 2 to the left side of the given equation, as follows:

\(\begin{aligned}{}{\left\| {{\rm{u}} + {\rm{v}}} \right\|^2} &= \left\langle {{\rm{u}} + {\rm{v,}}\,{\rm{u}} + {\rm{v}}} \right\rangle \\ &= \left\langle {{\rm{u,}}\,{\rm{u}} + {\rm{v}}} \right\rangle + \left\langle {{\rm{v,}}\,{\rm{u}} + {\rm{v}}} \right\rangle \\ &= \left\langle {{\rm{u,}}\,{\rm{u}}} \right\rangle + \left\langle {{\rm{u,v}}} \right\rangle + \left\langle {{\rm{v,}}\,{\rm{u}}} \right\rangle + \left\langle {{\rm{v,}}\,{\rm{v}}} \right\rangle \end{aligned}\)

Apply axiom 1

According to axiom 1, if \({\bf{u}}\) and \({\rm{v}}\) be pair of vectors in a vector space \(V\), then, the inner product on \(V\), relates a real number \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle \)and satisfies the axiom, \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle = \left\langle {{\rm{v}},{\rm{u}}} \right\rangle \)for all \({\bf{u}}\), \({\rm{v}}\), \({\rm{w}}\)and scalars \(c\).

Apply axiom 1 to the left side of the given equation, as follows:

\(\begin{aligned}{}{\left\| {{\rm{u}} + {\rm{v}}} \right\|^2} &= \left\langle {{\rm{u,}}\,{\rm{u}}} \right\rangle + \left\langle {{\rm{u,v}}} \right\rangle + \left\langle {{\rm{u,v}}} \right\rangle + \left\langle {{\rm{v,}}\,{\rm{v}}} \right\rangle \\ &= \left\langle {{\rm{u,}}\,{\rm{u}}} \right\rangle + {\rm{2}}\left\langle {{\rm{u,v}}} \right\rangle + \left\langle {{\rm{v,}}\,{\rm{v}}} \right\rangle \\ &= {\left\| {\rm{u}} \right\|^2} + 2\left\langle {{\rm{u,v}}} \right\rangle + {\left\| {\rm{v}} \right\|^2}\end{aligned}\)

Following the same steps, we can write that \({\left\| {{\rm{u}} - {\rm{v}}} \right\|^2} = {\left\| {\rm{u}} \right\|^2} - 2\left\langle {{\rm{u,v}}} \right\rangle + {\left\| {\rm{v}} \right\|^2}\).

Subtract the two results

Subtract the obtain equation for\({\left\| {{\rm{u}} + {\rm{v}}} \right\|^2}\)and\({\left\| {{\rm{u}} - {\rm{v}}} \right\|^2}\), as follows:

\(\begin{aligned}{}{\left\| {{\rm{u}} + {\rm{v}}} \right\|^2} - {\left\| {{\rm{u}} - {\rm{v}}} \right\|^2} &= \left( {{{\left\| {\rm{u}} \right\|}^2} + 2\left\langle {{\rm{u,v}}} \right\rangle + {{\left\| {\rm{v}} \right\|}^2}} \right) - \left( {{{\left\| {\rm{u}} \right\|}^2} - 2\left\langle {{\rm{u,v}}} \right\rangle + {{\left\| {\rm{v}} \right\|}^2}} \right)\\ &= 4\left\langle {{\rm{u,v}}} \right\rangle \end{aligned}\)

Now divide the resulting equation by 4 and simplify as shown below:

\(\begin{aligned}{}\frac{{{{\left\| {{\rm{u}} + {\rm{v}}} \right\|}^2} - {{\left\| {{\rm{u}} - {\rm{v}}} \right\|}^2}}}{4} &= \frac{{4\left\langle {{\rm{u,v}}} \right\rangle }}{4}\\\left\langle {{\rm{u,v}}} \right\rangle &= \frac{1}{4}{\left\| {{\rm{u}} + {\rm{v}}} \right\|^2} - \frac{1}{4}{\left\| {{\rm{u}} - {\rm{v}}} \right\|^2}\end{aligned}\)

Thus, the statement\(\left\langle {{\rm{u,v}}} \right\rangle = \frac{1}{4}{\left\| {{\rm{u}} + {\rm{v}}} \right\|^2} - \frac{1}{4}{\left\| {{\rm{u}} - {\rm{v}}} \right\|^2}\) is verified.