Question

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

16. If \(\left\{ {{\rm{u,}}\,{\rm{v}}} \right\}\) is an orthonormal set in \(V\), then \(\left\| {{\rm{u}} - {\rm{v}}} \right\| = \sqrt 2 \).

Short answer

The statement \(\left\| {{\rm{u}} - {\rm{v}}} \right\| = \sqrt 2 \) 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:

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

Use the given information

It is given that \(\left\{ {{\bf{u}},{\rm{v}}} \right\}\) is orthonormal. So, \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle = 0\)and \({\left\| {\bf{u}} \right\|^2} = {\left\| {\rm{v}} \right\|^2} = 1\).

Apply the obtained results into the resulting equation, as follows:

\(\begin{aligned}{}{\left\| {{\rm{u}} - {\rm{v}}} \right\|^2} &= {\left\| {\rm{u}} \right\|^2} - {\rm{2}}\left\langle {{\rm{u,v}}} \right\rangle + {\left\| {\rm{v}} \right\|^2}\\ &= 1 - 2\left( 0 \right) + 1\\ &= 2\end{aligned}\)

So,\(\left\| {{\rm{u}} - {\rm{v}}} \right\| = \sqrt 2 \)

Thus, the statement\(\left\| {{\rm{u}} - {\rm{v}}} \right\| = \sqrt 2 \) is verified.