Question

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1. Show that the Cauchy-Schwarz inequality holds for \({\bf{x}} = \left( {{\bf{3}}, - {\bf{2}}} \right)\) and \({\bf{y}} = \left( { - {\bf{2}},{\bf{1}}} \right)\). (Suggestion: Study \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).)

Short answer

The Cauchy-Schwartz inequality is true.

Step-by-step solution

Find the values \(\left\| {\bf{x}} \right\|\) and \(\left\| {\bf{y}} \right\|\)

Find the value of \(\left\| {\bf{x}} \right\|\).

\(\begin{aligned}\left\| {\bf{x}} \right\| &= \sqrt {\left\langle {{\bf{x}},{\bf{x}}} \right\rangle } \\ &= \sqrt {\left\langle {\left( {3, - 2} \right),\left( { - 2,1} \right)} \right\rangle } \\ &= \sqrt {4\left( 3 \right)\left( 3 \right) + 5\left( { - 2} \right)\left( { - 2} \right)} \\ &= \sqrt {36 + 20} \\ &= \sqrt {56} \end{aligned}\)

Thus, the value of \(\left\| {\bf{x}} \right\|\) is \(\sqrt {56} \).

Find the value of \(\left\| {\bf{y}} \right\|\).

\(\begin{aligned}\left\| {\bf{y}} \right\| &= \sqrt {\left\langle {{\bf{y}},{\bf{y}}} \right\rangle } \\ &= \sqrt {\left\langle {\left( { - 2,1} \right),\left( { - 2,1} \right)} \right\rangle } \\ &= \sqrt {4\left( { - 2} \right)\left( { - 2} \right) + 5\left( 1 \right)\left( 1 \right)} \\ &= \sqrt {21} \end{aligned}\)

Thus, the value of \(\left\| {\bf{y}} \right\|\) is \(\sqrt {21} \).

Find the inner product and \({\left\| {\bf{x}} \right\|^{\bf{2}}}{\left\| {\bf{y}} \right\|^{\bf{2}}}\)

The inner product \(\left\langle {{\bf{x}},{\bf{y}}} \right\rangle \)can be calculated as follows:

\(\begin{aligned}\left\langle {{\bf{x}},{\bf{y}}} \right\rangle & = \left\langle {\left( {3, - 2} \right),\left( { - 2,1} \right)} \right\rangle \\ &= 4\left( 3 \right)\left( { - 2} \right) + 5\left( { - 2} \right)\left( 1 \right)\\ &= - 34\end{aligned}\)

Find the value of \({\left\| {\bf{x}} \right\|^2}{\left\| {\bf{y}} \right\|^2}\).

\(\begin{aligned}{\left\| {\bf{x}} \right\|^2}{\left\| {\bf{y}} \right\|^2} &= 56 \times 21\\ &= 1176\end{aligned}\)

Check for Cauchy-Schwartz inequality

By the Cauchy-Schwartz inequality:

\(\begin{aligned}\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right| \le \left\| {\bf{x}} \right\|\left\| {\bf{y}} \right\|\\{\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^2} \le {\left\| {\bf{x}} \right\|^2}{\left\| {\bf{y}} \right\|^2}\\1156 < 1176\end{aligned}\)

Thus, the Cauchy-Schwartz inequality is true.