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Chapter 6: Q14E (page 331)

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

Expert verified

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

01

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.

02

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.

03

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.

04

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.

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Most popular questions from this chapter

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram–Schmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{array}{*{20}{c}}3\\{-2}\\1\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{-1}\\3\\{-3}\\4\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\8\\7\\0\end{array}} \right]\)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

5. \(\left( {\frac{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}} \right){\mathop{\rm v}\nolimits} \)

In Exercises 7–10, let\[W\]be the subspace spanned by the\[{\bf{u}}\]’s, and write y as the sum of a vector in\[W\]and a vector orthogonal to\[W\].

10.\[y = \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\0\\1\\1\end{aligned}} \right]\],\[{{\bf{u}}_3} = \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

In Exercises 11 and 12, find the closest point to \[{\bf{y}}\] in the subspace \[W\] spanned by \[{{\bf{v}}_1}\], and \[{{\bf{v}}_2}\].

12. \[y = \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right]\], \[{{\bf{v}}_1} = \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\], \[{{\bf{v}}_2} = \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\]
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