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Chapter 6: Q15E (page 331)

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

15.\(\left\langle {{\rm{u,}}\,c{\rm{v}}} \right\rangle = c\left\langle {{\rm{u,}}\,{\rm{v}}} \right\rangle \) for all scalars \(c\).

Short Answer

Expert verified

The statement \(\left\langle {{\rm{u}},c{\rm{v}}} \right\rangle = c\left\langle {{\rm{u}},{\rm{v}}} \right\rangle \) is verified.

Step by step solution

01

Apply axiom 1

According to axiom 1, if\({\bf{u}}\)and\({\rm{v}}\)bepair 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:

\(\left\langle {{\rm{u}},c{\rm{v}}} \right\rangle = \left\langle {c{\rm{v}},{\rm{u}}} \right\rangle \)

02

Apply axiom 3

According to axiom 3, if\({\bf{u}}\)and\({\rm{v}}\)bepair 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,\(\left\langle {c{\rm{u}},{\rm{v}}} \right\rangle = c\left\langle {{\rm{u}},{\rm{v}}} \right\rangle \)for all \({\bf{u}}\),\({\rm{v}}\),\({\rm{w}}\)and scalars\(c\).

Apply axiom 3 to the resulting equation, as follows:

\(\begin{aligned}\left\langle {{\rm{u}},c{\rm{v}}} \right\rangle = \left\langle {c{\rm{v}},{\rm{u}}} \right\rangle \\ = c\left\langle {{\rm{v}},{\rm{u}}} \right\rangle \end{aligned}\)

Again, apply axiom 1 to the resulting equation:

\(\begin{aligned}\left\langle {{\rm{u}},c{\rm{v}}} \right\rangle = c\left\langle {{\rm{v}},{\rm{u}}} \right\rangle \\ = c\left\langle {{\rm{u}},{\rm{v}}} \right\rangle \end{aligned}\)

Thus, the statement \(\left\langle {{\rm{u}},c{\rm{v}}} \right\rangle = c\left\langle {{\rm{u}},{\rm{v}}} \right\rangle \) is verified.

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Most popular questions from this chapter

Find an orthonormal basis of the subspace spanned by the vectors in Exercise 4.

Find a \(QR\) factorization of the matrix in Exercise 11.

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

\(\left[ {\begin{align}1\\{ - 2}\\1\end{align}} \right]\), \(\left[ {\begin{align}0\\1\\2\end{align}} \right]\), \(\left[ {\begin{align}{ - 5}\\{ - 2}\\1\end{align}} \right]\)

Exercises 19 and 20 involve a design matrix \(X\) with two or more columns and a least-squares solution \(\hat \beta \) of \({\bf{y}} = X\beta \). Consider the following numbers.

(i) \({\left\| {X\hat \beta } \right\|^2}\)鈥攖he sum of the squares of the 鈥渞egression term.鈥� Denote this number by .

(ii) \({\left\| {{\bf{y}} - X\hat \beta } \right\|^2}\)鈥攖he sum of the squares for error term. Denote this number by \(SS\left( E \right)\).

(iii) \({\left\| {\bf{y}} \right\|^2}\)鈥攖he 鈥渢otal鈥� sum of the squares of the \(y\)-values. Denote this number by \(SS\left( T \right)\).

Every statistics text that discusses regression and the linear model \(y = X\beta + \in \) introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume that the mean of the -values is zero. In this case, \(SS\left( T \right)\) is proportional to what is called the variance of the set of -values.

19. Justify the equation \(SS\left( T \right) = SS\left( R \right) + SS\left( E \right)\). (Hint: Use a theorem, and explain why the hypotheses of the theorem are satisfied.) This equation is extremely important in statistics, both in regression theory and in the analysis of variance.

Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

4.\(y = \left[ {\begin{aligned}{\bf{6}}\\{\bf{3}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{3}}\\{\bf{4}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{4}}}\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right]\)
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