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Chapter 6: Q6.2-26E (page 331)

Question: Suppose W is a subspace of \({\mathbb{R}^n}\) spanned by a nonzero orthogonal vectors. Explain why \(W = {\mathbb{R}^n}\).

Short Answer

Expert verified

By Theorem 4, the orthogonal vectors must be linearly independent, so the set spans W and it is the basis for W. Thus, \(W = {\mathbb{R}^n}\).

Step by step solution

01

Write the given information

It is given that W is a subspace of \({\mathbb{R}^n}\) spanned by a nonzero orthogonal vector.

02

Write why \(W = {\mathbb{R}^n}\)

Let subspace \(S = \left\{ {{{\bf{u}}_1},.....,{{\bf{u}}_p}} \right\}\) is the set of orthogonal vectors in \({\mathbb{R}^n}\), then S is linearly independent, and hence it is the basis for the subspace spanned by S. So, the set spans W, and it is the basis for W.

Therefore, the set S is linearly independent.

So, the n vectors in the matrix are linearly independent.

Thus, the equation \(W = {\mathbb{R}^n}\) is true.

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Most popular questions from this chapter

In Exercises 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).

2. \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{1}}\\{ - {\bf{2}}}&{\bf{0}}\\{\bf{2}} {\bf{3}}\end{aligned}} \right)\), \(b = \left( {\begin{aligned}{{}{}}{ - {\bf{5}}}\\{\bf{8}}\\{\bf{1}}\end{aligned}} \right)\)

In Exercises 9-12, find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

9. \(A = \left[ {\begin{aligned}{{}{}}{\bf{1}}&{\bf{5}}\\{\bf{3}}&{\bf{1}}\\{ - {\bf{2}}}&{\bf{4}}\end{aligned}} \right]\), \({\bf{b}} = \left[ {\begin{aligned}{{}{}}{\bf{4}}\\{ - {\bf{2}}}\\{ - {\bf{3}}}\end{aligned}} \right]\)

Show that if \(U\) is an orthogonal matrix, then any real eigenvalue of \(U\) must be \( \pm 1\).

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