In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\)as\(n \times 1\)matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is a \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

28. If u and v are in \({\mathbb{R}^n}\), how are \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \) related? How are \({{\mathop{\rm uv}\nolimits} ^T}\) and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) related?

Short Answer

Expert verified

The relation of inner and outer product is \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} = {{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \) and \({\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T} = {\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\), respectively.

Step by step solution

01

Determine the relation of the inner product

Theorem 3states that \(A\) and \(B\) denotes matrices whose sizes are appropriate for the following sums and products.

  1. \({\left( {{A^T}} \right)^{^T}} = A\).
  2. \({\left( {A + B} \right)^T} = {A^T} + {B^T}\).
  3. For any scalar \(r\), \({\left( {rA} \right)^T} = r{A^T}\).
  4. \({\left( {AB} \right)^T} = {B^T}{A^T}\).

The inner product \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) is a real number since it equals its transpose.

Use theorem 3 to obtain the relation of the inner product as follows:

\(\begin{aligned}{c}{{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} = {\left( {{{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} } \right)^T}\\ = {{\mathop{\rm v}\nolimits} ^T}{\left( {{{\mathop{\rm u}\nolimits} ^T}} \right)^T}\\ = {{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \end{aligned}\)

02

Determine the relation of the outer product

A \(n \times n\) matrix is an outer product \({\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T}\).

Use theorem 3 to obtain the relation of the outer product as follows:

\(\begin{aligned}{c}{\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T} = {\left( {{\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T}} \right)^T}\\ = {\left( {{{\mathop{\rm v}\nolimits} ^T}} \right)^T}{{\mathop{\rm u}\nolimits} ^T}\\ = {\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\end{aligned}\)

Thus, the relation of inner and outer products is \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} = {{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \) and \({\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T} = {\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\), respectively.

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Most popular questions from this chapter

In Exercise 10 mark each statement True or False. Justify each answer.

10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.

b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.

d. If A can be row reduced to the identity matrix, then A must be invertible.

e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).

In exercise 11 and 12, mark each statement True or False.Justify each answer.

a. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}&{{A_{\bf{2}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), with \({A_{\bf{1}}}\) and \({A_{\bf{2}}}\) the same sizes as \({B_{\bf{1}}}\) and \({B_{\bf{2}}}\), respectively then \(A + B = \left[ {\begin{array}{*{20}{c}}{{A_1} + {B_1}}&{{A_{\bf{2}}} + {B_{\bf{2}}}}\end{array}} \right]\).

b. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_{\bf{2}}}}\end{array}} \right]\), then the partitions of \(A\) and \(B\) are comfortable for block multiplication.

Explain why the columns of an \(n \times n\) matrix Aare linearly independent when Ais invertible.

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{1}}}\\{\bf{5}}&{ - {\bf{2}}}\end{aligned}} \right)\). Compute \({\bf{3}}{I_{\bf{2}}} - A\) and \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\).

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