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Chapter 2: Q27Q (page 93)

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 an \(n \times n\) 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.

27. Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\) and \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\). Compute \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \), \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \),\({{\mathop{\rm uv}\nolimits} ^T}\), and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

Short Answer

Expert verified

\({{\mathop{\rm u}\nolimits} ^T}v = {{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} = - 2a + 3b - 4c\), \({\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{ - 2b}&{ - 2c}\\{3a}&{3b}&{3c}\\{ - 4a}&{ - 4b}&{ - 4c}\end{aligned}} \right)\), and \(v{{\mathop{\rm u}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{3a}&{ - 4a}\\{ - 2b}&{3b}&{ - 4b}\\{ - 2c}&{3c}&{ - 4c}\end{aligned}} \right)\).

Step by step solution

01

Determine the product \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \)

It is known that the transpose of Ais denoted by \({A^T}\).

The matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, commonly represented by a real number and written without the matrix brackets.

\(\begin{aligned}{c}{{\mathop{\rm u}\nolimits} ^T}v = \left( {\begin{aligned}{*{20}{c}}{ - 2}&3&{ - 4}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\\ = - 2a + 3b - 4c\\{{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a&b&c\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\\ = - 2a + 3b - 4c\end{aligned}\)

02

Determine the product \({\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T}\) and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\)

\(\begin{aligned}{c}{{\mathop{\rm uv}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}a&b&c\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{ - 2b}&{ - 2c}\\{3a}&{3b}&{3c}\\{ - 4a}&{ - 4b}&{ - 4c}\end{aligned}} \right)\\v{{\mathop{\rm u}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 2}&3&{ - 4}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{3a}&{ - 4a}\\{ - 2b}&{3b}&{ - 4b}\\{ - 2c}&{3c}&{ - 4c}\end{aligned}} \right)\end{aligned}\)

Thus, the products is \({{\mathop{\rm u}\nolimits} ^T}v = {{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} = - 2a + 3b - 4c\), \({\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{ - 2b}&{ - 2c}\\{3a}&{3b}&{3c}\\{ - 4a}&{ - 4b}&{ - 4c}\end{aligned}} \right)\), and \(v{{\mathop{\rm u}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{3a}&{ - 4a}\\{ - 2b}&{3b}&{ - 4b}\\{ - 2c}&{3c}&{ - 4c}\end{aligned}} \right)\).

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Most popular questions from this chapter

Suppose Ais an \(m \times n\) matrix and there exist \(n \times m\) matrices C and D such that \(CA = {I_n}\) and \(AD = {I_m}\). Prove that \(m = n\) and \(C = D\). (Hint: Think about the product CAD.)

Let \(X\) be \(m \times n\) data matrix such that \({X^T}X\) is invertible., and let \(M = {I_m} - X{\left( {{X^T}X} \right)^{ - {\bf{1}}}}{X^T}\). Add a column \({x_{\bf{0}}}\) to the data and form

\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

Compute \({W^T}W\). The \(\left( {{\bf{1}},{\bf{1}}} \right)\) entry is \({X^T}X\). Show that the Schur complement (Exercise 15) of \({X^T}X\) can be written in the form \({\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}\). It can be shown that the quantity \({\left( {{\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}} \right)^{ - {\bf{1}}}}\) is the \(\left( {{\bf{2}},{\bf{2}}} \right)\)-entry in \({\left( {{W^T}W} \right)^{ - {\bf{1}}}}\). This entry has a useful statistical interpretation, under appropriate hypotheses.

In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

\(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the 鈥渋nput鈥� to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the 鈥渙utput鈥� and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the 鈥渟tate鈥� vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)

Suppose P is invertible and \(A = PB{P^{ - 1}}\). Solve for Bin terms of A.

Suppose the second column of Bis all zeros. What can you

say about the second column of AB?

Suppose Ais \(n \times n\) and the equation \(A{\bf{x}} = {\bf{0}}\) has only the trivial solution. Explain why Ahas npivot columns and Ais row equivalent to \({I_n}\). By Theorem 7, this shows that Amust be invertible. (This exercise and Exercise 24 will be cited in Section 2.3.)
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