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Chapter 2: Q2.9-7Q (page 93)

Let \({b_1} = \left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right]\), \({b_2} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\end{array}} \right]\), \({\bf{w}} = \left[ {\begin{array}{*{20}{c}}{\bf{7}}\\{ - {\bf{2}}}\end{array}} \right]\), \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{1}}\end{array}} \right]\), and \(B = \left\{ {{b_1},{b_2}} \right\}\). Use the figure to estimate \({\left[ {\bf{w}} \right]_B}\) and \({\left[ {\bf{x}} \right]_B}\). Confirm your estimate of \({\left[ {\bf{x}} \right]_B}\) by using it and \(\left\{ {{b_1},{b_2}} \right\}\)to compute x.

Short Answer

Expert verified

The vectors are \({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{.5}\end{array}} \right]\), and \[{\left[ {\bf{w}} \right]_B} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\]. The estimation of \({\left[ {\bf{x}} \right]_B}\) is confirmed, which is \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}4\\1\end{array}} \right]\).

Step by step solution

01

Definition of coordinate systems

For the basis of asubspace H, let the set be\(B = \left\{ {{{\bf{b}}_1},...,{{\bf{b}}_p}} \right\}\). For theweights \({c_1},...,{c_p}\), the coordinate of x is represented as\({\bf{x}} = {c_1}{{\bf{b}}_1} + \cdots + {c_p}{{\bf{b}}_p}\).

The\(B\)-coordinate vector of x is represented as shown below:

\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\ \vdots \\{{c_p}}\end{array}} \right]\)

02

Compute the B coordinate vector of w and x

From the figure, the following are the steps required to reach vector w, as shown below:

  • From the origin, move 2 units in the direction of\({{\bf{b}}_1}\).
  • Then, move 1 step in the negative direction of\({{\bf{b}}_2}\).

In the vector form, it is represented as\[{\bf{w}} = {\bf{2}}{{\bf{b}}_1}--{{\bf{b}}_2}\].

The following are the steps required to reach vector x as shown below:

  • From the origin, move 1.5 units in the direction of\({{\bf{b}}_1}\).
  • Then, move .5 step in the positive direction of\({{\bf{b}}_2}\).

In the vector form, it is represented as\[{\bf{x}} = {\bf{1}}.{\bf{5}}{{\bf{b}}_1} + .{\bf{5}}{{\bf{b}}_2}\].

The\(B\)-coordinate vectorof x is represented as shown below:

\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{.5}\end{array}} \right]\)

The\(B\)-coordinate vector of w is represented as shown below:

\[{\left[ {\bf{w}} \right]_B} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\]

03

Confirm the estimate

Compare\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right]\)with \({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{.5}\end{array}} \right]\). So,\({c_1} = 1.5\), and\({c_2} = .5\).

For theweights\({c_1}\)and\({c_2}\), and the vector x in H, it must satisfy the equation\({\bf{x}} = {c_1}{{\bf{b}}_1} + {c_2}{{\bf{b}}_2}\).

Consider the vectors\({b_1} = \left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right]\)and\[{{\bf{b}}_2} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\].

Then, it can be represented as shown below:

\(\begin{array}{l}{\bf{x}} = 1.5\left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right] + .5\left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\\{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{4.5}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - .5}\\1\end{array}} \right]\\{\bf{x}} = \left[ {\begin{array}{*{20}{c}}4\\1\end{array}} \right]\end{array}\)

Thus, the estimation of \({\left[ {\bf{x}} \right]_B}\) is confirmed.

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Most popular questions from this chapter

[M] Suppose memory or size restrictions prevent your matrix program from working with matrices having more than 32 rows and 32 columns, and suppose some project involves \(50 \times 50\) matrices A and B. Describe the commands or operations of your program that accomplish the following tasks.

a. Compute \(A + B\)

b. Compute \(AB\)

c. Solve \(Ax = b\) for some vector b in \({\mathbb{R}^{50}}\), assuming that \(A\) can be partitioned into a \(2 \times 2\) block matrix \(\left[ {{A_{ij}}} \right]\), with \({A_{11}}\) an invertible \(20 \times 20\) matrix, \({A_{22}}\) an invertible \(30 \times 30\) matrix, and \({A_{12}}\) a zero matrix. [Hint: Describe appropriate smaller systems to solve, without using any matrix inverse.]

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.)

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?

3. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{5}}\\{ - {\bf{7}}}&{ - {\bf{5}}}\end{aligned}} \right)\).

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)
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