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Chapter 2: Q2.9-6Q (page 93)

In Exercises 3鈥�6, the vector x is in a subspace Hwith a basis \(B = \left\{ {{b_{\bf{1}}},{{\bf{b}}_{\bf{2}}}} \right\}\). Find the B-coordinate vector of x.

6. \[{b_1} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\{ - 4}\end{array}} \right]\], \[{b_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}7\\5\\{ - 6}\end{array}} \right]\], \[x = \left[ {\begin{array}{*{20}{c}}{{\bf{11}}}\\0\\7\end{array}} \right]\]

Short Answer

Expert verified

TheB-coordinate vector of x is \({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 5/2}\\{1/2}\end{array}} \right]\).

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 vectorof 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

Construct the augmented matrix

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 thevectors \[{b_1} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\{ - 4}\end{array}} \right]\],\[{b_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}7\\5\\{ - 6}\end{array}} \right]\], and\[x = \left[ {\begin{array}{*{20}{c}}{11}\\0\\7\end{array}} \right]\].

Then, it can be represented as shown below:

\({c_1}\left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\{ - 4}\end{array}} \right] + {c_2}\left[ {\begin{array}{*{20}{c}}7\\5\\{ - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{11}\\0\\7\end{array}} \right]\)

Write theaugmented matrix \(\left[ {\begin{array}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{\bf{x}}\end{array}} \right]\)as shown below:

\(\left[ {\begin{array}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{\bf{x}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 3}&7&{11}\\1&5&0\\{ - 4}&{ - 6}&7\end{array}} \right]\)

03

Compute the weights

Exchange row 1 with row 3.

\(\left[ {\begin{array}{*{20}{c}}{ - 3}&7&{11}\\1&5&0\\{ - 4}&{ - 6}&7\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&5&0\\{ - 3}&7&{11}\\{ - 4}&{ - 6}&7\end{array}} \right]\)

Use the\({x_1}\)term in the first equation to eliminate the\( - 3{x_1}\)term from the second equation. Add 3 times row 1 to row 3.

\(\left[ {\begin{array}{*{20}{c}}1&5&0\\{ - 3}&7&{11}\\{ - 4}&{ - 6}&7\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&5&0\\0&{22}&{11}\\{ - 4}&{ - 6}&7\end{array}} \right]\)

Multiply row 2 by \(\frac{1}{{22}}\).

\(\left[ {\begin{array}{*{20}{c}}1&5&0\\0&{22}&{11}\\{ - 4}&{ - 6}&7\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&5&0\\0&1&{1/2}\\{ - 4}&{ - 6}&7\end{array}} \right]\)

Use the\({x_2}\)term in the second equation to eliminate the\(5{x_2}\)term from the first equation. Add\( - 5\)times row 2 to row 1.

\(\left[ {\begin{array}{*{20}{c}}1&5&0\\0&1&{1/2}\\{ - 4}&{ - 6}&7\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - 5/2}\\0&1&{1/2}\\0&0&0\end{array}} \right]\)

Thus, the weights are \({c_1} = - \frac{5}{2}\) and \({c_2} = \frac{1}{2}\).

Obtain the B-coordinate vector of x.

\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 5/2}\\{1/2}\end{array}} \right]\)

Thus, the B-coordinate vector of x is \({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 5/2}\\{1/2}\end{array}} \right]\).

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Most popular questions from this chapter

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]

Suppose A, B,and Care invertible \(n \times n\) matrices. Show that ABCis also invertible by producing a matrix Dsuch that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4

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.
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