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

Exercises 9–12 display a matrix Aand an echelon form of A. Find bases for Col Aand Nul A, and then state the dimensions of these subspaces.

\(A = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&{ - 4}\\{ - 3}&9&{ - 1}&5\\2&{ - 6}&4&{ - 3}\\{ - 4}&{12}&2&7\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&{ - 4}\\0&0&5&{ - 7}\\0&0&0&5\\0&0&0&0\end{array}} \right]\)

Short Answer

Expert verified

The bases for Col A are \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\2\\{ - 4}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\4\\2\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\5\\{ - 3}\\7\end{array}} \right]} \right\}\). The bases for Nul A are \(\left[ {\begin{array}{*{20}{c}}3\\1\\0\\0\end{array}} \right]\). The dimension of Col A is 3. The dimension of Nul A is 1.

Step by step solution

01

Bases for Nul A and Col A

The set of all linear combinations of the columns of matrix A is Col A, or it is called the column space of A. Pivot columns are the bases for Col A.

The set of all homogeneous equation solutions\(A{\bf{x}} = 0\)is Nul A, or it is called the null space of A.

02

Write the bases for Col A

To identify the pivot and the pivot position, observe the matrix’s leftmost column (nonzero column), which is the pivot column. At the top of this column, 1 is the pivot.

It is observed that the first, third, and fourth columns have pivot elements.

The corresponding columns of matrix A are shown below:

\(\left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\2\\{ - 4}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\4\\2\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\5\\{ - 3}\\7\end{array}} \right]\)

The column space is given as shown below:

\({\rm{Col }}A = \left\{ {\left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\2\\{ - 4}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\4\\2\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\5\\{ - 3}\\7\end{array}} \right]} \right\}\)

Thus, the bases for Col A are \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\2\\{ - 4}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\4\\2\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\5\\{ - 3}\\7\end{array}} \right]} \right\}\).

03

Write the bases for Nul A

It is given that there are 4 columns in the given matrix, which means there should be 4 entries in vector x.

Thus, the equation \(A{\bf{x}} = 0\) can be written as shown below:

\(\begin{array}{c}Ax = 0\\\left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&{ - 4}\\0&0&5&{ - 7}\\0&0&0&5\\0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\end{array}\)

The augmented matrix is as shown below:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&{ - 4}&0\\0&0&5&{ - 7}&0\\0&0&0&5&0\\0&0&0&0&0\end{array}} \right]\)

Multiply row 3 by 5. Then, add 4 times row 3 to row 1.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&{ - 4}&0\\0&0&5&{ - 7}&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&0&0\\0&0&5&{ - 7}&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\)

Add 7 times row 3 to row 2, and \( - 2\) times row 2 to row 1.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&0&0\\0&0&5&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 3}&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\)

So, the system of equations is as shown below:

\(\begin{array}{c}{x_1} = 3{x_2}\\{x_3} = 0\\{x_4} = 0\end{array}\)

From the above equations, \({x_1}\), \({x_3}\), and \({x_4}\) correspond to the pivot positions. So, \({x_1}\), \({x_3}\), and \({x_4}\) are the basic variables, and \({x_2}\) is the free variable.

Let \({x_2} = a\).

Substitute the value \({x_2} = a\) in the equation \({x_1} = 3{x_2}\) to obtain the general solution.

\(\begin{array}{l}{x_1} = 3\left( a \right)\\{x_1} = 3a\end{array}\)

Obtain the vector in the parametric form by using \({x_1} = 3a\), \({x_2} = a\), \({x_3} = 0\), and \({x_4} = 0\).

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3a}\\a\\0\\0\end{array}} \right]\\ = a\left[ {\begin{array}{*{20}{c}}3\\1\\0\\0\end{array}} \right]\\ = {x_2}\left[ {\begin{array}{*{20}{c}}3\\1\\0\\0\end{array}} \right]\end{array}\)

Nul A is shown below:

\({\rm{Nul }}A = \left[ {\begin{array}{*{20}{c}}3\\1\\0\\0\end{array}} \right]\)

Thus, the bases for Nul A are \(\left[ {\begin{array}{*{20}{c}}3\\1\\0\\0\end{array}} \right]\).

04

Dimensions of subspaces

It is observed that matrix A has three pivot columns; so the dimension of Col A is 3. Thus, Col A= 3.

Also, it is observed that the homogeneous equation \(A{\bf{x}} = 0\) has only one free variable; so the dimension of Nul A is 1. Thus, Nul A= 1.

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Most popular questions from this chapter

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

Use the inverse found in Exercise 1 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)

Show that if the columns of Bare linearly dependent, then so are the columns of AB.

Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).

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 “input” to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the “output” and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the “state” 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.)
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