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Chapter 1: Q29E (page 1)

Construct a \(3 \times 3\) matrix, not in echelon form, whose columns span \({\mathbb{R}^3}\). Show that the matrix you construct has the desired property.

Short Answer

Expert verified

The required matrix is \(A = \left[ {\begin{array}{*{20}{c}}0&0&3\\0&4&0\\5&0&0\end{array}} \right]\).

Step by step solution

01

Writing the conditions for the echelon form

The matrix is in the echelon form if it satisfies the following conditions:

  • Non-zero rows should be positioned above zero rows.
  • Each row's leading entry should bein the column to the right of the row above its leading item.
  • In each column, all items below the leading entry must be zero.
02

Constructing a matrix in the non-echelon matrix

Assume that the matrix is \(A = \left[ {\begin{array}{*{20}{c}}0&0&3\\0&4&0\\5&0&0\end{array}} \right]\).

Here, each row's leading entry is not in the column to the right of the row above its leading item. So, the matrix is in non-echelon form.

03

Constructing an arbitrary vector in \({\mathbb{R}^3}\)

Suppose \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\\{{b_3}}\end{array}} \right]\) is the required vector.

Re-arrange the assumed vector as shown below:

\(\begin{array}{c}{\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\\{{b_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{0 + 0 + {b_1}}\\{0 + {b_2} + 0}\\{{b_3} + 0 + 0}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}0\\0\\{{b_3}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\{{b_2}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{b_1}}\\0\\0\end{array}} \right]\end{array}\)

04

Checking if the column matrix span is \({\mathbb{R}^3}\)

Simplify vector \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}0\\0\\{{b_3}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\{{b_2}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{b_1}}\\0\\0\end{array}} \right]\) further.

\(\begin{array}{c}{\bf{b}} = \left[ {\begin{array}{*{20}{c}}0\\0\\{\frac{{5{b_3}}}{5}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\{\frac{{4{b_2}}}{4}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{\frac{{3{b_1}}}{3}}\\0\\0\end{array}} \right]\\ = \frac{{{b_3}}}{5}\left[ {\begin{array}{*{20}{c}}0\\0\\5\end{array}} \right] + \frac{{{b_2}}}{4}\left[ {\begin{array}{*{20}{c}}0\\4\\0\end{array}} \right] + \frac{{{b_1}}}{3}\left[ {\begin{array}{*{20}{c}}3\\0\\0\end{array}} \right]\end{array}\)

Thus, it shows that the columns of the assumed matrix span are \({\mathbb{R}^3}\).

Therefore, the required matrix is \(A = \left[ {\begin{array}{*{20}{c}}0&0&3\\0&4&0\\5&0&0\end{array}} \right]\).

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Most popular questions from this chapter

In Exercise 22, mark each statement True or False. Justify each answer.

22. a. Every matrix transformation is a linear transformation.

b. The codomain of the transformation \({\bf{x}} \mapsto {\bf{Ax}}\) is the set of all linear combinations of the columns of \({\bf{A}}\).

c. If \({\bf{T}}:{\mathbb{R}^{\bf{n}}} \to {\mathbb{R}^{\bf{m}}}\) is a linear transformation and if \({\bf{c}}\) is in \({\mathbb{R}^{\bf{m}}}\), then a uniqueness is 鈥淚s c in the range of T?鈥�

d. A linear transformation preserves the operations of vector addition and scalar multiplication.

e. The superposition principle is a physical description of a linear transformation.

Suppose the coefficient matrix of a linear system of three equations in three variables has a pivot position in each column. Explain why the system has a unique solution.

In Exercises 7鈥�10, the augmented matrix of a linear system has been reduced by row operations to the form shown. In each case, continue the appropriate row operations and describe the solution set of the original system.

10. \(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\0&1&0&{ - 4}&7\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

In Exercises 21 and 22, mark each statement True or False. Justify each answer on the basis of a careful reading of the text.

21.

a. The columns of a matrix \(A\) are linearly independent if the equation \(Ax = 0\) has a trivial solution.

b. If \(S\) is a linearly dependent set, then each vector is a linear combination of the other vectors in \(S\).

c. The columns of any \(4 \times 5\) matrix are linearly dependent.

d. If \({\mathop{\rm x}\nolimits} \) and \(y\) are linearly independent, and if \(\left\{ {x,y,z} \right\}\) is linearly dependent, then \(z\) is in Span\(\left\{ {x,y} \right\}\).

In Exercises 3 and 4, display the following vectors using arrows

on an \(xy\)-graph: u, v, \( - {\bf{v}}\), \( - 2{\bf{v}}\), u + v , u - v, and u - 2v. Notice thatis the vertex of a parallelogram whose other vertices are u, 0, and \( - {\bf{v}}\).

3. u and v as in Exercise 1
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