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Chapter 4: Q14E (page 191)

If A is a \({\bf{4}} \times {\bf{3}}\) matrix, what is the largest possible dimension of the row space of A? If Ais a \({\bf{3}} \times {\bf{4}}\) matrix, what is the largest possible dimension of the row space of A? Explain.

Short Answer

Expert verified

If A is a \(4 \times 3\) matrix, then the largest possible dimension of the row space of A is 3.

If Ais a \(3 \times 4\) matrix, then the largest possible dimension of the row space of A is 3.

Step by step solution

01

Use the rank theorem

Note that the number of pivots in A gives the dimension of its column space.

By the rank theorem, \(\dim {\rm{Col}}\,A = \dim {\rm{Row}}\,A = {\rm{rank}}\,A\).

02

Compute the dimension of the row space for \({\bf{4}} \times {\bf{3}}\) matrix

If A is a \(4 \times 3\) matrix, then the number of pivots cannot exceed the number of columns. Here, the number of columns is minimum. Hence, the largest possible dimension of the row space of A is 3.

03

Compute the dimension of the row space for \({\bf{3}} \times {\bf{4}}\) matrix

If A is a \(3 \times 4\) matrix, then the number of pivots cannot exceed the number of rows. Here, the number of rows is minimum. Hence, the largest possible dimension of the row space of A is 3.

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Most popular questions from this chapter

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

19. \(A = \left( {\begin{array}{*{20}{c}}{.9}&1&0\\0&{ - .9}&0\\0&0&{.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}0\\1\\1\end{array}} \right)\).

Consider the following two systems of equations:

\(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 1\\4{x_1} + {x_2} - 6{x_3} = 9\end{array}\) \(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 5\\4{x_1} + {x_2} - 6{x_3} = 45\end{array}\)

It can be shown that the first system of a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.)

In Exercise 6, find the coordinate vector of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

6. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{6}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{0}}\end{array}} \right)\)
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