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Chapter 4: Q6E (page 191)

If a\({\bf{6}} \times {\bf{3}}\)matrix A has a rank 3, find dim Nul A, dim Row A, and rank\({A^T}\).

Short Answer

Expert verified

0, 3, and 3

Step by step solution

01

Find dim Nul A

Using the rank theorem,you get:

\(\begin{aligned} {\rm{rank}}\,A + \dim \,{\rm{Nul}}A &= n\\3 + \dim \;{\rm{Nul}}\,A &= 3\\\dim \;{\rm{Nul}}\,A &= 3 - 3\\ &= 0\end{aligned}\)

02

Find dim row A

The dim row A is equal to the rank of A i.e., 3.

03

Find the rank of \({A^T}\)

\(\dim \,\;{\rm{Row}}\;A = \dim \,{\rm{Col}}\,{A^T} = 3\)

The rank of \({A^T}\) is equal to dim Col \({A^T}\); so the rank of \({A^T}\) is 3.

Thus, dim Nul A =0, dim row A=3, and the rank of \({A^T}\)=3.

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Most popular questions from this chapter

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

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

(M) Show thatis a linearly independent set of functions defined on. Use the method of Exercise 37. (This result will be needed in Exercise 34 in Section 4.5.)

Explain what is wrong with the following discussion: Let \({\bf{f}}\left( t \right) = {\bf{3}} + t\) and \({\bf{g}}\left( t \right) = {\bf{3}}t + {t^{\bf{2}}}\), and note that \({\bf{g}}\left( t \right) = t{\bf{f}}\left( t \right)\). Then, \(\left\{ {{\bf{f}},{\bf{g}}} \right\}\) is linearly dependent because g is a multiple of f.

[M] Repeat Exercise 35 for a random integer-valued matrixwhose rank is at most 4. One way to makeis to create a random integ\(6 \times 7\)er-valued \(6 \times 4\) matrix \(J\) and a random integer-valued \(4 \times 7\) matrix \(K\), and set \(A = JK\). (See supplementary Exercise 12 at the end of the chapter; and see the study guide for the matrix-generating program.)

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.)
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