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Chapter 4: Q5E (page 191)

If a \({\bf{3}} \times {\bf{8}}\) matrix A has a rank 3, find dim Nul A, dim Row A, and rank \({A^T}\).

Short Answer

Expert verified

5, 3, and 3

Step by step solution

01

Find dim Nul A

Using the rank theorem,you get:

\(\begin{aligned} {\rm{rank}}\,A + \dim \,{\rm{Nul}}A &= n\\3 + \dim \;{\rm{Nul}}\,A &= 8\\\dim \;{\rm{Nul}}\,A &= 8 - 3\\ &= 5\end{aligned}\)

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 =5, dim row A=3, and the rank of \({A^T}\)=3.

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Most popular questions from this chapter

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that a subset \(\left\{ {{{\bf{u}}_1},...,{{\bf{u}}_p}} \right\}\) in V is linearly independent if and only if the set of coordinate vectors \(\left\{ {{{\left( {{{\bf{u}}_{\bf{1}}}} \right)}_B},.....,{{\left( {{{\bf{u}}_p}} \right)}_B}} \right\}\) is linearly independent in \({\mathbb{R}^n}\)(Hint: Since the coordinate mapping is one-to-one, the following equations have the same solutions, \({c_{\bf{1}}}\),鈥�.,\({c_p}\))

\({c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p} = {\bf{0}}\) The zero vector V

\({\left( {{c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p}} \right)_B} = {\left( {\bf{0}} \right)_B}\) The zero vector in \({\mathbb{R}^n}\)a

Use Exercise 28 to explain why the equation\(Ax = b\)has a solution for all\({\rm{b}}\)in\({\mathbb{R}^m}\)if and only if the equation\({A^T}x = 0\)has only the trivial solution.

The null space of a \({\bf{5}} \times {\bf{6}}\) matrix A is 4-dimensional, what is the dimension of the column space of A.

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

14. Show that if \(Q\) is an invertible, then \({\mathop{\rm rank}\nolimits} AQ = {\mathop{\rm rank}\nolimits} A\). (Hint: Use Exercise 13 to study \({\mathop{\rm rank}\nolimits} {\left( {AQ} \right)^T}\).)

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

21. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 2}&{ - 4.2}&{ - 4.8}&{ - 3.6}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).
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