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Suppose a \({\bf{5}} \times {\bf{6}}\) matrix A has four pivot columns. What is dim Nul A? Is \({\bf{Col}}\,A = {\mathbb{R}^{\bf{3}}}\)? Why or why not?

Short Answer

Expert verified

dim Nul A=2, \({\rm{Col}}\,A \ne {\mathbb{R}^4}\)

Step by step solution

01

Find dim Nul A

A has four pivot columns. By using therank theorem,you get:

\(\begin{aligned}{c}{\rm{rank}}\,A + \dim \,{\rm{Nul}}\,A &= 6\\4 + \dim \,{\rm{Nul}}\,A &= 6\\\dim \,{\rm{Nul}}\,A &= 2\end{aligned}\)

02

Check for Nul A

The dim Col A =4, but Col A is a subspace of \({\mathbb{R}^5}\). Therefore, \({\rm{Col}}\,A \ne {\mathbb{R}^4}\),

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Most popular questions from this chapter

Consider the polynomials \({{\bf{p}}_{\bf{1}}}\left( t \right) = {\bf{1}} + t\), \({{\bf{p}}_{\bf{2}}}\left( t \right) = {\bf{1}} - t\), \({{\bf{p}}_{\bf{3}}}\left( t \right) = {\bf{4}}\), \({{\bf{p}}_{\bf{4}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}}\), and \({{\bf{p}}_{\bf{5}}}\left( t \right) = {\bf{1}} + {\bf{2}}t + {t^{\bf{2}}}\), and let H be the subspace of \({P_{\bf{5}}}\) spanned by the set \(S = \left\{ {{{\bf{p}}_{\bf{1}}},\,{{\bf{p}}_{\bf{2}}},\;{{\bf{p}}_{\bf{3}}},\,{{\bf{p}}_{\bf{4}}},\,{{\bf{p}}_{\bf{5}}}} \right\}\). Use the method described in the proof of the Spanning Set Theorem (Section 4.3) to produce a basis for H. (Explain how to select appropriate members of S.)

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation.

a. What is the dimension of range of T if T is one-to-one mapping? Explain.

b. What is the dimension of the kernel of T (see section 4.2) if T maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^m}\)? Explain.

(M) Determine whether w is in the column space of \(A\), the null space of \(A\), or both, where

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\2\\1\\0\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}{ - 8}&5&{ - 2}&0\\{ - 5}&2&1&{ - 2}\\{10}&{ - 8}&6&{ - 3}\\3&{ - 2}&1&0\end{array}} \right)\)

If A is a \({\bf{6}} \times {\bf{8}}\) matrix, what is the smallest possible dimension of Null A?

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.)

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