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Chapter 4: Q30E (page 191)

Suppose \(A\) is \(m \times n\)and \(b\) is in \({\mathbb{R}^m}\). What has to be true about the two numbers rank \(\left[ {A\,\,\,{\rm{b}}} \right]\) and \({\rm{rank}}\,A\) in order for the equation \(Ax = b\) to be consistent?

Short Answer

Expert verified

The condition \({\rm{rank}}\,\left[ {A\,\,\,{\rm{b}}} \right] = {\rm{rank}}\,A\) must be true about the two numbers rank \(\left[ {A\,\,\,{\rm{b}}} \right]\) and \({\rm{rank}}\,A\) in order for the equation \(Ax = b\) to be consistent.

Step by step solution

01

Assume an arbitrary system and relate the given statement with it

Consider the nonhomogeneous system \(Ax = b\) with the joint matrix \(\left[ {A\,\,\,{\rm{b}}} \right]\). If \(b\) is not a pivot column of the matrix \(\left[ {A\,\,\,{\rm{b}}} \right]\), then the rank of matrix \(A\) and the joint matrix \(\left[ {A\,\,\,{\rm{b}}} \right]\) are the same, that is, \({\rm{rank}}\,\left[ {A\,\,\,{\rm{b}}} \right] = {\rm{rank}}\,A\).

02

Use the existence and uniqueness theorem

By the existence and uniqueness theorem, a linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot. So, for the system to be consistent, \(b\) must not be a pivot column of the matrix \(\left[ {A\,\,\,{\rm{b}}} \right]\).

03

Draw a conclusion

If \(b\) is not a pivot column of the matrix \(\left[ {A\,\,\,{\rm{b}}} \right]\), then \({\rm{rank}}\,\left[ {A\,\,\,{\rm{b}}} \right] = {\rm{rank}}\,A\). So, the condition \({\rm{rank}}\,\left[ {A\,\,\,{\rm{b}}} \right] = {\rm{rank}}\,A\) is true about the two numbers rank \(\left[ {A\,\,\,{\rm{b}}} \right]\) and \({\rm{rank}}\,A\) in order for the equation \(Ax = b\) to be consistent.

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Most popular questions from this chapter

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.

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, \({c_{\bf{1}}}\),….,\({c_p}\))

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

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 the coordinate mapping is one to one. (Hint: Suppose \({\left( {\bf{u}} \right)_B} = {\left( {\bf{w}} \right)_B}\) for some u and w in V, and show that \({\bf{u}} = {\bf{w}}\)).

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

13. Show that if \(P\) is an invertible \(m \times m\) matrix, then rank\(PA\)=rank\(A\).(Hint: Apply Exercise12 to \(PA\) and \({P^{ - 1}}\left( {PA} \right)\).)

Given \(T:V \to W\) as in Exercise 35, and given a subspace \(Z\) of \(W\), let \(U\) be the set of all \({\mathop{\rm x}\nolimits} \) in \(V\) such that \(T\left( {\mathop{\rm x}\nolimits} \right)\) is in \(Z\). Show that \(U\) is a subspace of \(V\).
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