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Prove theorem 3 as follows: Given an \(m \times n\) matrix A, an element in \({\mathop{\rm Col}\nolimits} A\) has the form \(Ax\) for some x in \({\mathbb{R}^n}\). Let \(Ax\) and \(A{\mathop{\rm w}\nolimits} \) represent any two vectors in \({\mathop{\rm Col}\nolimits} A\).

  1. Explain why the zero vector is in \({\mathop{\rm Col}\nolimits} A\).
  2. Show that the vector \(A{\mathop{\rm x}\nolimits} + A{\mathop{\rm w}\nolimits} \) is in \({\mathop{\rm Col}\nolimits} A\).
  3. Given a scalar \(c\), show that \(c\left( {A{\mathop{\rm x}\nolimits} } \right)\) is in \({\mathop{\rm Col}\nolimits} A\).

Short Answer

Expert verified

a. The zero vector is in \({\mathop{\rm Col}\nolimits} A\).

b. It is proved that the vector \(A{\mathop{\rm x}\nolimits} + A{\mathop{\rm w}\nolimits} \) is in \({\mathop{\rm Col}\nolimits} A\).

c. It is proved that \(c\left( {A{\mathop{\rm x}\nolimits} } \right)\) is in \({\mathop{\rm Col}\nolimits} A\)

Step by step solution

01

Explain that the zero vector is in \({\mathop{\rm Col}\nolimits} A\)

a)

It is given that in an \(m \times n\) matrix, an element in \({\mathop{\rm Col}\nolimits} A\) has the form \(A{\mathop{\rm x}\nolimits} \) for some x in \({\mathbb{R}^n}\).

Thezero vector is in \({\mathop{\rm Col}\nolimits} A\) because of the equation \(A0 = 0\).

Thus, the zero vector is in \({\mathop{\rm Col}\nolimits} A\).

02

Show that the vector \(A{\mathop{\rm x}\nolimits}  + A{\mathop{\rm

b)

It is easy to find vectors in \({\mathop{\rm Col}\nolimits} A\). Thecolumnsof A are displayed; others are formed from them.

The vector \(A{\mathop{\rm x}\nolimits} + A{\mathop{\rm w}\nolimits} \) is in \({\mathop{\rm Col}\nolimits} A\) because\(A{\mathop{\rm x}\nolimits} + A{\mathop{\rm w}\nolimits} = A\left( {{\mathop{\rm x}\nolimits} + {\mathop{\rm w}\nolimits} } \right)\).

Thus, it is proved that the vector \(A{\mathop{\rm x}\nolimits} + A{\mathop{\rm w}\nolimits} \) is in \({\mathop{\rm Col}\nolimits} A\).

03

Show that \(c\left( {A{\mathop{\rm x}\nolimits} } \right)\) is in \({\mathop{\rm Col}\nolimits} A\)

c)

The vector \(c\left( {A{\mathop{\rm x}\nolimits} } \right)\) is in \({\mathop{\rm Col}\nolimits} A\) because \(c\left( {A{\mathop{\rm x}\nolimits} } \right) = A\left( {c{\mathop{\rm x}\nolimits} } \right)\).

Thus, it is proved that \(c\left( {A{\mathop{\rm x}\nolimits} } \right)\) is in \({\mathop{\rm Col}\nolimits} A\).

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Most popular questions from this chapter

Is it possible for a nonhomogeneous system of seven equations in six unknowns to have a unique solution for some right-hand side of constants? Is it possible for such a system to have a unique solution for every right-hand side? Explain.

If the null space of A \({\bf{7}} \times {\bf{6}}\) 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 \(m \times n\).

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.

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

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

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 3}\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1\\{ - 5}&{ - 1}&0&{ - 2}\\9&{ - 11}&7&{ - 3}\\{19}&{ - 9}&7&1\end{array}} \right)\)

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