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Chapter 4: Q15E (page 191)

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

Short Answer

Expert verified

The smallest possible dimension of Null A is 2.

Step by step solution

01

Describe the given data

From the given \(6 \times 8\) matrix, the number of pivotscannot exceed 6. That is

\({\rm{rank}}\,A \le 6\).

02

Use the rank theorem

By the rank theorem, you get

\(\begin{aligned} n &= {\rm{rank}}\,A + \dim \,{\rm{Null}}\,A\\8 &\le 6 + \dim \,{\rm{Null}}\,A\\8 - 6 &\le \dim \,{\rm{Null}}\,A\\2 &\le \dim \,{\rm{Null}}\,A.\end{aligned}\)

03

Draw a conclusion

Hence, the smallest possible dimension of Null A is 2.

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Most popular questions from this chapter

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

  1. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\). (Hint: Explain why every vector in the column space of \(AB\) is in the column space of \(A\).
  2. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\). (Hint: Use part (a) to study rank\({\left( {AB} \right)^T}\).)

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

22. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 1}&{ - 13}&{ - 12.2}&{ - 1.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

In Exercise 17, Ais an \(m \times n\] matrix. Mark each statement True or False. Justify each answer.

17. a. The row space of A is the same as the column space of \({A^T}\].

b. If B is any echelon form of A, and if B has three nonzero rows, then the first three rows of A form a basis for Row A.

c. The dimensions of the row space and the column space of A are the same, even if Ais not square.

d. The sum of the dimensions of the row space and the null space of A equals the number of rows in A.

e. On a computer, row operations can change the apparent rank of a matrix.

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

Given vectors, \({u_{\bf{1}}}\),鈥�.,\({u_p}\) and w in V, show that w is a linear combination of \({u_{\bf{1}}}\),鈥�.,\({u_p}\) if and only if \({\left( w \right)_B}\) is a linear combination of vectors \({\left( {{{\bf{u}}_{\bf{1}}}} \right)_B}\),鈥�.,\({\left( {{{\bf{u}}_p}} \right)_B}\).

(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.)
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