031 Review Part 2, Problem 10

(a) Suppose a  $6\times 8$  matrix  $A$  has 4 pivot columns. What is  ${\text{dim Nul }}A?$  Is  ${\text{Col }}A=\mathbb {R} ^{4}?$  Why or why not?

(b) If  $A$  is a  $7\times 5$  matrix, what is the smallest possible dimension of  ${\text{Nul }}A?$

Foundations:
1. The dimension of  ${\text{Col }}A$  is equal to the number of pivots in  $A.$
2. By the Rank Theorem, if  $A$  is a  $m\times n$  matrix, then
${\text{rank }}A+{\text{dim Col }}A=n.$

Solution:

(a)

Step 1:
Since  $A$  has 4 pivot columns,
${\text{dim Col }}A=4.$

Step 2:
Since  $A$  is a  $6\times 8$  matrix,  ${\text{Col }}A$  contains vectors in  $\mathbb {R} ^{6}.$
Since a vector in  $\mathbb {R} ^{6}$  is not a vector in  $\mathbb {R} ^{4},$  we have
${\text{Col }}A\neq \mathbb {R} ^{4}.$

(b)

Step 1:

Step 2:

Final Answer:
(a)     ${\text{dim Col }}A=4$  and  ${\text{Col }}A\neq \mathbb {R} ^{4}$
(b)

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