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
${\text{dim Nul }}A+{\text{dim Col }}A=n.$

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:
By the Rank Theorem, we have
${\text{dim Nul }}A+{\text{dim Col }}A=5.$
Thus,
${\text{dim Nul }}A=5-{\text{dim Col}}A.$

Step 2:
If we want to minimize  ${\text{dim Nul }}A,$  we need to maximize  ${\text{dim Col }}A.$
To do this, we need to find out the maximum number of pivots in  $A.$
Since  $A$  is a  $7\times 5$  matrix, the maximum number of pivots in  $A$  is 5.
Hence, the smallest possible value for  ${\text{dim Nul }}A$  is
${\begin{array}{rcl}\displaystyle {{\text{dim Nul }}A}&=&\displaystyle {5-{\text{dim Col }}A}\\&&\\&\geq &\displaystyle {5-5}\\&&\\&\geq &\displaystyle {0.}\end{array}}$

Final Answer:
(a)     ${\text{dim Col }}A=4$  and  ${\text{Col }}A\neq \mathbb {R} ^{4}$
(b)     $0$

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