031 Review Part 2, Problem 1

Consider the matrix  $A={\begin{bmatrix}1&-4&9&-7\\-1&2&-4&1\\5&-6&10&7\end{bmatrix}}$  and assume that it is row equivalent to the matrix

$B={\begin{bmatrix}1&0&-1&5\\0&-2&5&-6\\0&0&0&0\end{bmatrix}}.$

(a) List rank  $A$  and  ${\text{dim Nul }}A.$

(b) Find bases for  ${\text{Col }}A$  and  ${\text{Nul }}A.$  Find an example of a nonzero vector that belongs to  ${\text{Col }}A,$  as well as an example of a nonzero vector that belongs to  ${\text{Nul }}A.$

Foundations:
1. For a matrix  $A,$  the rank of  $A$  is
${\text{rank }}A={\text{dim Col }}A.$
2.  ${\text{Col }}A$  is the vector space spanned by the columns of  $A.$
3.  ${\text{Nul }}A$  is the vector space containing all solutions to  $Ax=0.$

Solution:

(a)

Step 1:
From the matrix  $B,$  we see that  $A$  contains two pivots.
Therefore,
${\begin{array}{rcl}\displaystyle {{\text{rank }}A}&=&\displaystyle {{\text{dim Col }}A}\\&&\\&=&\displaystyle {2.}\end{array}}$

Step 2:
By the Rank Theorem, we have
${\begin{array}{rcl}\displaystyle {4}&=&\displaystyle {{\text{rank }}A+{\text{dim Nul }}A}\\&&\\&=&\displaystyle {2+{\text{dim Nul }}A.}\end{array}}$
Hence,  ${\text{dim Nul }}A=2.$

(b)

Step 1:
From the matrix  $B,$  we see that  $A$  contains pivots in Column 1 and 2.
So, to obtain a basis for  ${\text{Col }}A,$  we select the corresponding columns from  $A.$
Hence, a basis for  ${\text{Col }}A$  is
${\Bigg \{}{\begin{bmatrix}1\\-1\\5\end{bmatrix}},{\begin{bmatrix}-4\\2\\-6\end{bmatrix}}{\Bigg \}}.$

Step 2:
To find a basis for  ${\text{Nul }}A$  we translate the matrix equation  $Bx=0$  back into a system of equations
and solve for the pivot variables.
Hence, we have
${\begin{array}{rcl}\displaystyle {x_{1}-x_{3}+5x_{4}}&=&\displaystyle {0}\\&&\\\displaystyle {-2x_{2}+5x_{3}-6x_{4}}&=&\displaystyle {0.}\end{array}}$
Solving for the pivot variables, we have
${\begin{array}{rcl}\displaystyle {x_{1}}&=&\displaystyle {x_{3}-5x_{4}}\\&&\\\displaystyle {x_{2}}&=&\displaystyle {{\frac {5}{2}}x_{3}-3x_{4}.}\end{array}}$
Hence, the solutions to  $Ax=0$  are of the form
$x_{3}{\begin{bmatrix}1\\{\frac {5}{2}}\\1\\0\end{bmatrix}}+x_{4}{\begin{bmatrix}-5\\-3\\0\\1\end{bmatrix}}.$
Therefore, a basis for  ${\text{Nul }}A$  is
${\Bigg \{}{\begin{bmatrix}1\\{\frac {5}{2}}\\1\\0\end{bmatrix}},{\begin{bmatrix}-5\\-3\\0\\1\end{bmatrix}}{\Bigg \}}.$

Final Answer:
(a)     ${\text{rank }}A=2$  and  ${\text{dim Nul }}A=2$
(b)     A basis for  ${\text{Col }}A$  is  ${\Bigg \{}{\begin{bmatrix}1\\-1\\5\end{bmatrix}},{\begin{bmatrix}-4\\2\\-6\end{bmatrix}}{\Bigg \}}$
and a basis for  ${\text{Nul }}A$  is  ${\Bigg \{}{\begin{bmatrix}1\\{\frac {5}{2}}\\1\\0\end{bmatrix}},{\begin{bmatrix}-5\\-3\\0\\1\end{bmatrix}}{\Bigg \}}.$

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