Let $F_2$ denote the field of scalars {0,1} with operations done using mod 2 arithmetic.
Consider the matrix:
$$\begin{bmatrix} 1&1&1&0 \\ 1&1&0&1 \\ 0&0&1&1 \end{bmatrix} \in M_{3,4}(F_2)$$
(a) Determine a basis for the column space of A, and list all the elements in the column space.
(b) Determine a basis for the row space of A, and list all the elements in the row space.
(c) Determine a basis for the solution space of A, and list all the elements in the solution space.
From my understanding, the first step is calculating the RREF, here is the RREF:
$$\begin{bmatrix}1&1&0&1\\0&0&1&1\\0&0&0&0\end{bmatrix}$$
Part a:
Leading entries are in column 1, 3 therefore the column space has a basis $\{(1,1,0), (1,0,1)\}$.
Hence the span is: $\{\alpha_1(1,1,0) + \alpha_2(1,0,1)\}$
There are four points in the space:
$\alpha_1 = \alpha_2 = 0 : (0,0,0)$
$\alpha_1 = 1, \alpha_2 = 0 : (1,1,0)$
$\alpha_1 = 0,\alpha_2 = 1 : (1,0,1)$
$\alpha_1 = 1, \alpha_2 = 1 : (0,1,1)$
Part b:
Basis of row space of A is non zero rows in RREF, therefore the basis is $\{(1,1,1,0), (1,1,0,1)\}$. Hence, the span is : $\{\alpha_1(1,1,1,0) + \alpha_2(1,1,0,1)\}$
There are four points in the space:
$\alpha_1 = \alpha_2 = 0 : (0,0,0)$
$\alpha_1 = 1, \alpha_2 = 0 : (1,1,1,0)$
$\alpha_1 = 0,\alpha_2 = 1 : (1,1,0,1)$
$\alpha_1 = 1, \alpha_2 = 1 : (0,0,1,1)$
Part c:
A$\begin{bmatrix}x_1 \\ x_2 \\x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$
$\begin{bmatrix}1&1&0&1\\0&0&1&1\\0&0&0&0\end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$
$x_2 = s, x_4 = t$
$x_1+x_2+x_4=0 => x_1 +s+t+s+t = s+t => x_1 = s+t$
$x_3 + t = 0 => x_3 +t+t= t => x_3=t$
Can someone please verify this solution?
Best Answer
Yes, it's all correct.
You still need to list the elements of the solution space: just substitute all 4 possible $(s,t)$ pairs in your solution.