# Column, Row and solution space of a matrix

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?

• @Bernard the operations are only in mod 2 arithmetic, thus the values can only be 1 or 0. – jessica Oct 5 '19 at 10:10
• Sorry, I didn't see that. I'll remove my comment. – Bernard Oct 5 '19 at 10:12

You still need to list the elements of the solution space: just substitute all 4 possible $$(s,t)$$ pairs in your solution.