I'm having trouble with a linear algebra problem that asks me to find a basis for the column space and null space of the matrix A:
$$ A=\left[\begin{array}{cccc} 2 & 4 & 5 & -1 \\ 1 & 2 & 1 & 1 \\ 3 & 6 & 0 & 6 \end{array}\right] $$
Here's what I've tried so far:
To find the column space of A, I reduced the matrix to echelon form using row operations and found that the first three columns are linearly independent, so they form a basis for the column space.
However, I'm not sure how to find the null space. I know that the null space is the set of all solutions to the equation Ax = 0, but I'm not sure how to find those solutions.
If anyone could offer some guidance or help me understand how to approach this problem, I would really appreciate it.
Thanks in advance!
First put the matrix in reduced-row echelon form:
$\begin{bmatrix} 2&4&5&-1 \\ 1&2&1&1 \\3&6&0&6 \end{bmatrix} \overrightarrow{\substack{r_3^*=\frac{1}{6}(r_3-r_1-r_2)\\r_2^*=\frac{1}{3}(2r_2-r_1)}}\begin{bmatrix} 2&4&5&-1 \\ 0&0&-1&1 \\0&0&-1&1 \end{bmatrix}\overrightarrow{\substack{r_3^*=r_3-r_2\\r_1^*=\frac{1}{2}(r_1+5r_2)\\r_2^*=-r_2}}\begin{bmatrix} 1&2&0&2 \\ 0&0&1&-1 \\0&0&0&0 \end{bmatrix}$
Pivot columns are those whose first non-zero entry has no non-zero entries to it's left. Here there are two pivots--one in column $1$ and one in column $3$. So the first and third columns of $A$ form a basis for the column space: $C(A)=\text{span}\left\{\begin{pmatrix} 2\\1\\3\end{pmatrix}, \begin{pmatrix} 5\\1\\0\end{pmatrix}\right\}$.
A general method for finding the null space is to go one-by-one and assign $1$ to the first free variable and zero to all other free variables to get a first basis vector, then assign $1$ to the second free variable and zero to all the others to get the second basis vector, etc.
Here, there are $4$ total columns and $2$ pivot columns in columns $1$ and $3$. This leave columns $2$ and $4$ as free columns. So we follow the method described above. In the first case ($x_4=1, x_2=0$), we want the solutions to $\begin{bmatrix} 1&2&0&2 \\ 0&0&1&-1 \\0&0&0&0 \end{bmatrix}\begin{bmatrix} x_1\\0\\x_3\\1\end{bmatrix}=\begin{bmatrix} 0\\0\\0\end{bmatrix}$. By the rules of matrix multiplication, we get $x_3=1$ from the second row $([0]x_1+[0]0+[1]x_3+[-1]1=0)$, and thus $x_1=-2$ from the first row $([1]x_1+[2]0+[0]1+[2]1=0)$; a basis vector for the null space is $(-2,0,1,1)^t$.
In the second case ($x_4=0, x_2=1$), we want the solutions to $\begin{bmatrix} 1&2&0&2 \\ 0&0&1&-1 \\0&0&0&0 \end{bmatrix}\begin{bmatrix} x_1\\1\\x_3\\0\end{bmatrix}=\begin{bmatrix} 0\\0\\0\end{bmatrix}$. By the rules of matrix multiplication, we get $x_3=0$ from the second row $([0]x_1+[0]1+[1]x_3+[-1]0=0)$, and thus $x_1=-2$ from the first row $([1]x_1+[2]1+[0]0+[2]0=0)$; a second basis vector for the null space is $(-2,1,0,0)^t$.
So the null space is $N(A)=\text{span}\left\{\begin{pmatrix} -2\\0\\1\\1\end{pmatrix}, \begin{pmatrix} -2\\1\\0\\0\end{pmatrix}\right\}$