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I need help computing $\mathrm{null}(A)$ and $\mathrm{col}(A)$

$$A = \begin{bmatrix} 1 & 2 & 1 & 4 \\ 1 & 1 & 1 & 3 \\ -1 & 1 & -1 & \ -1 \end{bmatrix}$$

I think I did $\mathrm{null}(A)$ correctly. I found the row reduced echelon form and found that $x_{1} = -x_{3} - 2x_{4}$ and that $x_{2} = -x_{4}$. I also found that $x_3, x_4$ are free.

So my basis for $\mathrm{null}(A)$ was

$$\left\{\begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \\ \end{bmatrix}, \begin{bmatrix} -2 \\ -1 \\ 0 \\ 1 \\ \end{bmatrix}\right\}.$$

I do not know how to proceed.

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    You calculated the null-space correctly. – 2017-02-20
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    ok. Thanks. I do not know how to get Col A . another question I have is what's the difference between this and span? – 2017-02-20
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    I am not sure what Col A is. Is it the image of the transpose of $A$ ? – 2017-02-20
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    The span is the vector space spanned by the row-vectors. It has dimension $2$, since the kernel has dimension $2$. (The sum of the dimensions has to be $4$ because the row vectors have $4$ components) – 2017-02-20
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    how can i find col A ? (basis of column space) – 2017-02-20
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    Show that the third and the fourth column are linear combinations of the first and second column. The first two columns form a basis for the column-space. – 2017-02-20
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    i do not understand i tried reading online but can not find anything about what the final answer should look like – 2017-02-20

1 Answers 1

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For the column space: note that the reduced row-echelon form of $A$ is given by $$\mathrm{RREF}(A)=\begin{bmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}.$$

To find the column space of $A$, we want to look at the columns in $A$ which have leading $1$'s in $\mathrm{RREF}(A)$. Since columns one and two have leading $1$'s, then that tells us that $$\left\{\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}\right\}$$ forms a basis for $\mathrm{col}(A)$.

Moreover, the entries in columns three and four tell us exactly how to write said columns as a linear combination of the basis:

$$\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}=1\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}+0\begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix},$$ $$\begin{bmatrix} 4 \\ 3 \\ -1 \end{bmatrix}=2\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}+1\begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}.$$ Thus, columns three and four are indeed in $$\mathrm{span}\left\{\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}\right\},$$ and $\left\{\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}\right\}$ is a linearly independent set (if not, $\mathrm{RREF}(A)$ would not have a leading $1$ in both of the first two columns).