Why not just reduce the matrix ? its not more work and often makes things easier.
\begin{bmatrix}
1&0&0&0
\\0&1&0&0
\\0&0&1&0
\end{bmatrix}
1) We see that the row and column rank (they are always equal) are $3$.
The nullity $=n-$rank$=4-3=1$.
To find the null space we have to solve $A\mathbf{x}=0$, and this is easy now in row reduced form $\mathbf{x}=\begin{bmatrix}
0\\0\\0\\a
\end{bmatrix}$.
2) The row space has dimension $3$ as mentioned, for the basis one can take:
$$(1,0,0,0)$$
$$(0,1,0,0)$$
$$(0,0,1,0)$$
Or one could take the rows of the original matrix, since the rank is $3$.
3) The column rank is also $3$ row reduction has not changed the column vectors, just expressed them in a different basis so a basis for the column space will be the first $3$ vectors of the original matrix (corresponding to the pivot position of the reduced matrix):
$$\begin{bmatrix}
1\\3\\-1
\end{bmatrix},
\begin{bmatrix}
1\\-1\\2
\end{bmatrix},
\begin{bmatrix}
-1\\2\\-4\\
\end{bmatrix}. $$
HINT
To start with, notice the columns are proportional. So that $\text{rank}(A) = 1$.
In order to find out a basis for the kernel, you have to solve the system of linear equations:
\begin{align*}
\begin{bmatrix}
a & b\\
a & b
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix} =
\begin{bmatrix}
0\\
0
\end{bmatrix} & \Longleftrightarrow ax + by = 0 \Longleftrightarrow x = -\frac{by}{a}
\end{align*}
Hence we conclude that $\ker(A) = \text{span}\left\{\left(-\dfrac{b}{a},1\right)\right\}$, whose dimension equals one.
Similarly, in order to find a basis for the row space, notice the rows are LD.
Consequently, we conclude the row space equals $\text{span}\left\{\left(1,\dfrac{b}{a}\right)\right\}$.
Based on such results, can you take it from here?
Best Answer
Let $q_i$ be a null vector of $A A^\top$, i.e. $ A A^\top q_i =0 $, then $ 0 = q_i^\top A A^\top q_i = \vert\vert A^\top q_i \vert\vert_2$, and thus $q_i$ is also a null vector of $A^\top$. Thus $\mathrm{rank}(A A^\top) = \mathrm{rank}(A^\top) = \mathrm{rank}(A)$.