Determining row space of a $n$ by $n$ matrix (Linear Algebra) Prove the following statements or provide a counterexample if it is false
For two $n$ by $n$ matrices $A$ and $U$  the row space of $UA$ is contained in the row space of $A$
What i tried
I started by trying out some simple 2 by 2 matrices to determine whether this statement is true of false. i took the matrices 
$
A=\begin{pmatrix}
1 &  1\\
0  &  0\\
\end{pmatrix}
$ and $
U=\begin{pmatrix}
1 &  0\\
1  &  0\\
\end{pmatrix}
$ Multiplying both statements together, i got   $
\begin{pmatrix}
2 &  0\\
0  &  0\\
\end{pmatrix}
$ And i observe that the row space of $A$ and $UA$ are the same for this example. And thus the statement is true for this example.(If the statement is false, the the above example would immediately provide a counterexample) Thus i began trying to prove the general case. First i notice that the row space of the matrix $A$ if is $1$ for example which means that the row containing the row space(The pivot row) cannot be expressed as a linear combination of the other rows, hence linearly independent. So by doing a matrix multiplication, essentially multiplying the pivot row by a scalar, the pivot row will still remain linearly independent. Hence the row space is preserved in the product $UA$. Am i correct, could anyone explain. Thanks  

The elements of the row space of the $m\times n$ matrix $B$ can be obtained by computing all products
$$
\begin{bmatrix}x_1 & x_2 & \dots & x_m\end{bmatrix}B
$$
where $x=\begin{bmatrix}x_1 & x_2 & \dots & x_m\end{bmatrix}$ is a row with $m$ columns.
So, if $r$ is an element of the row space of $UA$, there is a row $x$ such that $r=xUA$. In particular you see that
$$
r=(xU)A
$$
belongs to the row space of $A$.
It doesn't necessarily belong to the row space of $U$, though. Consider, for instance,
$$
U=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix},\qquad
A=\begin{bmatrix}1 & 2 \\ 0 & 0\end{bmatrix}
$$
Then
$$
UA=\begin{bmatrix}1 & 2\\ 0 & 0\end{bmatrix}
$$
The first row of $U$ doesn't belong to the row space of $UA$.
