Finding a specific base for matrices space I need to find a base that includes only matrices with Rank=1 for the following space: $$V=\left \{  \right.(\begin{smallmatrix}
x-y &2x+3y+3z \\ 
-14x-7y-21z & -8x+8y
\end{smallmatrix}\bigr):x,y,z\in\mathbb{R}\left.  \right \}$$
What I did is:
1) Extracted $x,y,x$ to find the $span$: $$V=sp\left \{ x\begin{pmatrix}
1 &2 \\ 
 -14&8 
\end{pmatrix},y\begin{pmatrix}
-1 &1 \\ 
-7 &8 
\end{pmatrix},z\begin{pmatrix}
0 &1 \\ 
-7 &0 
\end{pmatrix} \right \}$$
2) Find the base of this $span$ using Gauss-Elimination: $$\Rightarrow V=sp\left \{ \bigl(\begin{smallmatrix}
1 &0 \\ 
0 &-8 
\end{smallmatrix}\bigr),\bigl(\begin{smallmatrix}
0 &1 \\ 
-7 &0 
\end{smallmatrix}\bigr) \right \}$$
The base I found includes matrices with Rank=2 therefore doesn't comply with the question terms.
I would be really glad to hear some insights about this one!
 A: A good start is to see which matrices out of
$$
e_1=\begin{pmatrix}
1&0\\0&0
\end{pmatrix},~
e_2=\begin{pmatrix}
0&1\\0&0
\end{pmatrix},~
e_3=\begin{pmatrix}
0&0\\
1&0
\end{pmatrix}
\text{ and }
e_4=\begin{pmatrix}
0&0\\0&1
\end{pmatrix}
$$
are contained in $V$.
You can directly check that $e_1,e_4\notin V$. But we still get 
$$
e_2=\begin{pmatrix}
x-y &2x+3y+3z \\ 
-14x-7y-21z & -8x+8y
\end{pmatrix}
\text{ where }x=y=-z=\frac12
$$
and
$$
e_3=\begin{pmatrix}
x-y &2x+3y+3z \\ 
-14x-7y-21z & -8x+8y
\end{pmatrix}
\text{ where }x=y=-\frac35z=\frac1{14}
$$
From your first step, we deduce $\dim V=3$ and the last one is a little difficult to find. We observe, that a matrix with rank $1$ has to have the form 
$$
\begin{pmatrix}a & b\\\lambda a & \lambda b\end{pmatrix}.
$$
for some $a,b,\lambda\in\mathbb R$. Now we compare and get
$$
\begin{pmatrix}a & b\\\lambda a & \lambda b\end{pmatrix}
=\begin{pmatrix}
x-y &2x+3y+3z \\ 
-14x-7y-21z & -8x+8y
\end{pmatrix}
\Leftrightarrow \begin{matrix}(i)~ a=x-y\\(ii)~ b=2x+3y+3z\\(iii)~\lambda a=-14x-7y-21z\\(iv)~\lambda b=-8x+8y\end{matrix}
$$
Here starts the tricky part. If $b=0$, we can use $(i)$ and $(iv)$ to deduce that $a=0$ and we get the null matrix, which is bad. But for $b\neq 0$, we get $\lambda=-8\frac{a}b$ and consider
$$
\begin{pmatrix}a & b\\-8\frac{a^2}b & -8a\end{pmatrix}
=\begin{pmatrix}
x-y &2x+3y+3z \\ 
-14x-7y-21z & -8x+8y
\end{pmatrix}\\
\Leftrightarrow\begin{pmatrix}x-y\\2x+2y+3z\\-14x-7y-21z\\-8x+8y\end{pmatrix}=\begin{pmatrix}a\\b\\-8\frac{a^2}b\\-8a\end{pmatrix}\\
\Leftrightarrow \begin{pmatrix}1 & -1 & 0\\2 & 3 & 3\\-14& -7 & -21\end{pmatrix}\cdot\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}a\\b\\-8\frac{a^2}b\end{pmatrix}
$$
You see that this linear equation has always a solution. To simplify it, we choose $a=b=1$ and solve the system to get
$$
\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}
 1 &1/2 &1/14  \\
 0&1/2&1/14\\  
-2/3&-1/2&-5/42 
\end{pmatrix}\cdot\begin{pmatrix}1\\1\\-8\end{pmatrix}=\begin{pmatrix}13/14\\-1/14\\-3/14\end{pmatrix}
$$
Next, we get
$$
f:=\begin{pmatrix}1&1\\-8&-8\end{pmatrix}=\begin{pmatrix}
x-y &2x+3y+3z \\ 
-14x-7y-21z & -8x+8y
\end{pmatrix}
$$
where $x=\frac{13}{14}$,$y=-\frac1{14}$ and $z=-\frac3{14}$. Finally, we see that $\{e_2,e_3,f\}\subseteq V$ is a linearly independent set of matrices with rank $1$ and the base of $V$, since $\dim V=3$.
A: First off, you’ve got some errors in your initial decomposition of the defining equation for $V$. The space actually consists of all linear combinations $$x\begin{bmatrix}1&2\\-14&-8\end{bmatrix}+y\begin{bmatrix}-1&3\\-7&8\end{bmatrix} + z\begin{bmatrix}0&3\\-21&0\end{bmatrix}.$$ You can use any of the usual methods to find a simpler basis for $V$. From one such basis we find that $V$ consists of matrices of the form $$\alpha \begin{bmatrix}1&0\\0&-8\end{bmatrix} + \beta \begin{bmatrix}0&1\\0&0\end{bmatrix} + \gamma \begin{bmatrix} 0&0\\1&0 \end{bmatrix} = \begin{bmatrix} \alpha & \beta \\ \gamma & -8\alpha \end{bmatrix}.\tag{*}$$ Now, a rank-one $2\times2$ matrix is a non-zero matrix with zero determinant, so you need to find a linearly-independent set of three matrices of the form given by (*) for which $8\alpha^2+\beta\gamma=0$, with at least one of these parameters non-zero. You can do this by inspection: setting $\alpha=0$ and one of $\beta$, $\gamma$ to zero as well gives two such matrices, and for the third, you can take any non-zero values with different signs for $\beta$ and $\gamma$, and $\pm\sqrt{-\beta\gamma/8}$ for $\alpha$. A simple example of such a set is $$\left\{ \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\1&0\end{bmatrix}, \begin{bmatrix}1&-1\\8&-8\end{bmatrix} \right\}.$$ Another approach, once you have the basis in (*), is to recall that in a rank-one matrix, all the columns are scalars multiples of each other (as are the rows). The second two matrices in the basis—the standard basis elements $E_{12}$ and $E_{21}$—already satisfy this condition, but the first one doesn’t. However, you can add multiples of $E_{12}$ and $E_{21}$ to it to set the off-diagonal elements to any desired value without disturbing the diagonal elements, and the resulting matrix will also be linearly independent of $E_{12}$ and $E_{21}$. Taking $\beta=1$ and $\gamma=8$ produces $\tiny{\begin{bmatrix}1&-1\\8&-8\end{bmatrix}}$, which obviously has rank one.
