Matrix representation of a linear operator given an arbitrary basis 
Prove or give a counterexample.
Let $L:\Bbb R^3 \to \Bbb R^3$ be a linear operator. For each basis $\alpha$ of $\Bbb R^3$ there always exists a basis $\beta$ of $\Bbb R^3$ such that the matrix $L_\alpha^\beta$ is given by
$$\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\\\end{pmatrix}$$

Let's analyse it a bit. We see that all elements of the $\alpha$ basis of $\Bbb R^3$ are going to be mapped to themselves except for $v_3$, which is mapped to zero (i.e. $v_3 \in \ker(L)$).
Let's work with the standard basis $\alpha=\{(1,0,0),(0,1,0),(0,0,1)\}$.
We of course have that $L(1,0,0)=(1,0,0), L(0,1,0)=(0,1,0)$ and $L(0,0,1)=(0,0,0)$. If we write these vectors as a linear combination of $\beta=\alpha$ we clearly get the stated matrix.
Let's work with another basis; $\alpha=\{(2,0,0),(0,5,0),(0,0,6)\}$.
We get $L(2,0,0)=(2,0,0), L(0,5,0)=(0,5,0)$ and $L(0,0,6)=(0,0,0)$. The linear combinations yield information on what $\beta$ equals to
$$\begin{aligned}(2,0,0)&= 1(2,0,0)+0(0,5,0)+0(a,b,c)\\(0,5,0)&=0(2,0,0)+1(0,5,0)+0(a,b,c)\\(0,0,0)&=0(2,0,0)+0(0,5,0)+0(a,b,c)\end{aligned}$$
So we indeed get the stated matrix if $\beta = \{(2,0,0), (0,5,0), (a,b,c)\}$, where $(a,b,c) \notin \operatorname{span}((2,0,0), (0,5,0))$.
So I'd say the statement is true. We pick $\alpha = \{v_1, v_2, v_3\}$, and there's always going to exist $\beta = \{v_1, v_2, c\}$ where $c \notin \operatorname{span}(v_1, v_2)$.
However, the above does not look like a proof to me. Is what I have done OK? If not, what am I missing?
EDIT
OK The statement is false. User1551 and TheSilverDoe proposed two counterexamples
$1)$ Take the zero map (i.e. $\ker(L)=\Bbb R^3$)
In this case the matrix representation of the linear map is of course
$$\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\\\end{pmatrix}$$
Which does not match the provided matrix.
$2)$ Take the identity map (i.e. $\ker(L)=\{0\}$)
In this case the matrix representation of the linear map (wrt the standard basis) is of course
$$\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\\\end{pmatrix}$$
Which does not match the provided matrix.
 A: $L_\alpha^\beta=\operatorname{diag}(1,1,0)$ iff $L$ has rank $2$. Hence every $L$ whose rank $\ne2$ serves as a counterexample.
Let $E$ be the standard basis of $\mathbb R^3$. Since every nonsingular matrix can be viewed as a basis transition matrix,
\begin{aligned}
\operatorname{rank}(L)=2
&\Leftrightarrow \operatorname{rank}(L_E^E)=2\\
&\Leftrightarrow L_E^E=U\operatorname{diag}(1,1,0)V
\ \text{ for some nonsingular $U$ and $V$}\\
&\Leftrightarrow L_E^E=I_\beta^E\operatorname{diag}(1,1,0)I_E^\alpha
\quad (U=I_\beta^E,\ V=I_E^\alpha)\\
&\Leftrightarrow \operatorname{diag}(1,1,0)=I_E^\beta L_E^E I_\alpha^E=L_\alpha^\beta.
\end{aligned}
Alternatively, suppose $L_\alpha^\beta=\operatorname{diag}(1,1,0)$. Let $\alpha=\{u,v,w\}$ and $\beta=\{x,y,z\}$. Then $L(u)=x,\,L(v)=y$ and $L(w)=0$. Hence $L(\mathbb R^3)=\operatorname{span}\{x,y\}$ and $\operatorname{rank}(L)=2$.
Conversely, suppose $\operatorname{rank}(L)=2$. Then $\dim L(\mathbb R^3)=2$ and $\dim\ker(L)=1$. Pick a basis $\{x,y\}$ of $L(\mathbb R^3)$ and pick a nonzero vector $w\in\ker(L)$. Since $x$ and $y$ are two linearly independent images of $L$, we have $x=L(u)$ and $y=L(v)$ for some linearly independent vectors $u$ and $v$. Hence $\alpha=\{u,v,w\}$ is linearly independent and it is a basis of $\mathbb R^3$. Now complete $\{x,y\}$ to a basis $\beta=\{x,y,z\}$ of $\mathbb R^3$. Then $L_\alpha^\beta=\operatorname{diag}(1,1,0)$.
