Define a linear operator which has as Kernel the line $y=-x$ and as image the line $y=$x So the question asks basically to determine a linear operator $F: \mathcal{R}^2 \rightarrow \mathcal{R}^2$ which has as Kernel the line $y=-x$ and as image the line $y=x$.
Here is what i tried: i supposed i could write the operator in the form $F(x,y)=(ax+by,cx+dy)$. Thus
$Ker(F)$={$(x,y) \in \mathcal{R}^2 : (ax+by,cx+dy)=(0,0)$}
and i defined a=b=1 and c=d=0, so that i would have
$Ker(F)$={$(x,y) \in \mathcal{R}^2 : (x+y)=(0,0)$}
and the solution is x=-y if i'm not wrong. But for the image with these same guesses i got
$Im(F)$={$(x+y,0) : (x,y) \in \mathcal{R}^2$}
={$(x+y)(1,0) : (x,y) \in \mathcal{R}^2$}
But idk if i can conclude that the image is y=x. I don't have arguments to support it or i'm just wrong in my trial. Any help would be great, thank you
 A: Let $f: \mathbb{R}^{2} \to \mathbb{R}^{2}$ such that if $(x,y) \in \mathbb{R}^{2}$ so $\ker(f)=\text{ span}\{(-1,1)\}$ and $\text{im}(f)=\text{ span}\{(1,1)\}$.
Then we need a linear transformation such that
$$f(-1,1)=(0,0)$$ and $$f(0,1)=(1,1)$$
Note that a basis for $\mathbb{R}^{2}$ is $\beta_{\mathbb{R}^{2}}=\{(-1,1),(0,1)\}$.
Also by rank-nullity theorem:$$\dim(\text{im}(f))+\dim(\ker(f))=1+1=\dim(\mathbb{R}^{2})=2$$
Now, let $$(x,y)=\alpha_{1}(-1,1)+\alpha_{2}(0,1) \quad \implies \alpha_{1}=-x, \quad \wedge \quad \alpha_{2}=x+y$$
so $$f(x,y)=-xf(-1,1)+(x+y)f(0,1) \quad \implies \quad f(x,y)=(x+y,x+y)$$
Finally, you can see that $\ker(f)=\text{ span}\{(-1,1)\}$ and $\text{im}(f)=\text{ span}\{(1,1)\}$.
A: Let's figure out the matrix for $T$.
First, the column space of $T$ is a one-dimensional space spanned by
$$v = \begin{bmatrix} 1 \\ -1 \end{bmatrix}.$$ So the columns of $T$ must be scalar multiples of $v$:
$$ T = \left[\begin{array}{c|c} c_1v & c_2v \end{array}\right].$$
(Here $c_1$ and $c_2$ are scalars that cannot both be zero.)
Second, the nullspace of $T$ is spanned by $$w = \begin{bmatrix}1 \\ 1\end{bmatrix}.$$
Multiplying, we have
$$ Tw = \left[\begin{array}{c|c} c_1v & c_2v \end{array}\right]\begin{bmatrix} 1 \\ 1\end{bmatrix} = c_1w + c_2w = \begin{bmatrix} 0 \\ 0\end{bmatrix}$$
which implies that $c_1 = -c_2$.
Putting these together, the linear transformation given by
$$
T = \left[\begin{array}{c|c} cv & -cv \end{array}\right] = c\begin{bmatrix}1 & -1 \\ -1 & 1 \end{bmatrix}
$$
for any nonzero $c$ satisfies the requirements.
