I am given the kernel and image of an endomorphism in $\Bbb R^3 $, which are:
$\ker f =\langle(1,1,0)\rangle$
$\text{Im}\, f = \langle(0, 1, -1), (2, 1, 2)\rangle$
and I have to find the matrix associated with the function.
This is what I've done si far: $$ f(x,y,z) = ((0, 1, -1),(2,1,2),(a,b,c))*(x,y,z) = ((2y+az), (x+y+bz), (-x+2y+cz))$$ (I multiplied the two matrices) but then I obtain that $f(1,1,0) = (2,2,1)$ which cannot be since this is the base of the kernel linear subspace.
Thanks to everyone in advance 🙂
Best Answer
As @ArturoMagidin said there is not a unique solution, but it is easy to construct a map with this properties. It is well known that a linear map is determined by the image of the elements of any basis of the first vector space, so in your case you only have to determine the image of a basis of $\mathbb{R}^3$.
You want $\ker f =\langle(1,1,0)\rangle$, so $f(1,1,0)=(0,0,0)$ and one of these three vectors is already fixed. For the other two, since you want $\text{Im}\, f = \langle(0, 1, -1), (2, 1, 2)\rangle$, you know that you have to send one of the vectors to $(0,1,-1)$ and the other to $(2,1,2)$. Hence, to find a map $f$ you just have to add vectors to the set $\lbrace (1,1,0)\rbrace$ until you get a basis of $\mathbb{R}^3$ and then send one of the vectors you added to $(0,1,-1)$ and the other to $(2,1,2)$.
As an example, you can take $\lbrace (1,0,0),(0,0,1),(1,1,0)\rbrace$ as a basis of $\mathbb{R}^3$ and define $f$ as: $$f(1,0,0)=(0,1,-1),~f(0,0,1)=(2,1,2),~f(1,1,0)=(0,0,0)$$