Finding an endomorphism with given characteristics.

Question

I have been trying the following, but can't get my head around how to go on. I need to find an endomorphism $f$ in $\mathbb{R}^3$ (in the canonical basis) such that:

1. Its rank is 2.
2. Its trace is 4.
3. $(2,1,1)$ is one of its eigenvectors.
4. $U = \{ (x,y,z) \in \mathbb{R}^3 : x+y-z = 0\}$ is an eigenspace.

I can derive some information from this facts, but I can't come close to actually defining an endomorphism. So far I know:

1. If the rank of $f$ is 2, then $\dim(f(\mathbb{R}^3))=2$ and by the rank-nullity theorem, the nullity of $f$ must be 1.
2. If its trace is four, the sum of its eigenvalues is also four: $\lambda_1 + \lambda_2 + \lambda_3 = 4$

I don't know how to use the rest of the information provided. Could anyone guide me in the right direction?

Answer 1

Consider the vectors $v_1=(2,1,1)$, $v_2=(1,0,1)$, and $v_3=(0,1,1)$. Clearly, $\bigl\langle\{v_2,v_3\}\bigr\rangle=U$. So, take $f$ such that:

• $f(v_1)=0$;
• $f(v_2)=2v_2$;
• $f(v_3)=2v_3$

and you're done. Note that if $f$ is constructed this way, then$$f(x,y,z)=(-2y+2z,-x+y+z,-x-y+3z).$$