Let $f:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ be an endomorphism such as
$f(1,0,1)=(1,2,1)$
$f(1,1,0)=(2,0,0)$
and $v=(-1,1,1)$ is an eigenvector respect to the eigenvalue $-2$
$1)$ Describe the $Im(f)$ and $Ker(f)$
$2)$ Describe explicitly $f$
$3)$ Determine if $f$ is diagonalizable
I've been give this exercise and I'm wondering if it lacks information in order to be solved or if it's me who lacks of information. Since I don't have a basis of the transformation defined and I can't either arbitrarily finish to define the function $f$ because must be a function that satisfies that the condition that $v=(-1,1,1)$ is an eigenvector. For the other side, I know that since $-2$ is eigenvalue satisfies that $(A+2Id)v=0$ but I don't know how to deduce the matrix of the transformation with this data! What am I missing here?
I'll be thankful if someone helps me!
You know that$$f(1,0,1)=(1,2,1)\text{, that }f(1,1,0)=(2,0,0)\text{, and that }f(-1,1,1)=(2,-2,-2).$$Since $\{(1,0,1),(1,1,0),(-1,1,1)\}$ is a basis of $\Bbb R^3$, you know all that you need to know in order to work with $f$. For instance, since$$(1,0,0)=\frac13(1,0,1)+\frac13(1,1,0)-\frac13(-1,1,1),$$you know that$$f(1,0,0)=\frac13(1,2,1)+\frac13(2,0,0)-\frac13(2,-2,-2)=\left(\frac13,\frac43,1\right),$$an so on.