# Check of a problem about nullspace

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1. Feb 4, 2017

### Zero2Infinity

1. The problem statement, all variables and given/known data
Let $V\subset \mathbb{R}^3$ be the subspace generated by $\{(1,1,0),(0,2,0)\}$ and $W=\{(x,y,z)\in\mathbb{R}^3|x-y=0\}$. Find a matrix $A$ associated to a linear map $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ through the standard basis such that its nullspace is $V$.

2. Relevant equations

Nullspace definition

3. The attempt at a solution

Using the nullspace definition I get that $f(0,2,0)=f(1,1,0)=(0,0,0)$. Thus,
$$f(0,2,0) =(0,0,0)$$
$$f(1,1,0) = (0,0,0)$$
$$f(0,0,1)=(1,1,1)$$
Since the matrix has to be written wrt the standard basis of $\mathbb{R}^3$, which is $(1,0,0),(0,1,0),(0,0,1)$, I infer that
$$f(1,0,0)=f(1,1,0)-\frac{1}{2}f(0,2,0)=(0,0,0)-\frac{1}{2}(0,0,0)=(0,0,0)$$
$$f(0,1,0)=\frac{1}{2}f(0,2,0)=\frac{1}{2}(0,0,0)=(0,0,0)$$
while I know from the text of the problem that $f(0,0,1)=(1,1,1)$. In conclusion,
$$A=\begin{pmatrix} 0&0&1\\0&0&1\\0&0&1\end{pmatrix}$$
If I look for a basis of the kernel I effectively get the two vectors $(1,1,0),(0,2,0)$, so it should be ok. Is it correct?

2. Feb 6, 2017

### andrewkirk

Does this mean
1. Let $V\subset \mathbb{R}^3$ be the subspace generated by $\{(1,1,0),(0,2,0)\}$ and let $W=\{(x,y,z)\in\mathbb{R}^3|x-y=0\}$; or does it mean
2. Let $V\subset \mathbb{R}^3$ be the subspace generated by $\{(1,1,0),(0,2,0)\}\cup\{(x,y,z)\in\mathbb{R}^3|x-y=0\}$?
If it means the second one then $V=\mathbb R^3$, so $f$ must be the zero map. If it means the first one then $W$ is not used in the problem, which raises the question of why it has been defined.