- #1

Zero2Infinity

- 7

- 0

## Homework Statement

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##.

## Homework Equations

Nullspace definition

## The Attempt at a Solution

Using the nullspace definition I get that ##f(0,2,0)=f(1,1,0)=(0,0,0)##. Thus,

\begin{equation} f(0,2,0) =(0,0,0)\end{equation}

\begin{equation}f(1,1,0) = (0,0,0) \end{equation}

\begin{equation}f(0,0,1)=(1,1,1) \end{equation}

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

\begin{equation} 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) \end{equation}

\begin{equation}f(0,1,0)=\frac{1}{2}f(0,2,0)=\frac{1}{2}(0,0,0)=(0,0,0)\end{equation}

while I know from the text of the problem that ##f(0,0,1)=(1,1,1)##. In conclusion,

\begin{equation} A=\begin{pmatrix} 0&0&1\\0&0&1\\0&0&1\end{pmatrix}\end{equation}

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?