I'd like a help, dont know how to start: Find a linear operator T: R³---R³ that the kernel is generated by (1,2,-1) and (1,-1,0)

Question

I'd like a help, dont know how to start: Find a linear operator T: R³--->R³ that the kernel is generated by (1,2,-1) and (1,-1,0)

anonymous · Answer

You want to find a linear transformation/operator $L(x,y,z)$ such that the vectors (1,2,-1) and (1,-1,0) make $L(1,2,-1)=L(1,-1,0)=(0,0,0)$.

A nice generally linear transformation has the form $L(x,y,z)=ax+by+cz$ (for an operator $L:\mathbb{R}^3\to\mathbb{R}$, but you can extend it to $\mathbb{R}^3$ by making it $L(x,y,z)=(ax+by+cz,~ax+by+cz,~ax+by+cz)$).

Since all the components of the translation are identical, we can focus on just one. We want the given vectors to be the kernel, so we want to guarantee the following:
$$\begin{cases}a+2b-c=0&\text{for }(1,2,-1)\
a-b=0&\text{for }(1,-1,0)\end{cases}$$
Solving, you have $3a=3b=c$, which means you can plug in any variety of $a,b,c$ that satisfies this. This simplest nontrivial solution set would be $a=b=1$ and $c=3$.

So, one such linear transformation would be
$$L(x,y,z)=(x+y+3z,~x+y+3z,~x+y+3z)$$