# Problem from Hoffman's Linear Algebra

Just look at the picture, the problem says Is there a Linear transformation T:$\mathbb R ^3$ $\rightarrow$ $\mathbb R ^2$ , such that T(1,-1,1)=(1,0), T(1,1,1)=(0,1), and this is Prof's notes. I dont get it, because he never mention that determinant...

-
that determinant is vector (or cross) product of two vectors – Cortizol Feb 28 '13 at 22:52
@Dylan Zhu What don't you understand exactly? Is it how he concludes that the vectors form a basis from that "determinant"? – Git Gud Feb 28 '13 at 22:53
is he just wanna generate a vector which is linear independent from the other two, i think i get it, its brilliant – Dylan Zhu Feb 28 '13 at 22:55
@DylanZhu That's right. You're in for many, many more surprises. – Git Gud Feb 28 '13 at 22:56
..anyone knows how to delete a comment? – Dylan Zhu Feb 28 '13 at 23:27

First note that the two vectors $e_1=(1,-1,1)$ and $e_2=(1,1,1)$ are linearly independent.
Then pick any $e_3$ such that $(e_1,e_2,e_3)$ be a basis of $\mathbb{R}^3$. What your professore did is that he/she took $e_3$ to be the cross product of $e_1$ and $e_2$. Indeed, if $e_1$ and $e_2$ are linearly independent, then $(e_1,e_2,e_1\times e_2)$ is automatically a basis of $\mathbb{R}^3$. Then he/she divided the cross product by $2$ for aesthetical reasons. But this is still a basis of course.
For any choice of three vectors $f_1,f_2,f_3$ in $\mathbb{R}^2$, there exists a unique $T:\mathbb{R}^3\longrightarrow\mathbb{R}^2$ linear such that $$T(e_i)=f_i\qquad i=1,2,3.$$
Indeed, a linear operator is completely determined by the image of a basis via the formula: $$T(x_1e_1+x_2e_2+x_3e_3)=x_1T(e_1)+x_2T(e_2)+x_3T(e_3).$$
Just take $f_1$ and $f_2$ to be the two vectors considered in the problem. And $f_3=(1,2)$ like your professor, for instance.