Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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... problem

share|improve this question
    
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
1  
..anyone knows how to delete a comment? –  Dylan Zhu Feb 28 '13 at 23:27
show 3 more comments

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.