0
$\begingroup$

I came across this question in Linear Transformations.

Find all Linear Maps $L\colon\mathbb{R}^3\longrightarrow\mathbb{R}^3$ whose kernel is exactly the plane $\{(x_1,x_2,x_3)∈\mathbb{R}^3\,|\,x_1+2x_2-x_3=0\}$.

How do I find the required Linear Transformations and how to denote them without going into specifying the corresponding matrix associated with the Transformation(I want to understand the Transformation first than the associated matrix)?

I do understand the terminology and I want to know the method to go about.

One of the answers I have seen directly gave me the matrix.

$\endgroup$
0
$\begingroup$

Consider the basis $\{e_1,e_2,e_3\}$ of $\mathbb{R}^3$ such that $e_1=(1,0,1)$, $e_2=(0,1,2)$, and $e_3=(1,0,0)$. Note that $\{e_1,e_2\}$ is a basis of your plane. Therefore, given a linear map $f\colon\mathbb{R}^3\longrightarrow\mathbb{R}^3$, $\ker f$ is your plane if and only if $f(e_1)=f(e_2)=(0,0,0)$. On the other hand, $f(e_3)$ can be any vector $(a,b,c)$.

Now, note that $(0,1,0)=-2e_1+e_2+2e_3$ and that $(0,0,1)=e_1-e_3$. Therefore, if $(x,y,z)\in\mathbb{R}^3$, then\begin{align}f(x,y,z)&=f\bigl(x(1,0,0)+y(0,1,0)+z(0,0,1)\bigr)\\&=xf(1,0,0)+yf(0,1,0)+zf(0,0,1)\\&=x(a,b,c)+2y(a,b,c)-z(a,b,c)\\&=\bigl(a(x+2y-z),b(x+2y-z),c(x+2y-z)\bigr)\\&=(x+2y-z)(a,b,c).\end{align}

$\endgroup$
  • $\begingroup$ Could you please explain the steps in the solution in a more detailed way? What is the motivation behind choosing $e1,e2,e3$ in that way? Also, I did not understand how $e1$ & $e2$ form basis for the plane. Please forgive if these are simple ones to answer, but I am still learning... $\endgroup$ – Yaksha Oct 23 '17 at 10:52
  • $\begingroup$ @Yaksha Since $(1,0,1)$ and $(0,1,2)$ are linear independent vectors of your plane and since the plane is $2$-dimensional, they are a basis of the plane. Then I added a third vector (I chose $(1,0,0)$, more or less at random) outside the plane in order to get a basis of $\mathbb{R}^3$. $\endgroup$ – José Carlos Santos Oct 23 '17 at 10:56
  • $\begingroup$ Thank you. Now I understand it. In the third simplification step, shouldn't $yf(0,1,0)$ be replaced by $+2y(a,b,c)$ and $zf(0,0,1)$ be replaced by $-z(a,b,c)$? $\endgroup$ – Yaksha Oct 23 '17 at 11:31
  • $\begingroup$ @Yaksha Sure. My mistake. Is everything well now? $\endgroup$ – José Carlos Santos Oct 23 '17 at 11:35
  • $\begingroup$ Yeah! For future readers with doubts on basis of a plane, please refer this question math.stackexchange.com/q/251822 $\endgroup$ – Yaksha Oct 23 '17 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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