I came across this question whilst revising:

give the general expresion h(x,y,z) for the linear map h:$\mathbb{R}^3$ $\to$ $\mathbb{R}^3$ defined by h(1,1,1)=(2,2,0), h(1,2,1)=(3,3,0) and h(1,0,0)=(1,0,1)

in the answer it says:

(x,y,z)=(-y+2z)(1,1,1)+(y-z)(1,2,1)+(x-z)(1,0,0) then linearity implies h(x,y,z)=(x+y,y+z,x-z)

I dont understand how to arrive at the final answer. i found that (x+y,y+z,x-z) = (1,1,0),(0,1,1),(0,1,-1). is this a standard expression?

  • $\begingroup$ Try to write with LaTeX, otherwise is between very hard and utterly impossible to understand. Also, what is that r*3 thingy there...?? $\endgroup$ – DonAntonio Dec 28 '13 at 17:22
  • $\begingroup$ sorry \mathbb{R}^3 is r*3 $\endgroup$ – user107783 Dec 28 '13 at 17:25
  • $\begingroup$ Try enclosing the formula between dollar signs $ $\endgroup$ – DonAntonio Dec 28 '13 at 17:26
  • $\begingroup$ $\mathbb{R}^3$ $\to$ $\mathbb{R}^3$ $\endgroup$ – user107783 Dec 28 '13 at 17:28


$$h:\Bbb R^3\to\Bbb R^3\;\;,\;\;h(1,1,1)=(2,2,0)\;,\;h(1,2,1)=(3,3,0)\;,\;h(1,0,0)=(1,0,1)$$

Observe that $\;\Bbb B:=\{(1,1,1)\,,\,(1,2,1)\,,\,(1,0,0)\}\;$ is a basis of $\;\Bbb R^3\;$ , so every element $\;v\in\Bbb R^3\;$ can be uniquely expressed as a linear combination of that basis, say

$$v=a(1,1,1)+b(1,2,1)+c(1,0,0)\;,\;\;a,b,c\in\Bbb R$$

and now you get by linearity that


Take it from here...

| cite | improve this answer | |
  • $\begingroup$ thank you, does this then mean that, a= (-y+2x), b=(y-z) c=(x-z) $\endgroup$ – user107783 Dec 28 '13 at 17:44
  • $\begingroup$ @user107783 Do the math with $\;v=(x,y,z)\;$ $\endgroup$ – DonAntonio Dec 28 '13 at 17:48
  • $\begingroup$ im confused on why (-y+2z),(y-z),(x-z) are used. isit because when expanded they give (1,1,1) =(x,y,z) $\endgroup$ – user107783 Dec 28 '13 at 18:17
  • $\begingroup$ You take a vector $\;v=(x,y,z)\;$ which is written with respect to the usual, canonical basis $\;(1,0,0),(0,1,0), (0,0,1)\;$ , and in order to know how will $\;h\;$ work on it you must write this vector wrt the given basis $\;\Bbb B\;$ , so:$$(x,y,z)=a(1,1,1)+b(1,2,1)+c(1,0,0)=(a+b+c,a+2b,a+b)\iff \begin{cases}x=a+b+c\\y=a+2b\\z=a+b\end{cases}$$ so, for example, if you substract the second line from the sum of the first and the third you get $$a+c=x+z-y$$ and now substract twice the first from the second $$-a-2c=y-2x$$Add now add both eq's above and get:$$c=x-z$$ and etc. $\endgroup$ – DonAntonio Dec 28 '13 at 18:36

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.