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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?

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  • $\begingroup$ 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
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So

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

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

$$h(v)=ah(1,1,1)+bh(1,2,1)+ch(1,0,0)=a(2,2,0)+b(3,3,0)+c(1,0,1)$$

Take it from here...

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  • $\begingroup$ 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

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