Question in Isomorphism: T(x,y,z) = (x-2z, 2x-y+3z, 4x+y+8z)

[yossa]

Hello everyone, Need help with the following question please :

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There shall be a non-$\displaystyle T:R^3\rightarrow R^3$ linear copy defined by:

. . . . .$\displaystyle T(x,\, y,\, z)\, =\, (x\, -\, 2z,\, 2x\, -\, y\, +\, 3z,\, 4x\, +\, y\, +\, 8z)$

to all $\displaystyle (x,\, y,\, z)\, \in\, R^3$

Prove that T - isomorphism and thought - $\displaystyle .T^{-1}(x,\, y,\, z)$

 

[Dr.Peterson]

Hello everyone, Need help with the following question please :
There shall be a non - View attachment 9668linear copy defined by:
View attachment 9669 to all - View attachment 9670
Prove that T - isomorphism and thought - View attachment 9671.

Again, this is very hard to follow, both due to the pictures and the English. I read it as:

Let T:R³->R³ be a non-linear mapping defined by T(x,y,z) = (x-2z, 2x-y+3z, 4x+y+8z) for all (x,y,z) \in R³. Prove that T is an isomorphism and evaluate T^-1(x,y,z).
​

Is that correct? I'm not at all sure what you mean by "thought".

What is the definition of an isomorphism as you have been taught it? For that matter, an isomorphism is not just between two sets; you have to tell us what kind of structure you are assuming for R³ (vector space, or something else?). Finally, why do you say this is non-linear? (Or did you really mean it as you wrote it, "a non-T:R³->R³ linear mapping"? What does that mean?)

As always, please show work, so we can know what help you need.

 

[yossa]

Again, this is very hard to follow, both due to the pictures and the English. I read it as:

Let T:R³->R³ be a non-linear mapping defined by T(x,y,z) = (x-2z, 2x-y+3z, 4x+y+8z) for all (x,y,z) \in R³. Prove that T is an isomorphism and evaluate T^-1(x,y,z).
​

Is that correct? I'm not at all sure what you mean by "thought".

What is the definition of an isomorphism as you have been taught it? For that matter, an isomorphism is not just between two sets; you have to tell us what kind of structure you are assuming for R³ (vector space, or something else?). Finally, why do you say this is non-linear? (Or did you really mean it as you wrote it, "a non-T:R³->R³ linear mapping"? What does that mean?)

As always, please show work, so we can know what help you need.

again sorry abut english, i mean "calculation" where I wrote down "thought".
I do not have a starting direction so I need help ...
Thank you

 

[yossa]

again sorry abut english, i mean "calculation" where I wrote down "thought".
I do not have a starting direction so I need help ...
Thank you

I figured out how to calculate T ^ 1 just how to quantify isomorphism?

 

[Dr.Peterson]

I figured out how to calculate T ^ 1 just how to quantify isomorphism?

Please start by answering my main question: What field are you studying? Is this a question about vector spaces, or something else? Is it really linear or non-linear?

It will help particularly if you copy out the definition of isomorphism, which is what you need to think about anyway. That will tell you what to do.

And, like the other problem, if necessary you should show us the exact wording in the original language, so we can try to look it up. (Admittedly, there are some languages for which we probably couldn't do that, but let us decide that.)

 

[HallsofIvy]

A linear transformation between two spaces of the same dimension is an "isomorphism" if and only if the kernel is {0}. Here, T(x, y, z)= (x−2z,2x−y+3z,4x+y+8z). The kernel consists of (x, y, z) such that (x−2z,2x−y+3z,4x+y+8z)= (0, 0, 0). We have the three equations x- 2z= 0, 2x- y+ 3z= 0, and 4x+ y+ 8z= 0. Notice that the first equation has no "y" and the other two equations have "-y" and "+y" so if we add those two equations we eliminate y getting 6x+ 11z= 0. x- 2z= 0 then is the same as 6x- 12z= 0 and subtracting that to 6x+ 11z= 0 we 23z= 0 so z= 0. Then x= 2z= 0 and 2x- y+ 3z= -y= 0. x= y= z= 0 so (0, 0, 0) is the only member of the kernel is 0.

As for finding the inverse transformation write (x- 2z, 2x- y+ 3z, 4x+ y+ 8z)= (u, v, w) and solve for x, y, and z. That is almost the same as above. Adding the 2x- y+ 3z= v and 4x+ y+ 8z= w eliminates y giving 6x- 11z= v+ w. From x- 2z= u, 6x- 12z= 6u. Subtracting that from 6x- 11z= v+ w we have z= v+ w- 6u. x- 2z= x- 2(v+ w- 6u)= x- 2v- 2w+ 12u= u so x= -11u+ 2v+ 2w. Then finally, 2x- y+ 3z= 2(-11u+ 2v+ 2w)- y+ 3(v+ w- 6u)= -y -22u+ 4v+ 4w+ 3v+ 3w- 18u= -y- 40u+ 7v+ 7w= v so y= -40u+ 6v+ 6w. Replacing "u, v, w" with the generic variables "x, y, z" the inverse transformation is T(<x, y, z>)= (-11x+ 2y+ 2z, -40x+ 6y+ 6z, x+ y- 6).