Linear Algebra - Help Me or Shoot Me

[Eli]

A question, there is a small problem with translating stuff though.

T is a linear trasformation. T:R^3-->R^3.
T(1,1,1)=(7,7,7)
T(1,1,0)=(4,4,2)
T(0,1,1)=(2,4,4)

There is a type of linear transformations, I dont know how they are called in English so I'll call it [whatever].
I'll translate the simplest definition for [whatever] from my book...

A linear tranformation, T:V-->V is called [whatever] if there is a base for V in which T is represented by a diagonal matrix.

So I need to prove that T is [whatever].

All the stuff I already found, and a small problem I had:

T(x,y,z) = (5x-y+3z,3x+y+3z,3x-y+5z)
This should be correct, I think.

T(x,3x+3y,y)=2(x,3x+3y,y)
There is a special term for 2 in this case, I dont how it is called in English.

I also have a small problem:
We know that T(1,1,1)=(7,7,7)=7(1,1,1)
Which means that:
T(x,y,z) = (5x-y+3z,3x+y+3z,3x-y+5z) = 7(x,y,z) = (7x,7y,7z)
bla bla bla
5x-y+3z = 7x
3x+y+3z = 7y
3x-y+5z = 7z
bla bla bla
-2x-y+3z = 0
3x-6y+3z = 0
3x-y-2z = 0
After I start doing stuff on this matrix I find that there is only one solution:
x=y=z=0. But this is impossible!
So what did I do wrong?


And how do I solve the question...

------------------
<IMG SRC="http://forums.civfanatics.com/ubb/tank.gif" border=0> <IMG SRC="http://forums.civfanatics.com/ubb/tank.gif" border=0> <IMG SRC="http://forums.civfanatics.com/ubb/tank.gif" border=0> <IMG SRC="http://forums.civfanatics.com/ubb/tank.gif" border=0>

 

[GenghisK]

Email me the complete problem. I'll try to solve it later. Linear algaebra is something that requires all the data to be solved. Send me the problem.

 

[CornMaster]

Where would you like me to mail the bullet???

------------------
<IMG SRC="http://images.honesty.com/imagedata/h/207/85/32078598.gif" border=0> I AM CANADIAN! <IMG SRC="http://images.honesty.com/imagedata/h/207/85/32078598.gif" border=0>
CivFanatics Moderator and Tech Support
CornEmpire Owner/Operator
My Civ 2 Scenario Page.

 

[GenghisK]

After a little thought, I think there's a little mistake (but huge consequences...)

You said that:
We know that T(1,1,1)=(7,7,7)=7(1,1,1)
Which means that:
T(x,y,z) = (5x-y+3z,3x+y+3z,3x-y+5z) = 7(x,y,z) = (7x,7y,7z)

No, no, and probably not true.
T(1,1,1)=(7,7,7) by hypothesis, I agree. But you cannot generalise the situation to all the (x,y,z) triplet of R3...
Therefore, T(x,y,z) != 7(x,y,z) in all the situations, but one.... See? This fact being "false", all of your following reasonning falls too.


------------------
<IMG SRC="http://www.ifrance.com/genghisk/GKultima.gif" border=0>

 

[BlueMonday]

I've got a Glock 20 in my apartment. I think you should borrow it.

------------------
<IMG SRC="http://forums.civfanatics.com/image_uploads/goodbye3.jpg" border=0>
<FONT COLOR="blue">I take every day one beer at a time; every beer one sip at a time.</FONT c>

 

[Eli]

GK : I meant that I want to find the basis for the space of the Eigenvectors(at least that's how they called it on Apolyton) of 7. I just wrote it in a wrong mathematical way.

---

Anyway, thanks to Chowlett on Apolyton I managed to solve it. But I have more questions. All I need is to find the time to think about them first.