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### One more Linear Transformation (help please)

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Hey guys, I solved one more and final linear transformation function:

T(x1, x2, x3) = x1 + (x2 + x3) + (x1 + x3)

This to me tested linear, is my answer true? It's really hard to find a calculator for these, so annoying. I just want to make sure that my answer was true.
closed account (48T7M4Gy)
Here is my attempt (I followed UK Marine's 2nd method seeing as numeric proof may not be sufficient for the proof):

Let x1: a1, a2, a3 and x2: b1, b2, b3

u + v = a1 + b1, a2 + b2, a3 + b3
T(u + v) = a1 + b1, a2+b2 + a3 + b3, a1+b1 + a3 + b3 // Equal

T(u) = a1 + a2 + a3 + a1 + a3
T(v) = b1 + b2 + b3 + b1 + b3
T(u) + T(v) = a1 + b1, a2 + b2, a3 + b3, a1 + b1, a3 + b3 // Equal

Onward to 2nd condition...
T(cv) = cT(v)

cv = Ca1, Ca2, Ca3
T(cv) = Ca1 + Ca2 + Ca3 + Ca1 + Ca3 // Equal

T(v) = a1 + a2 + a3 + a1 + a3
cT(v) = c(a1 + a2 + a3 + a1 + a3) // Equal

Basically, I just want to know if this is correct, thank you very much.

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closed account (48T7M4Gy)
The second proof is the one helios gave IIRC, which is the mathematical proof and not just a specific numerical example.

In that light, I read your mathematical proofs and find them correct. Well done. :)

If I was to make any change I would suggest you use the LHS vs RHS convention and then show they are equal, like this:

To prove: T(u + v) = T(u) + T(v) then proced
LHS = ... as you have shown
RHS = ... as you have shown
Therefore LHS = RHS
QED

Cheers
Thank you! So glad I got it right ;)

Also by 2nd proof I meant UK Marine's arbitrary variable method as I found it pretty neat and easy to understand. I also appreciate Helios' method but it looked a bit cryptic for me, maybe cause I'm still just learning linear algebra so ^_^
Don't really pay much attention to Helios' posts, this individual doesn't know what they're posting.
Oh? I hope that was a joke UK :)

closed account (48T7M4Gy)
They're actually the same, so if UK was right second time around then that's all good, but the important thing not to lose sight of is a numerical example is not a mathematical proof. :)
closed account (48T7M4Gy)
https://www.mathway.com/examples/Algebra/Linear-Transformations/Proving-a-Transformation-is-Linear?id=266
@Kemort, I never got it wrong to begin with mate. Some professors are okay with numerical proofs, some are not. That doesn't necessarily mean that you should rule out my numerical proof as false.
closed account (48T7M4Gy)
 Some professors are okay with numerical proofs
Name them!
My professor, she didn't mind. So long as I include all the steps right.
closed account (48T7M4Gy)
Name her please, including subject and university.
That's confidential.
If a university professor accepts a single example as a proof, and this is a course you're paying for, you're getting ripped off. It doesn't matter what the rest of the classes are like if they're failing at such a fundamental level.

 That doesn't necessarily mean that you should rule out my numerical proof as false.
Don't confuse falseness with invalidity. "Today is Sunday, therefore all but one prime numbers are odd" is an invalid reasoning with a true conclusion.
"2 is even and 3 is odd and 5 is odd, therefore all but one prime numbers are odd" is an equally invalid reasoning with an equally true conclusion. This is not mathematics, this is empiricism.
 but the important thing not to lose sight of is a numerical example is not a mathematical proof. :)

 If a university professor accepts a single example as a proof,...

Single examples are valid proofs in many circumstances, e.g. proof by counter example, or proof of an existential statement.

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closed account (48T7M4Gy)
A counter example only proves something is wrong.

The test for the linearity property of a given transform is not an existential question.

The unidentified professor is an existential question with a definite cloud hanging over her mathematical ability and fee charging properties, all of which UK has decided to cloud in secrecy.

UK exists. The marine property? Doubtful. Maybe UK is a counter example. We may never know.
 A counter example only proves something is wrong. The test for the linearity property of a given transform is not an existential question.

I was just pointing out that what you and Helios wrote was wrong. Perhaps what was written wasn't what was meant, but how are we supposed to know?
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You know perfectly well what was meant. Don't be a smartass.
closed account (48T7M4Gy)
htrwin, it appears you have come into this too late. The bone of contention is whether a numerical example is or is not a mathematical proof. Neither helios nor I are incorrect and no real professor of mathematics would even remotely dispute that. That is our contention. UK contends otherwise based on an unidentified teacher.

I agree, this is not about you and I, or helios. The facts speak for themselves. Unfortunately you had nothing to point out, and what you did point out is not relevant.

Of course, you will want to dispute this, so your best avenue is to come up with a professor of mathematics to discuss it with. UK was given the same opportunity given that it was part of his claim, but he declined as I expect you will also do.

As for the property of smart-assedness is concerned, I'd prefer a counter example but so far your example is pointing pretty much in the direction helios describes.

A real professor at a real university wouldn't hide.
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