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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.

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.

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.

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

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 ^_^

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.

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.

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.

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.

That doesn't necessarily mean that you should rule out my numerical proof as false. |

"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.

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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closed account (*48T7M4Gy*)

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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