linear transformations

stan56

quick question
Let L : R^n --> R^n be a linear transformation. Is it true that if L(x1) = L(x2) then
x1 = x2?

Michael Collins

What do you think?

stan56

Michael Collins wrote: »

What do you think?

I can ask myself what i think without posting it on the internet, i post it up here for an answer, so let me ask it again what do YOU think?

Michael Collins

Sorry stan56, I wasn't trying to be smart. It just looks like a homework question & I was trying to engage you with it. It's not too hard to get your answer by thinking about it for a while.

What would your best guess be? You've a 50% chance of being right!

MathsManiac

It depends on how you've defined what is meant by a "linear transformation". (Or how your teacher/lecturer has defined it.) If you have the definition used by the vast majority, then the answer is "No".

Look at your definition and ask yourself this: "Is the function L:x-->0 a linear transformation?". If the answer to that is "yes", then you should be able to answer your original question (because it gives you a counter-example).

Jimbo

Michael Collins wrote: »

Sorry stan56, I wasn't trying to be smart. It just looks like a homework question & I was trying to engage you with it. It's not too hard to get your answer by thinking about it for a while.

What would your best guess be? You've a 50% chance of being right!

your taking the p**s man. you dont know either

Pherekydes

jimbo78 wrote: »

your taking the p**s man. you dont know either

If you've nothing to add to the thread then don't post.

stan56

Michael Collins wrote: »

Sorry stan56, I wasn't trying to be smart. It just looks like a homework question & I was trying to engage you with it. It's not too hard to get your answer by thinking about it for a while.

What would your best guess be? You've a 50% chance of being right!

thanks ya well i think the answer is yes they r the same but i dunno why its just a guess really

ZorbaTehZ

I think that people need to realise that this is not yahoo answers or some such homework site. Boardsies aren't required to help you - so why not show some respect for the advice you're getting, instead of acting like little winkers.

MathsManiac

Take the 2D case:

The transformation L(x,y)-->(0,0) is linear (by the standard definition). But ANY two points x1 and x2 get mapped to (0,0) under this transformation, and therefore it is NOT true to say that L(x1)=L(x2) implies x1=x2.

If you just think the zero transformation is just a quirky oddity, then try this one: L(x,y)-->(x+y, x+y). This sends both (1,0) and (0,1) to (1,1). (In fact it maps infinitely many points onto the point (1,1), and indeed it maps the whole plane onto the line x=y.

Basically, unless you specify in defining linear transformations that they are invertible (or specify something equivalent to this) then your assertion is not, in general, true.

(By the way, for any function, the property that: [f(x1)=f(x2) implies x1=x2], is called "surjectivity". Your question then is really "Are all linear transformations surjective?" The answer is "No", unless you decide to use a definition that forces them to be surjective. Such a definition is highly unusual but not unheard of; it would not allow my examples above to be called linear transformations.)

Edit!!! Yikes! the last bit above is wrong...See comments below. I should have said "injective", not "surjective". Sorry

stan56

ZorbaTehZ wrote: »

I think that people need to realise that this is not yahoo answers or some such homework site. Boardsies aren't required to help you - so why not show some respect for the advice you're getting, instead of acting like little winkers.

this is a maths discussion forum, i posted a question for discussion that appeared on last years exam paper, i haven't a clue how to do it so i posted it up here for help and i am greatful for the HELPFUL replies and not useless ones like yours

Thanks to everyone who helped

Michael Collins

MathsManiac wrote: »

By the way, for any function, the property that: [f(x1)=f(x2) implies x1=x2], is called "surjectivity".

Wouldn't the function need to be injective in order to say f(x1) = f(x2) => x1= x2. As in the surjective case, two serpate points in the domain could be mapped onto the same point in the co-domain? Take the surjective function

f(x) = x²

for example (obviously this is not a linear transform). Here both +2 and -2 map onto 4, but clearly +2 and -2 are not equal.

This is almost like asking if the transformation is invertable or not - i.e. can we take each point in our co-domain and go back to the original point in our domain. However, for invertability you need a stronger condition, namely a bijective transform i.e. one that is both injective and surjective at the same time - i.e. one that's both one-to-one (injective) and onto (surjective).

MathsManiac

Michael Collins wrote: »

Wouldn't the function need to be injective in order to say f(x1) = f(x2) => x1= x2.

Absolutely. My mistake. I said surjective when I should have said injective. Sincerest apologies.

So the OP's original question was basically "Are all linear transformations injective?" (to which the answer is still "No").