Linear Transformation: V --> W - True or False?

In summary: Remember, T is a transformation, so it takes in a vector and outputs another vector. In this case, T takes in any vector in R^2 and always outputs (0,0,0). So it's not 1-1 because multiple inputs can give the same output.
  • #1
jumbogala
423
4

Homework Statement


T: V --> W is a linear transformation where V and W are finite dimensional.

If dim V is less than or equal to dim W, then T is one-to-one. True or false?


Homework Equations





The Attempt at a Solution


First of all, I'm assuming that im(T) = W. Is that correct? If so, dim(im T) = dim(W).

Dimension thm says dim(V) = dim(ker T) + dim(im T).

So if dim(V) is less than or equal to dim(im T), then dim(ker T) = 0. Which means ker T is is as small as it can be, so T should be one-to-one.

But this is wrong. The answer is supposed to be false. Can anyone help?
 
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  • #2
W is the target space for the transformation, so the image of T is a subset of W. The image is equal to the target space if the function is onto.
 
  • #3
Hmm okay, and we don't know if this is onto. So my approach doesn't work =\

Can anyone give me a hint of what I should instead?
 
  • #4
If you already know it's false then try to construct a counterexample.
 
  • #5
Well I don't know it's false, I looked at the answers :) Haha. And a counterexample is given there... but is there a way to show that it's false without finding a specific counterexample?

(Just because on a test I wouldn't automatically know if it was true or false first, so I want to be able to understand why it's one or the other w/o a counterexample).
 
  • #6
A proof to show it's false might not work on all levels because there can be functions that satisfy the criteria. You could, as you pointed out, use the dimension theorem as well as that T is 1-1 implies that nullity T = 0 and try to reduce the problem into proving something simpler.

A good way to prove something is false is to find a counterexample. Can you find a linear transformation where dim V is less than or equal to dim W but T is not one-to-one?

For example, consider the mapping f(x,y) = x + y. Is f 1-1?
And consider g(x,y)=(f(x,y), 0, 0)
 
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  • #7
All we know about T is that it's a linear transformation.

T = 0
is a linear transformation, and it's about as far from being 1-1 as you can get...
 
  • #8
How are you getting T = 0? I understand that T = 0 is definitely not 1-1.

But if T = 0, I don't get why dim(V) is less than or equal to dim(W)...

I'm obviously missing something here =\
 
  • #9
Mathnerdmo isn't 'getting' T=0. It's proposing that T=0 is a counterexample for the proposition that T:V->W is necessarily 1-1 if dim(V)<dim(W).
 
  • #10
oh, okay. I still don't get what dim(V) and dim(W) actually are though.

V could have many vectors in it, but if T=0 then W only has one vector, 0, doesn't it? So dim(W) = 0. So wouldn't that mean that dim(V) > dim(W)? Obviously not, but I don't understand why.
 
  • #11
Put V=R^2 and W=R^3 and T:V->W=0. dim(V)=2 and dim(W)=3, you can agree with that, right? What T is doesn't affect what dim(V) and dim(W) are. Does it? And T is not 1-1.
 
  • #12
Ah I think I get it. Let me rephrase to make sure.

T: R^2 --> R^3. So T would like like T(x,y) = (0, 0, 0) if T = 0, right? (I'm a bit unsure about what T = 0 actually means, but it just means it takes any vector to zero, correct?)

So then dim(R^2) = 2 and dim(R^3) = 3, but T is not 1-1 because every vector in R^2 that map to (0,0,0), not just one.
 
  • #13
jumbogala said:
Ah I think I get it. Let me rephrase to make sure.

T: R^2 --> R^3. So T would like like T(x,y) = (0, 0, 0) if T = 0, right? (I'm a bit unsure about what T = 0 actually means, but it just means it takes any vector to zero, correct?)

So then dim(R^2) = 2 and dim(R^3) = 3, but T is not 1-1 because every vector in R^2 that map to (0,0,0), not just one.

Yes, that's it.
 

1. What is a linear transformation?

A linear transformation is a mathematical operation that maps one vector space, V, to another vector space, W, while preserving the basic algebraic structure of the original space.

2. How do you determine if a transformation is linear or not?

A transformation is linear if it satisfies two properties: 1) the transformation of a sum of vectors is equal to the sum of the individual transformations of each vector, and 2) the transformation of a scalar multiple of a vector is equal to the scalar multiple of the transformation of the original vector.

3. Can a linear transformation have multiple inputs or outputs?

Yes, a linear transformation can have multiple inputs and outputs. However, the number of inputs must be equal to the number of outputs for the transformation to be considered linear.

4. Is a linear transformation always a one-to-one mapping?

No, a linear transformation can be a one-to-one mapping, where each input vector maps to a unique output vector, or it can be a many-to-one mapping, where multiple input vectors map to the same output vector.

5. Can a linear transformation change the dimensionality of a vector space?

Yes, a linear transformation can change the dimensionality of a vector space. For example, a transformation from R³ to R² would reduce the dimensionality of the vector space from three dimensions to two dimensions.

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