Linear Algebra- Onto and One to One Linear Transformations

[Sasor]

Hey guys, I'm studying these concepts in linear algebra right now and I was wanting to confirm that my interpretation of it was correct.

One to one in algebra means that for every y value, there is only 1 x value for that y value- as in- a function must pass the horizontal line test (Even functions, trig functions would fail (not 1-1), for example, but odd functions would pass (1-1))

Onto means that in a function, every single y value is used, so again, trig and event functions would fail, but odd functions would pass- Any kind of function with a vertical asymptote would pass

So i tried to put these concepts in the context of linear functions and this is what I'm thinking-


Since transformations are represented by matrices,

Linearly independent transformation matrices would be considered one to one- because they have a unique solution. Linearly dependent transformations would not be one-to-one because they have multiple solutions to each y(=b) value, so you could have multiple x values for b

Now for onto, I feel like if a linear transformation spans the codomain it's in, then that means that all b values are used, so it is onto.

Examples:

1-1 but not onto

A linearly independent transformation from R3->R4 that ends up spanning only a plane in R4

Onto but not 1-1

A linearly dependent transformation from R3->R2 that's spans R2

1-1 AND onto

A linearly independent transformation from R3->R3 that spans R3

Neither 1-1 nor onto

A linearly dependent transformation from R2->R2 that spans a line


Is this interpretation correct?

 

[micromass]

Hey guys, I'm studying these concepts in linear algebra right now and I was wanting to confirm that my interpretation of it was correct.

One to one in algebra means that for every y value, there is only 1 x value for that y value- as in- a function must pass the horizontal line test (Even functions, trig functions would fail (not 1-1), for example, but odd functions would pass (1-1))

So an odd function such as sin(x) would be one-to-one?

Onto means that in a function, every single y value is used, so again, trig and event functions would fail, but odd functions would pass-

So sin(x) is onto?

Any kind of function with a vertical asymptote would pass

So |tan(x)| is onto??

So i tried to put these concepts in the context of linear functions and this is what I'm thinking-


Since transformations are represented by matrices,

Linearly independent transformation matrices would be considered one to one- because they have a unique solution. Linearly dependent transformations would not be one-to-one because they have multiple solutions to each y(=b) value, so you could have multiple x values for b

And what exactly is a "linear independent matrix" or "linearly dependent transformation"??

Now for onto, I feel like if a linear transformation spans the codomain it's in, then that means that all b values are used, so it is onto.

How does a linear transformation span the codomain exactly?? What does that mean?

Examples:

1-1 but not onto

A linearly independent transformation from R3->R4 that ends up spanning only a plane in R4

Onto but not 1-1

A linearly dependent transformation from R3->R2 that's spans R2

1-1 AND onto

A linearly independent transformation from R3->R3 that spans R3

Neither 1-1 nor onto

A linearly dependent transformation from R2->R2 that spans a line


Is this interpretation correct?

 

[Sasor]

What are you getting at? If I'm incorrect, then just tell me

 

[micromass]

What are you getting at? If I'm incorrect, then just tell me

I just gave feedback on your post and I gave you some things to think about.

Furthermore, I honestly do not understand what you mean with linearly dependent matrix or a transformation spanning the codomain. So you should really say what you mean with that.

 

[Sasor]

A linearly dependent matrix is a matrix that is linearly dependent matrix...I don't know how I can really explain this...you understand what linear dependence means, right?

Also, when I say spans the codomain, I mean that the b in T(x)=b could be any vector in the codomain...

 

[micromass]

A linearly dependent matrix is a matrix that is linearly dependent matrix...I don't know how I can really explain this...you understand what linear dependence means, right?

I know very well what linear dependence means, but not in the context you are talking about. To me a set [itex]\{v_1,...,v_n\}[/itex] is linearly independent if for all [itex]\alpha_1,...,\alpha_n\in \mathbb{R}[/itex] holds:
[tex]\alpha_1v_1+...\alpha_nv_n=0~\Rightarrow \alpha_1=...=\alpha_n=0[/tex]

So, we are talking about a set of vectors here that is linearly indepenent. What you mean with a linearly independent matrix is a mystery to me. How does your book define it?

Also, when I say spans the codomain, I mean that the b in T(x)=b could be any vector in the codomain...

What is b, what is x?? Can you look the definition up in your textbook and quote it here?

 

[Sasor]

Ok, well linear dependence in context of a matrix is just like linear dependence with a set of vectors...

for example

[1 4 8 3]
[2 4 1 7]
[3 2 6 7]

If this^ matrix is linearly dependent, then it is equivalent to saying that these vectors:

[1] [4] [8] [3]
[2] [4] [1] [7]
[3],[2],[6],[7]

are linearly independent...
(The formal definition is that if you can set a matrix Ax=0 and your only solution is x=0, then the matrix/set of vectors is linearly independent)
Here is what I mean by b

[A|b] is the augmented matrix

so that

Ax=b

where
A is the coefficient matrix
x is the solution or kernel of solutions
and b is the vector in question to be found

Analogous to regular algebra

Ax=b

is to

mx=y

 

[micromass]

OK, so you define a matrix to be linearly (in)dependent if their column vectors are linearly (in)dependent? I've never really seen this definition before, but ok.

This is indeed equivalent to

[tex]Ax=0~\Rightarrow x=0[/tex]

for all x.

And indeed, if a matrix satisfies that, then it is one-to-one. So any "linearly independent matrix" is one-to-one.

With "A spanning the codomain" you seem to mean that Ax=b has a solution for every b in the codomain. This is indeed equivalent to onto.

About your examples:

Examples:

1-1 but not onto

A linearly independent transformation from R3->R4 that ends up spanning only a plane in R4

I don't think it is possible for a transformation to be both "linearly independent" and only spanning a plane. So while your example does imply that the transformation is 1-1 but not onto, I fear that there are no such transformations.

Onto but not 1-1

A linearly dependent transformation from R3->R2 that's spans R2

OK

1-1 AND onto

A linearly independent transformation from R3->R3 that spans R3

Neither 1-1 nor onto

A linearly dependent transformation from R2->R2 that spans a line

Also ok.
A nice fact in linear algebra is the following: Let T is a transformation from [itex]\mathbb{R}^n\rightarrow \mathbb{R}^n[/itex] (so very important: domain and codomain must have the same dimension). If T is 1-1, then it is onto. And if T is onto, then it is 1-1.
This is sometimes called the alternative theorem.

 

[Sasor]

Ok cool, and that alternative theorem is very convenient and it makes sense...thanks for the help!

 

[Sasor]

Well about that 1-1 but not onto thing, I just did an example-

If you want to transform

[x1]
[x2]

->

[3*x1]
[x1+4*x2]
[x1+5]You'd get a transformation matrix

[3 0] ... ...[0]
[1 4] *[x1] +[0]
[1 0] *[x2] [5]

right?

in this case, it'd be linearly independent, but it'd only span a plane in r3

 

[micromass]

Right, but the domain of that is not [itex]\mathbb{R}^3[/itex], as you claimed before.

 

[Sasor]

well the domain of the matrix is indeed in R3...

[3 0]
[1 4]
[1 0]

 

[micromass]

well the domain of the matrix is indeed in R3...

[3 0]
[1 4]
[1 0]

Really? So what is the image of (1,1,1) then?

 

[Sasor]

oh wait, nevermind...domain is in r2...but either way...

it wouldn't be onto because you're still only spanning a plane in r3

but if you look at the transformation, matrix, it's linearly independent