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Linear Algebra help please

The function goes from R3 to R3, so at max the image can be 3 dimensional, is this correct?

I had to obtain a matrix using the change of basis theorem which I did and now I have to show how the image could be 1,2 and 3 dimensional by picking values for a,b,c,d,e.

Can someone check this?

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Reply 1
This is the full question, I've 6a and 6b and now need someone to check if my answer to 6c is correct.
Original post by e^x
This is the full question, I've 6a and 6b and now need someone to check if my answer to 6c is correct.


Would you be able to repost the question? It is very blurry/impossible to read.

Also have you come across the rank-nullity theorem in your course? This sounds like a standard example of using/understanding it. For your first question, yes the image can be at most 3 dimensional. This is because the image as it is defined, will be a subset of the vector space you are mapping to (in this case R3\mathbb{R} ^3). Have a look at the definition of the image and try and see why this is the case.
Reply 3
Q6c
Reply 4
Original post by Star-girl
Would you be able to repost the question? It is very blurry/impossible to read.

Also have you come across the rank-nullity theorem in your course? This sounds like a standard example of using/understanding it. For your first question, yes the image can be at most 3 dimensional. This is because the image as it is defined, will be a subset of the vector space you are mapping to (in this case R3\mathbb{R} ^3). Have a look at the definition of the image and try and see why this is the case.


^^^
Original post by e^x
^^^


You made a typo in your original post. The question actually says that f:R2R3f: \mathbb{R} ^2 \mapsto \mathbb{R} ^3 , so by rank-nullity theorem, the image space can have at most dimension 2. Also don't forget that a dimension of 0 is possible for the image space (in this case, this would happen if ff mapped everything in R2\mathbb{R} ^2 to (0,0,0)R3 (0,0,0) \in \mathbb{R} ^3 ), so the only vector in the image would be (0,0,0)(0,0,0) , which you can't put in a basis).

Like I asked before, have you come across rank-nullity theorem? It is a very useful/important theorem in Linear Algebra.

If not, there is another way you can see that you can't have the image space being dimension 3 in this case:

Suppose you could. What does this mean? It means you can find 3 linearly independent vectors in the image that span the image space. Since they are in the image, the three vectors will be of the form: f(u), f(v), f(w)f(\mathbf{u} ), \ f(\mathbf{v} ), \ f(\mathbf{w} ) with u, v, wR2\mathbf{u} , \ \mathbf{v} , \ \mathbf{w} \in \mathbb{R} ^2 by definition of the image.

However R2\mathbb{R} ^2 has dimension 2 and so at least one of the vectors u, v, w\mathbf{u} , \ \mathbf{v} , \ \mathbf{w} is a linear combination of the others. Say it's w\mathbf{w} , then α, βR\exists \alpha, \ \beta \in \mathbb{R} with w=αu+βv\mathbf{w} = \alpha \mathbf{u} + \beta \mathbf{v} . This means that:

f(w)=f(αu+βv)f( \mathbf{w} ) =f( \alpha \mathbf{u} + \beta \mathbf{v} )

=αf(u)+βf(v)= \alpha f( \mathbf{u} )+ \beta f( \mathbf{v} ) - since f is a linear map. This contradicts the linear independence assumption for the vectors f(u), f(v), f(w)f(\mathbf{u} ), \ f(\mathbf{v} ), \ f(\mathbf{w} ) , so the image really can't have dimension 3.

With these things in mind, look up the rank-nullity theorem and have another go at part (c) and post what you get this time around. It'll be good for you to do so.
(edited 8 years ago)
Reply 6
Original post by Star-girl
You made a typo in your original post. The question actually says that f:R2R3f: \mathbb{R} ^2 \mapsto \mathbb{R} ^3 , so by rank-nullity theorem, the image space can have at most dimension 2. Also don't forget that a dimension of 0 is possible for the image space (in this case, this would happen if ff mapped everything in R2\mathbb{R} ^2 to (0,0,0)R3 (0,0,0) \in \mathbb{R} ^3 ), so the only vector in the image would be (0,0,0)(0,0,0) , which you can't put in a basis).

Like I asked before, have you come across rank-nullity theorem? It is a very useful/important theorem in Linear Algebra.

If not, there is another way you can see that you can't have the image space being dimension 3 in this case:

Suppose you could. What does this mean? It means you can find 3 linearly independent vectors in the image that span the image space. Since they are in the image, the three vectors will be of the form: f(u), f(v), f(w)f(\mathbf{u} ), \ f(\mathbf{v} ), \ f(\mathbf{w} ) with u, v, wR2\mathbf{u} , \ \mathbf{v} , \ \mathbf{w} \in \mathbb{R} ^2 by definition of the image.

However R2\mathbb{R} ^2 has dimension 2 and so at least one of the vectors u, v, w\mathbf{u} , \ \mathbf{v} , \ \mathbf{w} is a linear combination of the others. Say it's w\mathbf{w} , then α, βR\exists \alpha, \ \beta \in \mathbb{R} with w=αu+βv\mathbf{w} = \alpha \mathbf{u} + \beta \mathbf{v} . This means that:

f(w)=f(αu+βv)f( \mathbf{w} ) =f( \alpha \mathbf{u} + \beta \mathbf{v} )

=αf(u)+βf(v)= \alpha f( \mathbf{u} )+ \beta f( \mathbf{v} ) - since f is a linear map. This contradicts the linear independence assumption for the vectors f(u), f(v), f(w)f(\mathbf{u} ), \ f(\mathbf{v} ), \ f(\mathbf{w} ) , so the image really can't have dimension 3.

With these things in mind, look up the rank-nullity theorem and have another go at part (c) and post what you get this time around. It'll be good for you to do so.


Thank you

I'll post what I get tomorrow. I have learned the R-N theorem but I didn't use it, but I understand how by using it you get that the dimension of the image can be at most 2.


Is my method correct though, I mean the way I'm trying to show that dimension can be 1,2 or 3 (can't be 3) in the original post?
Original post by e^x
Thank you

I'll post what I get tomorrow. I have learned the R-N theorem but I didn't use it, but I understand how by using it you get that the dimension of the image can be at most 2.


Is my method correct though, I mean the way I'm trying to show that dimension can be 1,2 or 3 (can't be 3) in the original post?


Sure. Ah OK. Whenever you see a question asking about the dimensions of images/kernels then r-n theorem is probably going to feature somewhere.

Sort of, even ignoring the lack of 0 dimensional case and the 3 dimensional case that is impossible. The question requires you to give specific values for a,b,c,d,e in each case. You have only given some of them for each case, i.e.you haven't actually done what the question asked. Plus yours is not technically a correct solution since for each of your answers I can give at least one set of values for the ones you haven't given that make the whole thing fail.

What you have given for 1,2 dimensional cases is a good start but it is not enough.
(edited 8 years ago)
Reply 8
Is this okay?
Reply 9
Original post by Star-girl
Sure. Ah OK. Whenever you see a question asking about the dimensions of images/kernels then r-n theorem is probably going to feature somewhere.

Sort of, even ignoring the lack of 0 dimensional case and the 3 dimensional case that is impossible. The question requires you to give specific values for a,b,c,d,e in each case. You have only given some of them for each case, i.e.you haven't actually done what the question asked. Plus yours is not technically a correct solution since for each of your answers I can give at least one set of values for the ones you haven't given that make the whole thing fail.

What you have given for 1,2 dimensional cases is a good start but it is not enough.


^
For 1 dimensional by saying e=a=c=0 I'm trying to imply that b can be anything, is that correct?
(edited 8 years ago)
Original post by e^x
^
For 1 dimensional by saying e=a=c=0 I'm trying to imply that b can be anything, is that correct?


No, it can't be anything because b=d=0 reduces it to the 0 - dimensional case. The question wants you to give explicit values for each of a, b, c, d, e. You need an a= ,b= ,c= ,d= ,e= for each case that work or you won't get all the marks.
Reply 11
I made a mistake when forming the equations from the matrix I forgot the d in the second equation:
(edited 8 years ago)
Reply 12
Original post by Star-girl
No, it can't be anything because b=d=0 reduces it to the 0 - dimensional case. The question wants you to give explicit values for each of a, b, c, d, e. You need an a= ,b= ,c= ,d= ,e= for each case that work or you won't get all the marks.


^^^
Original post by e^x
^^^


OK hang on a sec. Let me just think about your 2D case for a min. The rest are definitely correct.

I wasn't checking the equations though as you don't need them for part (c) - you only need the maps f(e1), f(e2)f(\mathbf{e_1} ) , \ f(\mathbf{e_2} ) to work out values for the 3 cases.
(edited 8 years ago)
Original post by e^x
^^^


For your "2D" case, taking linear combinations of (1,1,0){(1,1,0)} forms a line in R3\mathbb{R} ^3 i.e. this would give a 1D, not a 2D space. You need to pick different values of a,b,c,d,e.
(edited 8 years ago)
Reply 15
Original post by Star-girl
For your "2D" case, taking linear combinations of (1,1,0){(1,1,0)} forms a line in R3\mathbb{R} ^3 i.e. this would give a 1D, not a 2D space. You need to pick different values of a,b,c,d,e.


What about b=d=1 and a=c=e=0?

Also can u explain why u only need the maps f(e1) and f(e2)?
(edited 8 years ago)
Original post by e^x
What about b=d=1 and a=c=e=0?

Also can u explain why u only need the maps f(e1) and f(e2)?


That still only gives a 1D image space: see reasoning below.

We know that e1, e2R2\mathbf{e_1} , \ \mathbf{e_2} \in \mathbb{R} ^2 are linearly independent since they are the standard basis vectors. So it is possible to construct a 2-dimensional image space with f(e1), f(e2)f(\mathbf{e_1} ), \ f(\mathbf{e_2} ) as the basis vectors.

So all we need to do for part (c) for the 2-dimensional case is to choose a,b,c,d,e make sure that f(e1)=(a,b,0), f(e2)=(c,d,e)R3f(\mathbf{e_1} ) = (a,b,0), \ f(\mathbf{e_2} ) = (c,d,e) \in \mathbb{R} ^3 are linearly independent.

As analogues, for the 1-dimensional case we have two options:

1. Choose a,b,c,d,e such that either one of f(e1)=(a,b,0), f(e2)=(c,d,e)f(\mathbf{e_1} ) = (a,b,0), \ f(\mathbf{e_2} ) = (c,d,e) is the zero vector and the other one isn't.
2. Choose a,b,c,d,e such that f(e1)=(a,b,0), f(e2)=(c,d,e)f(\mathbf{e_1} ) = (a,b,0), \ f(\mathbf{e_2} ) = (c,d,e) are scalar multiples of each other and neither is the zero vector.

So your suggestion b=d=1, a=c=e=0 would make f(e1)=(0,1,0), f(e2)=(0,1,0)f(\mathbf{e_1} ) = (0,1,0), \ f(\mathbf{e_2} ) = (0,1,0) - they are the same vector so are scalar multiples of each other (option 2 above) and hence linearly dependent so you can't have both in the basis.

For the 0-dimensional case, we only have one option: choose a,b,c,d,e such that f(e1)=(a,b,0), f(e2)=(c,d,e)f(\mathbf{e_1} ) = (a,b,0), \ f(\mathbf{e_2} ) = (c,d,e) are both the zero vector, since we don't want any vectors in the basis.

This is the fastest way I can think of for doing part (c).
(edited 8 years ago)
Reply 17
Original post by Star-girl
That still only gives a 1D image space: see reasoning below.

We know that e1, e2R2\mathbf{e_1} , \ \mathbf{e_2} \in \mathbb{R} ^2 are linearly independent since they are the standard basis vectors. So it is possible to construct a 2-dimensional image space with f(e1), f(e2)f(\mathbf{e_1} ), \ f(\mathbf{e_2} ) as the basis vectors.

So all we need to do for part (c) for the 2-dimensional case is to choose a,b,c,d,e make sure that f(e1)=(a,b,0), f(e2)=(c,d,e)R3f(\mathbf{e_1} ) = (a,b,0), \ f(\mathbf{e_2} ) = (c,d,e) \in \mathbb{R} ^3 are linearly independent.


As analogues, for the 1-dimensional case we have two options:

1. Choose a,b,c,d,e such that either one of f(e1)=(a,b,0), f(e2)=(c,d,e)f(\mathbf{e_1} ) = (a,b,0), \ f(\mathbf{e_2} ) = (c,d,e) is the zero vector and the other one isn't.
2. Choose a,b,c,d,e such that f(e1)=(a,b,0), f(e2)=(c,d,e)f(\mathbf{e_1} ) = (a,b,0), \ f(\mathbf{e_2} ) = (c,d,e) are scalar multiples of each other and neither is the zero vector.

So your suggestion b=d=1, a=c=e=0 would make both f(e1)=(0,1,0), f(e2)=(0,1,0)f(\mathbf{e_1} ) = (0,1,0), \ f(\mathbf{e_2} ) = (0,1,0) - they are the same vector so are scalar multiples of each other and hence linearly dependent so you can't have both in the basis.

For the 0-dimensional case, we only have one option: that f(e1)=(a,b,0), f(e2)=(c,d,e)f(\mathbf{e_1} ) = (a,b,0), \ f(\mathbf{e_2} ) = (c,d,e) are both the zero vector, since we don't want any vectors in the basis.

This is the fastest way I can think of for doing part (c).


Okay so if I pick b=d= 1 and everything else 0 I just get f(e1)=f(e2)=(0,1,0)

So I can pick a=1 and e=1 to get f(e1)= (1,0,0) and f(e2)=(0,0,1) ?
(edited 8 years ago)
Original post by e^x
Okay so if I pick b=d= 1 and everything else 0 I just get f(e1)=f(e2)=(0,1,0)

So I can pick a=1 and e=1 to get f(e1)= (1,0,0) and f(e2)=(0,0,1) ?


Yes that works.
Reply 19
I need help with 6)c)

For 6)a) I showed the 4 conditions hold for inner product.

For 6)b) I use the Gram-Schimdt process to construct an orthonormal set.
(edited 8 years ago)

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