Find the linear transformation

[Physicsissuef]

dim(f(U))=dim(U)-dim(kernel(f)). No, that's not different from what Halls is saying. It's not a definition, it's a theorem.

kernel(f)?
Is it kernel(f(U))?
Why dim(kerF)=1
f(x,y)=(x,0)
(x,0)=(0,0)
x=0
so kerF=0

 

[Dick]

(0,1) is in the kernel of f.

 

[Physicsissuef]

(0,1) is in the kernel of f.

How did you find it?

 

[Dick]

It found me. f(0,1)=(0,0). I just looked at the definition of f.

 

[Physicsissuef]

It found me. f(0,1)=(0,0). I just looked at the definition of f.

Sorry, and how will you find it for my example?
f(x,y)=(x+y,y)??
(x+y,y)=(0,0)
or?

 

[Dick]

(x+y,y)=(0,0) means x+y=0 and y=0. Solve those linear equations. What's the solution?

 

[Physicsissuef]

(x+y,y)=(0,0) means x+y=0 and y=0. Solve those linear equations. What's the solution?

yes
x=0
y=0
so the kernel is 0.

I am saying that for (x,0)
(x,0)=(0,0)
so the kernel is 0
cos x=0

 

[Dick]

If you are saying for f(x,y)=(x,0) that the kernel is (0,0), then you are wrong. Yes, the equations you get are x=0 and 0=0. There's no y in those equations. That means y can be anything.

 

[Physicsissuef]

If you are saying for f(x,y)=(x,0) that the kernel is (0,0), then you are wrong. Yes, the equations you get are x=0 and 0=0. There's no y in those equations. That means y can be anything.

R² tells me that there should be 2 variables, that's why, right?
can I write (x,0), just like x?
in that case R²-> R, right?

 

[Dick]

R² tells me that there should be 2 variables, that's why, right?
can I write (x,0), just like x?
in that case R²-> R, right?

You want to figure out for what values of x and y that f(x,y)=(x,0)=(0,0). x=0 and y=anything. Period.

 

[Physicsissuef]

Can I write x instead of (x,0)
so f(x,y)=x instead of f(x,y)=(x,0)?

 

[Dick]

You can write anything you please. As long as you know f(x,y)=x and f(x,y)=(x,0) are two different linear maps. Their 'codomains' are different.

 

[Physicsissuef]

You can write anything you please. As long as you know f(x,y)=x and f(x,y)=(x,0) are two different linear maps. Their 'codomains' are different.

and f(x,0)=(x,0)
f(0,y)=(0,y)
Is not correct right? (about f(x,y)=(x,0) )
it is correct that
f(x,0)=(x,0)
f(0,y)=(0,0)
right?

 

[Dick]

Yes, that's right.

 

[Physicsissuef]

and dim(f(U))=dim(kerf), just in this case, or always??

 

[Dick]

Just in this case. Not always. If f(x,y,z)=(x,0,0), dim(f(U))=1 and dim(ker(f))=2.

 

[Physicsissuef]

Just in this case. Not always. If f(x,y,z)=(x,0,0), dim(f(U))=1 and dim(ker(f))=2.

Ok, thank you very much for the help. I appreciate your efforts to help me. Thanks again.