1. Image and Kernel

Where T : R^4 goes to R^2 is defined by
$T(x1,x2,x3,x4) = (x3, 5x1 + 3x4)$

for image do you just put

(1,0,0,0)
(0,1,0,0)
(0,0,1,0)
(0,0,0,1)

into the RHS

which equals (0,5) (0,0) (1,0) (0,3)

and for the kernel x3 = 0

5x1 +3x4 = 0
so they all = 0

so the basis is 0

2. T is a linear transformation, because given vectors x,y and constant a and b, you get [LaTeX ERROR: Convert failed] .

What you do next is take the basis vectors for [LaTeX ERROR: Convert failed] which are just [LaTeX ERROR: Convert failed] which is just 1 for entry i and 0 otherwise. Then you map [LaTeX ERROR: Convert failed] to [LaTeX ERROR: Convert failed] . The image then is all linear combinations of those, i.e. [LaTeX ERROR: Convert failed] . The image is all these

Another way to look at it is this. Since the [LaTeX ERROR: Convert failed] 's form a basis, all vectors in [LaTeX ERROR: Convert failed] are of the form [LaTeX ERROR: Convert failed] . Applying the linear transformation gives precisely [LaTeX ERROR: Convert failed] . Now since a,b,c and d can be anything, then you get that the image is the whole 2D plane or [LaTeX ERROR: Convert failed] .

As for the Kernel, the kernel is all vectors in [LaTeX ERROR: Convert failed] that map to 0. From the above you can see that this is true when c = 0 and 5a + 3d = 0. in other words, if a = -3d/5, then the vector x is in the kernel.

So in essence, you are right about the first part. You have to plug each basis vector in the transformation to get the the image. You have to remember to add the arbitrary constants to each basis vector to make sure you are give all points of [LaTeX ERROR: Convert failed] a chance.

3. Originally Posted by Vlasev
T is a linear transformation, because given vectors x,y and constant a and b, you get [LaTeX ERROR: Convert failed] .

What you do next is take the basis vectors for [LaTeX ERROR: Convert failed] which are just [LaTeX ERROR: Convert failed] which is just 1 for entry i and 0 otherwise. Then you map [LaTeX ERROR: Convert failed] to [LaTeX ERROR: Convert failed] . The image then is all linear combinations of those, i.e. [LaTeX ERROR: Convert failed] . The image is all these

Another way to look at it is this. Since the [LaTeX ERROR: Convert failed] 's form a basis, all vectors in [LaTeX ERROR: Convert failed] are of the form [LaTeX ERROR: Convert failed] . Applying the linear transformation gives precisely [LaTeX ERROR: Convert failed] . Now since a,b,c and d can be anything, then you get that the image is the whole 2D plane or [LaTeX ERROR: Convert failed] .

As for the Kernel, the kernel is all vectors in [LaTeX ERROR: Convert failed] that map to 0. From the above you can see that this is true when c = 0 and 5a + 3d = 0. in other words, if a = -3d/5, then the vector x is in the kernel.

So in essence, you are right about the first part. You have to plug each basis vector in the transformation to get the the image. You have to remember to add the arbitrary constants to each basis vector to make sure you are give all points of [LaTeX ERROR: Convert failed] a chance.
So how do i do the kernel as the sum of the dimension of the kernel and the image must equal 4. The dimension of the image = 4 so the kernel must be 0.

But 5x1 + 3x4 = 0 so x1 and x4 can be anything so the dimension isn't 0, so i have done something wrong.

4. What im trying to say the image is (0,5) (0,0) (1,0) (0,3) so the dimension is 4

and for the kernel x3 = 0

5x1 +3x4 = 0

x1 could equal 3 and x4 could equal -5
or x1 could = 6 and x4 could equal -10

So the dimension = 2

But Dimension of image + dimension of kernal = 4 as its R^4

But 4 + 2 = 6

So what have i done wrong?

5. The "kernel" of linear transformation T is all vectors u such that Tu= 0. Here, T(x1,x2,x3,x4) = (x3, 5x1 + 3x4) so if T(x1, x2, x3, x4)= (0, 0), we must have x3= 0, 5x1+ 3x4= 0. The second equation we can write as x4= -(5/3)x1. Then (x1, x2, x3, x4)= (x1, x2, 0, -(5/3)x1)= x1(1, 0, 0, -5/3)+ x2(0, 1, 0, 0). Those two vectors are a basis for the kernel. The dimension of the kernel is 2.

The "image" cannot have dimension 4 because the image is a subspace of the range space, $R^2$, and cannot have dimension larger than 2.
The image is the set of all (x, y) such that (x, y)= T(a, b, c, d)= (c, 5a+ 3d) for some a, b, c, d. Since x= c and c can be anything, x can be anything. Since y= 5a+ 3d and both a and d can be any real numbers, y can be any real number. That is, the image is all of $R^2$ and has dimension 2.

Again, if T is a linear transformation from vector space U to vector space V, the kernel is a subspace of U and must have dimension less than or equal to the dimension of U. The image, however, is a subspace of V and must have dimension less than or equal to the dimension of V.

Obviously, the vectors you got, (0,5) (0,0) (1,0) (0,3) are NOT independent and do not form a basis.

6. Originally Posted by HallsofIvy
Obviously, the vectors you got, (0,5) (0,0) (1,0) (0,3) are NOT independent and do not form a basis.
So the dimension is 0?

7. ??? I had just said "That is, the image is all of $R^2$ and has dimension 2." I have no idea how you conclude from the fact that a particular set of vectors is not a basis, that the dimension must be 0.

The set of vectors {(0, 5), (0, 0), (1, 0), (0, 3)} is not independent, first because a set of independent vectors cannot contain the 0 vector. Obviously, 0(0, 5)+ 1(0, 0)+ 0(1, 0)+ 0(0, 3)= 0 without all coefficients being 0. If you throw the 0 vector out, you have {(0, 5), (1, 0), (0, 3)} which is not independent because (0,5) and (0,3) are multiples of each other. Obviously, again, 3(0, 5)+ 0(1, 0)+ (-5)(0, 3)= (0, 0) without all coefficients being 0. If you throw out either (0, 5) or (0, 3) you get {(1, 0), (0, 3)} (or {(0, 5), (1, 0)}) which is an independent set and is a basis for the image.

8. Originally Posted by HallsofIvy
??? I had just said "That is, the image is all of $R^2$ and has dimension 2." I have no idea how you conclude from the fact that a particular set of vectors is not a basis, that the dimension must be 0.

The set of vectors {(0, 5), (0, 0), (1, 0), (0, 3)} is not independent, first because a set of independent vectors cannot contain the 0 vector. Obviously, 0(0, 5)+ 1(0, 0)+ 0(1, 0)+ 0(0, 3)= 0 without all coefficients being 0. If you throw the 0 vector out, you have {(0, 5), (1, 0), (0, 3)} which is not independent because (0,5) and (0,3) are multiples of each other. Obviously, again, 3(0, 5)+ 0(1, 0)+ (-5)(0, 3)= (0, 0) without all coefficients being 0. If you throw out either (0, 5) or (0, 3) you get {(1, 0), (0, 3)} (or {(0, 5), (1, 0)}) which [b]is[\b] an independent set and is a basis for the image.
Im lost

The Dimension of the image + the dimension of the kernel must = n

n is from R^n which is 4 in my case.

If you're saying there is no image because it is linearly dependent the dimension must be 0.

What is the dimension of the image and kernel?

9. ?? I have twice now said that the image is all of $R^2$ and so has dimension 2 and that the kernel has dimension 2.

If you're saying there is no image because it is linearly dependent the dimension must be 0.
I did NOT say that. YOU said that in post 6 and I said, in post 7, that that was wrong.

The set of four vectors that you claimed in post 4 was a basis was not because it was dependent. You immediately jumped to the conclusion that therefore the dimension must be 0. I showed that there is a subset of the vectors you gave, containing two vectors, that were independent and a basis.

10. Originally Posted by HallsofIvy
?? I have twice now said that the image is all of $R^2$ and so has dimension 2 and that the kernel has dimension 2.

I did NOT say that. YOU said that in post 6 and I said, in post 7, that that was wrong.

The set of four vectors that you claimed in post 4 was a basis was not because it was dependent. You immediately jumped to the conclusion that therefore the dimension must be 0. I showed that there is a subset of the vectors you gave, containing two vectors, that were independent and a basis.