Where T : R^4 goes to R^2 is defined by

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

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- August 1st 2010, 05:36 AMadam_leedsImage and Kernel
Where T : R^4 goes to R^2 is defined by

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 - August 1st 2010, 12:36 PMVlasev
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. - August 6th 2010, 02:44 AMadam_leeds
- August 7th 2010, 02:42 AMadam_leeds
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? - August 7th 2010, 05:17 AMHallsofIvy
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, , 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 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. - August 11th 2010, 04:18 AMadam_leeds
- August 11th 2010, 06:51 AMHallsofIvy
??? I had just said "That is, the image is all of 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. - August 11th 2010, 09:35 AMadam_leeds
- August 12th 2010, 04:32 AMHallsofIvy
?? I have twice now said that the image is all of and so has dimension 2 and that the kernel has dimension 2.

Quote:

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

**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.

I advise you to talk to your teacher about this. You seem to be very shaky on the basic definitions of "independent", "basis", and "dimension". - August 13th 2010, 04:08 AMadam_leeds
It's not in R^2 though its in R^4 its what it coming from not what it is going to.

Look at my lecture notes it goes from R^2 to R^3 and at the bottom the sum of the dimension of the image + kernel = 2 not 3

http://i34.tinypic.com/sy7b0o.jpg - August 13th 2010, 04:15 AMadam_leeds
Wait for my original post question, so the dimension of the image = 2

So the dimension of the kernel must = 2 - August 13th 2010, 04:36 AMadam_leeds
I will ask my teacher like you said. Thanks for your time.