Thread: Lin. Transformation

NOTE: [a, b, c, d] <-- form is a vector and BOLD denotes a vector, unless referring to R.

Suppose that T is a linear transformation from R^4 to R^3, defined by:

T(e_1) = [1, 5, 3], T(e_2) = [2, 6, 2], T(e_3) = [3, 7, 1], T(e_4) = [4, 8, 0].

1.) Find a standard Matrix A for the transformation.

2.) Find T([5, -1, 3, 2])

3.) Find an x such that T(x) = [-5, -9, 1]

4.) Is T 1-1? If yes, explain why it is. If not, give 2 different vectors that map to the same vector.

5.) Is T onto? If yes, explain why it is. If not, give a vector in the codomain that's not in the range.

WORK (PLEASE CHECK):

1.) The std. matrix A is:

[[1, 2, 3, 4], [5, 6, 7, 8], [3, 2, 1, 0]] (3x4 matrix)

2.) In order to find T([5, -1, 3, 2]), I would just take the matrix I formed in 1.) and multiply it by this vector, right?

3.) To find an x such that T(x) = [-5, -9, 1], I'd augment this vector to my matrix in 1.), row reduce, etc.

4.) NOT SURE!!

5.) It's clearly not onto, right? Since, we're taking a transformation from R^4 to R^3...everything can't possibly go from something bigger to something smaller.

Originally Posted by Ideasman

4.) Is T 1-1? If yes, explain why it is. If not, give 2 different vectors that map to the same vector.
.

Is the kernel trivial? Meaning is the trivial element in R^3 the only element mapped into the zero vector?

Originally Posted by ThePerfectHacker

Is the kernel trivial? Meaning is the trivial element in R^3 the only element mapped into the zero vector?

Um, I'm not sure. I don't even know what kernel means. I'm assuming the others are right..that method that is?

Hmm, these look wrong, so let me show you what I have, and I'll try use LaTeX to make it look good:

1.)



2.) For this one, I took my matrix above and multiplied it by:



to get:



3.)



Then I row-reduced it:



Now I'm not sure what "an x" means...

4.) Still need help on determining this.

5.) I can see why it is NOT onto makes sense..not sure how to say why..


Hope that helps make things look a little better.