consider the linear transformation T:R4--> R4 defined by

[w] [w+z]

T [x] = [y]

[y] [x-y]

[z] [w]

for each i have to justify the answer

a. is T injective?

b. is T surjective?

c.is T invertible?

can anybody please help?!

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- August 13th 2008, 02:41 PMlinearalgebraduncelinear transformations
consider the linear transformation T:R4--> R4 defined by

[w] [w+z]

T [x] = [y]

[y] [x-y]

[z] [w]

for each i have to justify the answer

a. is T injective?

b. is T surjective?

c.is T invertible?

can anybody please help?! - August 14th 2008, 01:31 AMflyingsquirrel
- August 14th 2008, 02:13 AMlinearalgebradunce
Ok I understand the definitions but am still struggling to decide if T is injective or not, for the kernel(T) obviously we know this equals the null space(T) so if I find one vector in R4 such that Tv=the zero vector does this mean it is injective.

eg. if I choose the vector

[1]

[0]

[0]

[-1] - August 14th 2008, 02:48 AMflyingsquirrel
To show that is injective you have to show that the only vector such that is ; that's not what you're doing. Anyway, this idea is often used to show that a linear application is

*not*injective. For if you find a non-zero vector such that then contains both 0 and hence and can't be injective.

EDIT: by the way, for :D - August 14th 2008, 03:07 AMlinearalgebradunce
Oh yeah I see that that is incorrect. So T(0) = 0 so T is injective!? Is that what you're saying?

- August 14th 2008, 04:30 AMflyingsquirrel
No! is true for

*any*linear application , even if is not injective (this result comes from ). A linear application is injective if . You are saying that a linear application is injective if . These two sentences are very different and only the former is correct.

Now, how can one show that ? Well, the only thing we have to do is solve for . Let .