Question

determine whether multiplication by A is one to one linear transformation
a)A=[ 1 -1
2 0
3 -4]
b) A=[ 1 2 3
-1 0 4]

Answer 1

One way to tell if a linear transformation is one-to-one is if the nullspace or kernel of the matrix is just the zero vector. We can find these by row-reducing
\[ \pmatrix{1&-1&0\\2&0&0\\3&-4&0}\]
to get
\[ \pmatrix{1&0&0\\0&1&0\\0&0&0}\]
This tells us that the only vector that goes to [0,0,0] when multiplied by A is [0,0]. Thus, A is one-to-one.

For the second one, do the same thing:
\[ \pmatrix{1&2&3&0\\-1&0&4&0} \rightarrow \pmatrix{0&1&7/2&0\\1&0&-4&0}\]
Thus if x=4z and y = -7z/2 for any z, when we multiply by A, we get [0,0]. We have infinitely many solutions, so A can't be one-to-one.

Answer 2

i salute you sir i got the concept

Answer 3

If the null space is just comprised of the zero vector, then it isn't enough for it to be injective transformation. If the linear transformation goes from vectorspace V to vectorspace W then for injectivity dim(v) has to be equal to dim(w). In your case this isn't true for both a and b.

Answer 4

so sir what is the correct solution

Answer 5

Not true; injectiveness implies dim(V) <= dim(W), but it's certainly not required for them to be equal. Embedding a line into a plane is a linear transformation from R^1 to R^2 and is clearly injective.

In fact, in any ring, if the kernel of a homomorphism is trivial, then the homomorphism is one-to-one. It's a basic but incredibly useful tidbit.

Answer 6

sir please i am not getting your point how i solve these question or can you solve one of them

Answer 7

What about points in R^2 not on that line, don't they have to have a counterpart in R^1?

Answer 8

Nope. I think you're mixing up injective and bijective.

Answer 9

Oh, I see, I mixed having a inverse function and invectiveness. Thanks.

Answer 10

(In fact, in any ring, if the kernel of a homomorphism is trivial, then the homomorphism is one-to-one.)
can you please explain this statement

Answer 11

Well, it's not necessary to know for a class in Linear Algebra. I was just pointing out that this technique is a special case of a more general one. A "ring" is a structure with two operations, and these vector spaces are examples. Another example would be the set of all integers with the usual addition and multiplication. A "homomorphism" is a function that preserve these operations: f(u+v) = f(u) + f(v) and f(rv) = r f(v), which you might recognize as the definition of a linear transformation.
(Actually the vector spaces are Modules, which are related.)

Answer 12

Sorry, I didn't mean to confuse you!

Answer 13

thanks sir i know these operation very well