Question

A few True and False.
If A is invertible then Ax=0 must have non-trivial solution
If columns of A are linearly dependent, the one cannot find a matrix C such that CA=I
If A is a 3x2 Matrix the transformation x to Ax cannot be one to one

Answer 1

I can sort of figure out the first two but I'm stumped on the third one

Answer 2

What are you answers to the first two?

The third: as A is 3x2, the column vector x must be 2x1, and the resulting column vector Ax is 3x1. Hence A takes something 2-dimensional x, and turns it into something 3-dimensional Ax. Hence A is a linear function R^2 --> R^3

Is possible to have a 1 to 1 linear function from R^2 --> R^3?
It certainly isn't possible to have a onto linear function R^2 --> R^3.

Answer 3

False and True.
Its the explanation that im missing for the first to. I know that an invertible matrix only has the trivial solution. I just dont know why.

For the second one, A inverse is the only thing you can multiply A in order to get I. And if A is invertible, the columns have to be linearly independent.

Third:
It's definitely not onto. But I guess it can be one to one.

Answer 4

First one: If A is invertible, then A^-1 exists and we and multiply on the left both sides of Ax = 0 by A^-1 to get

A^-1.Ax = A^-1 . 0
=> Ix = 0
=> x = 0

#3: Yes, for example
1 0
0 1
0 0

is an example of a 1:1 map R^2 to R^3. It takes (x,y)^T to (x,y,0)^T