Given an nxn matrix, if a b exists so Ax=b has no solutions, can A be one-to-one?
I understand that as a linear transformation, you need things such as (to be one-to-one as a linear trans)
1. n pivots
2. Only the trivial solution exists to Ax=0
Ax=b having no solutions is a bit of an oddball for me.
The Attempt at a Solution
If I set up an augmented matrix such as [I | b] (Identity matrix | some vector b)
By removing one pivot from the Ident. matrix I can see that I've created an Ax=b where b has no solution. This is akin to a linearly independent set (which contains the zero vector). in Rn space.
It technically doesn't span Rn space nor Rm space fully... I think? As it will have a row of zeroes and a column of zeroes.
But does this truly affect its ability to be one-to-one?
I don't see how it does........
For instance, let's say you have a 3x3 ID matrix A, but one pivot is missing.
If that's put into an augmented matrix[A|b], if a row with zeroes has a number to the right of it, it means there's no solution for this b. So doesn't this mean the situation is just n/a?