Linear algebra, can A be one-to-one given a case

  • #1

Homework Statement


Given an nxn matrix, if a b exists so Ax=b has no solutions, can A be one-to-one?

Homework Equations


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?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Given an nxn matrix, if a b exists so Ax=b has no solutions, can A be one-to-one?
<snip>

Hint: If Ax = b has no solution then the transformation is not onto. So the question could be rephrased as "Can a linear transformation that is 1-1 be not onto? Does that help? Think about ##R^2## and ##R^3##.

[Edit]: Never mind, I missed the nxn.
 
Last edited:
  • #3
Ray Vickson
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Homework Statement


Given an nxn matrix, if a b exists so Ax=b has no solutions, can A be one-to-one?

Homework Equations


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?

There is a very basic theorem in linear algebra: if A is nxn (which you say yours is), then either (1) the equation Ax = b has a unique solution for any n-vector b on the right; or (2) Ax = b has either no solution (for some b) or infinitely many solutions (for some other b).

Case (1) is the same as saying: Ax = 0 if and only if x = 0.
 
  • #4
Thanks everyone. I think that helped.
 

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