# Invertible Matrix Theorem

1. Apr 26, 2015

### henry3369

True or False:
If the linear combination x -> Ax maps Rn into Rn, then the row reduced echelon form of A is I.

I don't understand why this is False. My book says it is false because it is only true if it maps Rn ONTO Rn instead of Rn INTO Rn. What difference does the word into make?

2. Apr 26, 2015

### Staff: Mentor

Consider the transformation whose matrix looks like this:
$$A = \begin{bmatrix}1 & 0 & 0 & \dots & 0 \\ 0 & 0 & 0 & \dots & 0 \\ \vdots \\ 0 & 0 & 0 & \dots & 0 \end{bmatrix}$$
This transformation maps a vector x to its projection on the $x_1$ axis, a map from Rn into Rn (but not onto Rn). Does it seem likely to you that this matrix will row reduce to the identity matrix?

3. Apr 26, 2015

### henry3369

So if it maps onto Rn does that mean that it maps x into every position in Rn?

4. Apr 26, 2015

### henry3369

Also can you clarify why these two statements are equivalent in the Invertible Matrix Theorem::
1. The equation Ax = b has at least one solution for each b in Rn
2. The linear transformation x -> Ax is one-to-one.

What is throwing me off is the word least in statement 1. Shouldn't it have exactly one solution for every b in order for the transformation to be one-to-one?

5. Apr 26, 2015

### Staff: Mentor

Yes, I'm bothered by it as well. It's technically correct, but misleading, as it seems to imply that for some b there might two input x values. That can't happen if the transformation is one-to-one, though.

To me, it's sort of like saying 3 + 4 is at least 7.

6. Apr 26, 2015

### WWGD

Yes, this is rank-nullity, together with the fact that a linear map is injective iff it has a trivial kernel.