# Linear transformation, Linear algebra

1. Feb 24, 2013

### Nikitin

1. The problem statement, all variables and given/known data
Describe the possible echelon forms of the standard matrix for the linear transformation T.

T: |R3 --> |R4 is one to one.

3. The attempt at a solution

T(x)=Ax. Right? So A must be the standard matrix.

I got this: A =

| £ * * |
| 0 £ * |
| 0 0 £ |
| ? ? ? |

Where £ represents a pivot position, and * any random number.

The only problem is that I have no idea what to put in the lowest row.. Can you guys help?

2. Feb 24, 2013

### Karnage1993

First of all, do you know when a matrix is one-to-one? What conditions must be true?

3. Feb 24, 2013

### Nikitin

All the columns of A must be linearly independent. I don't understand why this is so, though.

4. Feb 24, 2013

### Karnage1993

Yes, that is true. Another way of saying that the columns are L.I. is that there is a pivot in each column of A. If you don't understand this, you should check out the proof of the theorem that relates one-to-one and linearly independence.

5. Feb 24, 2013

### Nikitin

It is linearly independent when only a trivial solution to the matrix being equal to [0,0,0,0] exists (ie, all the unknowns must be zero).

What I don't understand is why the matrix A needs to be linearly independent for the function T(x) being one-to-one.

I also still don't know what the lowest row should be.. Just zeros? But in that case the vector isn't really being transformed to the |R4 codomain...

Last edited: Feb 24, 2013
6. Feb 24, 2013

### Karnage1993

No, the trivial solution would not be [0,0,0,0]. There are only three variables in four equations.

The matrix A must have column vectors linearly independent for T to be one-to-one because that is the only instance where the solution to the system would consist only of the zero vector. Remember that by definition of one-to-one, the $\textrm{ker(T)} = \{\vec 0\}$. In other words, the null space is spanned by only the zero vector. If the vectors were not linearly independent, then the solution would not consist of solely the zero vector.

As to your answer to the question at hand, yes the last row consisting of zeroes would work. After all, you are showing the row-reduced echelon form, right?

Yes, the codomain is $\mathbb{R}^4$ but what does that have to do with T being one-to-one?

EDIT: Correction, you're right, I misspoke. I read it too quickly and thought you were referring to the three variables.

Last edited: Feb 24, 2013
7. Feb 24, 2013

### Nikitin

Yes. The matrix is linearly independent when there only exists a trivial solution for all the equations being equal to zero. In other words, x= only [0,0,0] for Ax=[0,0,0,0]. I obviously know this, you were just not understanding me.

The question is WHY T(x) is 1-to-1 over its whole domain if its standard matrix is linearly independent. Repeating the definition of l.i. and the rule of 1-to-1 over and over doesn't help me.

----

Can somebody tell me what should be in the bottom-row of my matrix (straight zeroes?) in the OP, and why? The matrix I posted is already in a linearly independent form, so can I just put whatever I like in the bottom row?

Last edited: Feb 24, 2013