# Linearly Independent R^3 -> Linearly Independent R^4

1. Feb 2, 2010

### svasudevan

Standard Matrix for R^3 -> R^4 where if domain is Linearly Independent so is codomain

1. The problem statement, all variables and given/known data

5. Suppose T : R^3 -> R^4 is a linear transformation with the following property:
For any linearly independent vectors v_1, v_2 and v_3 in R^3, the images T(v_1),
T(v_2) and T(v_3) are linearly independent in R4.
(a) Give an example of such a linear transformation. Give its standard matrix and the
reduced row echelon form of this matrix.
(b) Work out all possible shapes of the reduced row echelon form for such a matrix.
Use the symbols 0, 1, and * where * indicates an entry which may take on the value
of any real number.

2. Relevant equations

No relevant equations. The linear transformation must be given as a standard matrix with dimensions 3 * 4 since that would lead to a vector in R^4.

3. The attempt at a solution

So far I know that since the vectors are linearly independent, the reduced form of both vectors must have pivots in each column. I reasoned that the function must be injective because each vector in R^3 can only correspond to 1 vector in R^4 but there are more vectors in R^4 than R^3.

This question seems confusing and I can't put my thoughts together. This is not a very advanced class (it is just a standard freshman linear algebra course for scientists and scientists).

Last edited: Feb 2, 2010
2. Feb 3, 2010

If v = (a, b, c), try T(v) = (a, b, c, 0).

3. Feb 3, 2010

### svasudevan

Thank you for your response. After more thinking I was able to find that the matrix must be linearly independent (pivot in every column) and that when it is row reduced, the matrix would be:

[ 1 0 0 ]
[ 0 1 0 ]
[ 0 0 1 ]
[ 0 0 0 ]

4. Feb 3, 2010

### Staff: Mentor

Don't you means that the columns in the matrix must be linearly independent?