# Basis for Image and Kernel of matrix

1. Sep 19, 2009

### boneill3

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

Find an Basis for Image and Kernel of the matrix.
$$\left( \begin{array}{ccc} 2 & 1 & 3 \\ 0 & 2 & 5 \\ 1 & 1 & 1 \end{array} \right)$$

2. Relevant equations

3. The attempt at a solution
To find the kernel I solve the equation Ax = 0

I put the matrix in row reduced echelon form which is the identity matrix.
$$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

Therefore its the equation

x = 0
y = 0
z = 0

The kernel basis is just the unit basis, {(1,0,0),(0,1,0),(0,0,1)}

For the image basis I've seen that you can use the pivots of the rref matrix and use the corresponding column vectors of the original Matrix as the image basis.

So is that just

{(2,0,1),(1,2,1),(3,5,1)} ?

Last edited: Sep 19, 2009
2. Sep 19, 2009

### boneill3

Sorry wrong latex code

3. Sep 20, 2009

### Billy Bob

No. You found the kernel to be trivial, i.e. {(0,0,0)}. The basis for the kernel, then, is the empty set.

This is correct. You could also use {(1,0,0),(0,1,0),(0,0,1)} since the image is all of R3. You might have to write them as column vectors, depending on the conventions your instructor has adopted.

4. Sep 20, 2009

Thanks