# vectors that span the kernel of a matrix

• Sep 22nd 2009, 09:17 PM
noles2188
vectors that span the kernel of a matrix
Find vectors that span the kernel of matrix A.

A = [1 2 3 4 ; 0 1 2 3 ; 0 0 0 1] where ; is a new row.

I found that rref(A) = [1 0 -1 0 ; 0 1 2 0 ; 0 0 0 1] where ; is a new row.

Can anyone tell me where I go from here? Thanks
• Sep 23rd 2009, 04:18 AM
aman_cc
Solution space of AX = 0 will be equal to the the kernel of the mapping represented by A

If you solve the equation we get the solution space is of dimension 1 with basis = [1 -2 1 0]
• Sep 23rd 2009, 05:55 AM
HallsofIvy
Quote:

Originally Posted by noles2188
Find vectors that span the kernel of matrix A.

A = [1 2 3 4 ; 0 1 2 3 ; 0 0 0 1] where ; is a new row.

I found that rref(A) = [1 0 -1 0 ; 0 1 2 0 ; 0 0 0 1] where ; is a new row.

Can anyone tell me where I go from here? Thanks

$A= \begin{bmatrix}1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 1\end{bmatrix}$.
and
$rref(A)= \begin{bmatrix}1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$.
Actually you can work from either A itself of rref(A). The "kernel" of a matrix is the subspace of all vectors that it maps to 0. So if $v= \begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}$is a vector in the kernel, we must have $A(v)= \begin{bmatrix}1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$.

That is the same as the system of equation a+ 2b+ 3c+ 4d= 0, b+ 2c+ 3d= 0, d= 0. You could, of course, reduce those to a- c= 0, b+ 2c= 0, d= 0, the row reduced form. From the first equation, a= c. From the second b= -2c. c can by any number, d must be 0. So any vector satisfying those equations, any vector in the kernel of A, must be of the form $\begin{bmatrix}c \\ -2c\\ c\\ 0\end{bmatrix}= c\begin{bmatrix}1 \\ -2 \\ 1 \\ 0\end{bmatrix}$.

A basis for the kernel consists of that single vector $\begin{bmatrix}1 \\ -2 \\ 1 \\ 0\end{bmatrix}$, as aman_cc says.