Finding a Basis for the Kernel Space of a Matrix - Solving the RREF Method

[theRukus]

Homework Statement

Find a basis for the kernel space of the following matrix:
-1 -2 -1 2 2
-2 -4 -4 10 2
1 2 2 -5 2
-1 -2 0 -1 0

row reduce to

1 2 0 1 0
0 0 1 -3 0
0 0 0 0 1
0 0 0 0 0


Somehow read the solution as

{ [-2 1 0 0 0]^T, [-1 0 3 1 0]^T }

.. I don't understand how to read the basis from the RREF. Could someone shed some light for me? Thanks so much!

 

[lanedance]

The basis for the null space is orthogonal to each vector in the rowspace and so to each vector in the rowspace of the rref form

 

[HallsofIvy]

In more "simple-minded" terms, you can think of your row reduced matrix as referring to
$$\begin{bmatrix}1 & 2 & 0 & 1 & 0 \\ 0 & 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$

or simply the equations $x_2+ 2x_2+ 0x_3+ x_4+ 0x_5= 0$, $0x_1+ 0x_2+ x_3- 3x_4+ 0x_5= 0$, $0x_1+ 0x_2+ 0x_3+ 0x_4+ x_5= 0$, $0x_1+ 0x_2+ 0x_3+ 0x_4+ 0x_5= 0$.

If you interpret each of those equations as a "dot product" you can see how the vectors in the kernel must be "orthogonal" to the rows of the reduced matrix. Of course, you can also see, from the third equation that $x_5$ must be 0, from the second that $x_3- 3x_4= 0$ so that $x_3= 3x_4$, and from the first equation that $x_1+ 2x_2+ x_4= 0$ so that $x_1= -2x_2- x_4$. That is, we can write
$$\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\end{bmatrix}= \begin{bmatrix}-2x_2- x_4 \\ x_2 \\ 3x_4 \\ x_4 \\ 0\end{bmatrix}$$
$$= \begin{bmatrix}-2x_2 \\ x_2 \\ 0 \\ 0 \\ 0\end{bmatrix}+ \begin{bmatrix}-x_4 \\ 0 \\ 3x_4 \\ x_4 \\ 0\end{bmatrix}$$
$$= x_2\begin{bmatrix}- 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}+ x_4\begin{bmatrix}-1 \\ 0 \\ 3 \\ 1 \\ 0 \end{bmatrix}$$

 

[theRukus]

Thanks heaps hallsofivy, that cleared things up for me. Wish me luck on my exam!