Problem:

I can't find answer to a problem. The problem is like this:

Find the basis for $\displaystyle \text{kernel of}(T)$ where $\displaystyle T$ is a linear transformation.

Suppose $\displaystyle T:R^3 \to R^3,\quad T(x,y,z) = (x + 2y, y + 2z, z + 2x)$

Part of Solution:

The problem is solved like this:

$\displaystyle A = \left[{\begin{array}{ccc} 1 & 2 & 0\\ 0 & 1 & 2\\ 2 & 0 & 1 \end{array}}\right] $

The reduced row echelon form of augmented matrix $\displaystyle [A\vdots \mathbf{0}]$ is:

$\displaystyle \left[{\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{array}}\right] $

So the solutions are:

$\displaystyle x_1 = 0, x_2 = 0, x_3 = 0 $

My question:

How do you find the basis for $\displaystyle ker(T)$ now because $\displaystyle x_1 = x_2 = x_3 = 0$?