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I want to find the kernel basis of the transformation matrix:

\begin{pmatrix} 1 & -1 & 1 & -1\\ 1 & 0 & 0 & 0\\ 1 & 1 & 1 & 1\\ \end{pmatrix}

I have reduced the matrix to :

\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 0\\ \end{pmatrix}

I see we have one free variable (x4). How do I go on from here?

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1 Answer 1

Express the remaining variables $x_1,x_2,x_3$ in terms of your free veraiable $x_4$, using the fact that the reduced matrix must annihilate $(x_1,x_2,x_3,x_4)$.

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From the RREF I get the system: x1 = 0; x4 = -x2 -x3; x3 = 0. So i guess the final result would than be: x4 = - x2? –  Trom Nov 25 '12 at 9:49
    
Since you are solving for $x_1,x_2,x_3$ I woudlwrite the second equation as $x_2=-x_4$ instead. All in all $(x_1,x_2,x_3,x_4)=x_4(0,-1,0,1)$. By the way, you have just a row-echelon form, not a reduced row-echelon form, but that is OK, you don't need the latter. –  Marc van Leeuwen Nov 25 '12 at 11:14

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