I am still struggling to understand the relationship between the row echelon form and the basis of the nullspace. I have a questions that asks me to find the diagonalized normal form of a matrix A.
$$A=\begin{bmatrix}0 & 0 & 0\\\ 0 & b & 1\\\ 1 & 0 & 0\end{bmatrix}$$
The characteristic polynomial works out to be $P_{A}(\lambda)=\lambda^2(b-\lambda)$. Giving eigenvalues $\lambda=0$ with algebraic multiplicity 2 and $\lambda_{2}=2$ with algebraic multiplicity 1.
If I do gaussian elimination to get row reduced echelon form by swapping R1 with R3 then deviding R2 by b to get the pivot set up I get....
$$A=\begin{bmatrix} 1 & 0 & 0\\\ 0 & 1 & 1/b\\\ 0 & 0 & 0\end{bmatrix}$$
So does this mean $x_3=x_3$ being a free variable, then from there I am a bit confused.