I had a matrix which I solved for a characteristic poly = (1-λ)(2-λ)(2-λ). when I was solving for the eigenspace corresponding to λ=1, I got a 3x3 matrix in RREF of : \begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix} The solution manual states the solution "http://prntscr.com/jwh3x4"
My question is that why is my method of solving the eigenspace = t[1,0,0] + s[0,1,0] incorrect?
Thanks!
Remember that the eigenspace is the kernel of some matrix. What you gave is the span of that RREF, and not the kernel of the matrix.
You want to look at the solutions to the following system:
$$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}. $$
You should note that solutions to this equation must have $y = z = 0$, but $x$ is free to be whatever it wants to be. Hence, the kernel of this matrix (and thus the eigenspace associated to the eigenvalue $1$) is given by
$$\left\{t\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} : t \in \mathbb{R} \right\}.$$