I've got a matrix $A = \begin{bmatrix}1&-1\\ 2&-1\end{bmatrix}$ and wish to diagonalise it. I find the eigenvalues as below.
$\det(A - xI) = 0 = \det\begin{bmatrix}1-x&-1\\ 2&-1-x\end{bmatrix}$
$\det(A - xI) = (1-x)(-1-x) - (-2) = 1 + x^2$
Then $x =i, -i$. So, now I need to find the eigenvectors, which is where I'm a little confused. In the case that $x = i$:
$E_1 = ker\begin{bmatrix}1-i&-1\\ 2&-1-i\end{bmatrix}$
$\begin{bmatrix}1-i&-1\\ 2&-1-i\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$
At this point we usually row reduce and end up with a row of all zeros. However, in this case that isn't possible. So, I get
$2a + (-1-i)b \implies a = (b+ i)/2$ from the second row.
I want to write the kernel in terms of a spanning set, so get $E_1 = \{((1+i)/2, 1)\}$. Would this be a correct spanning set for $E_1$? Is the choice between which rows to use arbitrary? I'm just seeking clarification because all problems I've done before row reduce to only a single row.
Thanks