Eigenvalue and Matrix Reduction I was solving some linear algebra problems and I have a quick question about one problem. I'm given the matrix $A = \{a_1,a_2\}$ where $a_1=[1,1]$ and $a_2=[-1,1]$. I need to solve for the eigenvalues of the matrix $A$ over the complex numbers $\mathbb{C}$. I solved and got the eigenvalue $1-i$ and $1+i$. Now this is where I get confused. I was taught that after finding the eigenvalue you plug it into the equation $[A - \lambda I]$. However, the book is putting it into the this equation. $[\lambda I - A]$. I understand the equations are exactly the same because they are both set to zero. You just have to multiply one by negative one. However, why for this problem are they choosing this other equation. It is the first time I've seen it. Also, once I plug it in I obtain $B=\{a_1,a_2\}$ where $a_1=[i,1]$ and $a_2=[-1,i]$. How do I reduce this matrix so that I have one row completely zero?
 A: The $[A- \lambda I]v$ versus $[\lambda I - A]v$ is just a matter of choice. When we have $Av = \lambda v$, we can choose to subtract from either side, so just a convention. Some people hate negating each term of the matrix as this leaves more room for error. 
We are given:
$$A = \begin{bmatrix}1 & 1\\-1 & 1\end{bmatrix}$$
We find the eigenvalues as:
$$\lambda_{1,2} = 1 ~ \pm ~i$$
To find the eigenvectors, we setup and solve:
$$[A - \lambda I]v_i = 0$$
We have independent and complex conjugate eigenvalues, so finding eigenvectors, we can also use conjugates.
We have to find the RREF using $\lambda_1 = 1 + i$, yielding:
$$ [A -(1 + i)I]v_1 = \begin{bmatrix}-i & 1\\-1 & -i\end{bmatrix}v_1 = 0$$
Adding $i \times R_1$ to $R_2$, and dividing $R_1$ by $-i$  yields:
$$\begin{bmatrix}1 & i\\0 & 0\end{bmatrix}v_1 = 0$$
This gives us $v_1 = (-i, 1)$ and immediately using the complex conjugate $v_2 = (i, 1)$.
Update
If we had chosen the second eigenvalue, $\lambda_2 = 1 - i$, to find the first eigenvector, we have:
$$ [A -(1 - i)I]v_2 = \begin{bmatrix}i & 1\\-1 & i\end{bmatrix}v_2 = 0$$
Adding $-i \times R_1$ to $R_2$, and dividing $R_1$ by $i$  yields:
$$\begin{bmatrix}1 & -i\\0 & 0\end{bmatrix}v_2 = 0$$
This gives us $v_2 = (i, 1)$ and immediately using the complex conjugate $v_1 = (-i, 1)$.
Of course this matches with the result above.
A: Regarding your first question: as you have said, it's exactly the same. I have seen [xI-A] used mostly for finding the Eigenvalues, and the other way around for finding the Ker of a given eigenvalue (for the matching vectors, etc.). It's more comftorble to work with positive x's and negetive numbers than the other way around, I guess.
Regarding your second question: Hint: consider what happens if you double the first row with the scalar i over C. Then you can reduce rows from each other and get one row completely zero, as one is linearly dependent of the other.
