Complex eigenvectors... I can't get the right answer even though I'm using software I'm using software to calculate my eigenvectors, and I can't get the correct answers...
I have $$A = \begin{pmatrix} 1 & 2+i \\ 2+i & 1 \end{pmatrix}.$$ I solved the eigenvalues to be $3+i, -1-i$ (which is correct according to my software). Then I compute $\ker(A - (3+i)I)$ and $\ker(A-(-1-i)I)$, which gives me eigenvectors of $(1,)^T, (-1,0)^T$. But, plugging them in, they don't satisfy the requirement of an eigenvector. Can someone show me how my method is wrong and how to arrive at the correct answer?
 A: We all agree that given
$$
  \mathbf{A} = \left[
\begin{array}{cc}
 1 & 2+i \\
 2+i & 1 \\
\end{array}
\right]
$$
the eigenvalues are
$$
 \lambda \left( \mathbf{A} \right) = \left\{ 3 + i, -1 - i \right\}
$$
The eigenvectors are
$$
  v_{1} = 
\color{blue}{\left[
\begin{array}{c}
 1 \\
 1 \\
\end{array}
\right]}, \qquad 
  v_{2} = 
\color{red}{\left[
\begin{array}{r}
-1 \\
 1 \\
\end{array}
\right]}
$$
The eigenvector equations are
$$  
\begin{align}
%
  \left( \mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) \cdot v_{1} 
&= \mathbf{0} \\[5pt]
%
\left[
\begin{array}{rr}
 -2-i & 2+i \\
 2+i & -2-i \\
\end{array}
\right] 
\color{blue}{\left[
\begin{array}{r}
 1 \\
 1 \\
\end{array}
\right]}
&=
\left[
\begin{array}{r}
 0 \\
 0 \\
\end{array}
\right] \\[10pt]
\end{align}
$$
$$  
\begin{align}
  \left( \mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) \cdot v_{2} 
&= \mathbf{0} \\[5pt]
%
\left[
\begin{array}{rr}
 2+i & 2+i \\
 2+i & 2+i \\
\end{array}
\right] 
\color{red}{\left[
\begin{array}{r}
-1 \\
 1 \\
\end{array}
\right]}
&=
\left[
\begin{array}{r}
 0 \\
 0 \\
\end{array}
\right] \\[10pt]
%
\end{align}
$$

We don't know what software package the OP used. So to advance the discussion, here is Mathematica output.



Of course we can solve the eigenvalue equation directly.
$$
 \mathbf{A} v = \lambda v
$$
The answer is the same.


A: Here is an approach: Let an eigenvector be $X = \begin{pmatrix} x \\ y \end{pmatrix}$ and perform matrix multiplication $AX=\lambda{X}$ where $\lambda=3+i$. You are going to get two equations: $x+(2+i)y=(3+i)x$ and $(2+i)x+y=(3+i)y$. It is important to verify that these equations are equal to each other, as they should be if your eigenvalue is correct. Simplifying them gives $2x+ix=2y+iy$.(Check!) Here to verify actually isn't difficult. Removing brackets and bring terms over does the job. Sometimes you need to play around with some conjugate to show they are the same. Anyway, if you pick $x=1$ it follows that $y=1$ and thus an eigenvector is found. You can verify by multiplication that this eigenvector indeed works. Now you try to find an eigenvector for the other stated eigenvalue.
