a question about calculating eigenvectors It has to be the silliest question, but it’s not clear to me how to calculate eigenvectors quickly. I am just talking about a very simple 2-by-2 matrix.
When I have already calculated the eigenvalues from a characteristic polynomial, I can start to solve the equations with $A\mathbf{v}_1 = e_1\mathbf{v}_1$ and A\mathbf{v}_2 = e_2\mathbf{v}_2$, but in this case it always requires writing lines of equations and solving them.
On the other hand I figured out that just by looking at the matrix you can come up with the eigenvectors very quickly. But I'm a bit confused in this part.
When you have the matrix with subtracted $e_1$ values like this:
$$\left(\begin{array}{cc}
A&B\\
C&D
\end{array}\right).$$
Then for me, it always worked to use the eigenvector.
$$\left(\begin{array}{r}-B\\A\end{array}\right)$$
But in some guides I find that they are using A C as an eigenvector.
$$\left(\begin{array}{c}A\\C
\end{array}\right).$$
And when I check it, they are indeed multiples of each other. But this other method is not clear to me, how could $A$ and $C$ mean anything about the eigenvector, when both of them are connected to $x$, without having to do anything with $y$. But it’s still working. Was it just a coincidence?
So is the recommended method for calculating them is just to subtract the eigenvalues from the matrix and look at 
$$\left(\begin{array}{r}
-B\\A\end{array}\right)\qquad\text{or}\qquad\left(\begin{array}{r}
-D\\A
\end{array}\right).$$
 A: (Note that you are using $A$ for two things in your post: it is the original matrix, and then it's an entry of the matrix; that's very bad form. and likely to lead to confusion; never use the same symbol to represent two different things).
So, if I understand you: you start with a matrix $\mathscr{A}$,
$$\mathscr{A} = \left(\begin{array}{cc}
a_{11} & a_{12}\\
a_{21} & a_{22}
\end{array}\right).$$
Then, if you know that $e_1$ is an eigenvalue, then you look at the matrix you get when you subtract $e_1$ from the diagonal:
$$\left(\begin{array}{cc}
a_{11}-e_1 & a_{12}\\
a_{21} & a_{22}-e_1
\end{array}\right) = \left(\begin{array}{cc}A&B\\C&D
\end{array}\right).$$
Now, the key thing to remember is that, because $e_1$ is an eigenvalue, that means that the matrix is singular: an eigenvector corresponding to $e_1$ will necessarily map to $\mathbf{0}$. That means that the determinant of this matrix is equal to $0$, so $AD-BC=0$.
Essentially: one of the rows of the matrix is a multiple of the other; one of the columns is a multiple of the other. 
What this means is that the vector $\left(\begin{array}{r}-B\\A\end{array}\right)$ is mapped to $0$: because
$$\left(\begin{array}{cc}
A&B\\C&D\end{array}\right)\left(\begin{array}{r}-B\\A\end{array}\right) = \left(\begin{array}{c}-AB+AB\\-BC+AD \end{array}\right) = \left(\begin{array}{c}0\\0\end{array}\right)$$
because $AD-BC=0$. If $A$ and $B$ are not both zero, then this gives you an eigenvector.
If both $A$ and $B$ are zero, though, this method does not work because it gives you the zero vector. In that case, the matrix you are looking at is
$$\left(\begin{array}{cc}0&0\\C&D
\end{array}\right).$$
One vector that is mapped to zero is $\left(\begin{array}{r}-D\\C\end{array}\right)$; that gives you an eigenvalue unless $C$ and $D$ are both zero as well, in which case any vector will do. 
On the other hand, what about the vector $\left(\begin{array}{r}A\\C\end{array}\right)$? That vector is mapped to a multiple of itself by the matrix:
$$\left(\begin{array}{cc}
A&B\\C&D\end{array}\right)\left(\begin{array}{c}A\\C\end{array}\right) = \left(\begin{array}{c}A^2 + BC\\AC+DC\end{array}\right) = \left(\begin{array}{c}A^2+AD\\AC+DC\end{array}\right) = (A+D)\left(\begin{array}{c}A\\C\end{array}\right).$$
However, you are looking for a vector that is mapped to $\left(\begin{array}{c}0\\0\end{array}\right)$ by $\left(\begin{array}{cc}A&B\\C&D\end{array}\right)$, not for a vector that is mapped to a multiple of itself by this matrix. 
Now, if your original matrix has zero determinant already, and you haven't subtracted the eigenvalue from the diagonal, then here's why this will work: the sum of the two eigenvalues of the matrix equals the trace, and the product of the two eigenvalues equals the determinant. Since the determinant is $0$ under this extra assumption, then one of the eigenvalues is $0$, so the other eigenvalue equals the trace, $a_{11}+a_{22}$. In this case, the vector $\left(\begin{array}{c}A\\C\end{array}\right)$ is an eigenvector of $\mathscr{A}$ corresponding to $a_{11}+a_{22}$, unless $A=C=0$ (in which case $\left(\begin{array}{r}a_{22}\\a_{11}\end{array}\right)$ is an eigenvector unless $\mathscr{A}$ is the zero matrix). 
Added. As Robert Israel points out, though, there is another point here. Remember that $e_1+e_2 = a_{11}+a_{22}$, $A=a_{11}-e_1 = e_2-a_{22}$; and that $a_{11}a_{22}-a_{12}a_{21} = e_1e_2$; if we take the vector $\left(\begin{array}{c}A\\C\end{array}\right)$ with the original matrix $\mathscr{A}$, we have:
$$\begin{align*}
\left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right)\left(\begin{array}{c}A\\C\end{array}\right) &= \left(\begin{array}{c}
a_{11}A + a_{12}C\\a_{21}A + a_{22}C.\end{array}\right)\\
&= \left(\begin{array}{c}
a_{11}(e_2-a_{22}) + a_{12}a_{21}\\
a_{21}(e_2-a_{22}) + a_{22}a_{21}
\end{array}\right) = \left(\begin{array}{c}
e_2a_{11} + (a_{12}a_{21}-a_{11}a_{22})\\
e_2a_{21} + (a_{22}a_{21} - a_{22}a_{21}
\end{array}\right)\\
&= \left(\begin{array}{c}
e_2a_{11} - e_1e_2\\
e_2a_{21}
\end{array}\right) = \left(\begin{array}{c}
e_2(a_{11}-e_1)\\e_2a_{21}
\end{array}\right)\\
&= e_2\left(\begin{array}{c}
a_{11}-e_1\\a_{21}
\end{array}\right) = e_2\left(\begin{array}{c}A\\C
\end{array}\right).
\end{align*}$$
So if $A$ and $C$ are not both zero, then $\left(\begin{array}{c}A\\C\end{array}\right)$ is an eigenvector for the other eigenvalue of $\mathscr{A}$. 
To summarize:


*

*If you subtract the eigenvalue from the diagonal, and in the resulting matrix the first row is not equal to $0$, then your method will produce an eigenvector corresponding to the eigenvalue you subtracted.

*If you subtract the eigenvalue from the diagonal, and in the resulting matrix the first column is not equal to $0$, then taking that column will produce an eigenvector corresponding to the other eigenvalue of $\mathscr{A}$ (not the one you subtracted). 
A: Please goes through this list one by one. 


*

*Your method will work as long as either $A$ or $B$ is not zero. 

*When doesn't it work?  Try calculate the eigenvector of $\left(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix} \right )$.

*Does the method of taking $\left(\begin{smallmatrix} A\\C  \end{smallmatrix} \right )$ work?  Try calculate the eigenvector of $\left(\begin{smallmatrix} 3 & 3 \\ 4 & 7 \end{smallmatrix} \right )$ and see it for yourself.

*What is an obvious solution to the equation $Ax+By=0$? Hint: it has something to do with your method. 

*Say when either $A$ or $B$ is nonzero, what will happen to the second row when you do row reduction? Why? Remember, the determinant of $\left(\begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \right )$ is zero. 

*Can you see why your method works by now?

