Why can't I solve this set of equations for the unit characteristic vectors of $\begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}$?

Question

This is the problem:

50. (a) Determine the characteristic numbers $(\lambda_1, \lambda_2)$ and corresponding unit characteristic vectors $(e_1, e_2)$ of the matrix $$A = \begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}.$$

Ok, I guess I can solve it this way:

$$\begin{align}\begin{vmatrix} 5 & 2 \\ 2 & 2 \end{vmatrix} =& (5 - \lambda)(2 - \lambda) - 4 \\ &10 - 5\lambda - 2\lambda + \lambda^2 - 4 = \\ & \begin{array}{}\lambda^2 - 7\lambda + 6 & \implies & \lambda_1 = 1 \\ && \lambda_2 = 6 \end{array}\end{align}$$

And for finding e11 and e12, as you can see we must solve this set of equations: $$ \begin{bmatrix}4&2\\2&1\end{bmatrix} \begin{bmatrix}e_{11}\\e_{12}\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}.$$ $$\begin{align} 4e_{11} + 2e_{12} & = 0 \\ 2e_{11} + e_{12} & = 0 \end{align}$$

But we cannot solve this set of equations because the first equation is 2 times of the second equation and e11 and e12 become zero when you solve it.

Am I solving it correctly? if so, what does it mean?

(Original images: https://i.stack.imgur.com/maWPC.png, https://i.stack.imgur.com/MU6Bi.jpg, https://i.stack.imgur.com/yRchp.jpg)

Answer 1

Because the first equation is 2 times the second, we can choose to solve either equation 1 or equation 2 and will get the same answer. So lets take equation 1: Solve $e_{11}$ in terms of $e_{12}$

$4e_{11} +2e_{12} = 0$ $\implies e_{11} = -\frac{2}{4}e_{12}$ $\implies e_{11} = -\frac{1}{2}e_{12}$

Now pick a value (any real value!) for $e_{12}$. We can do this because an eigenvector and a scalar multiple of it are considered the same. Let's say $e_{12} = 1$ Then $e_{11} =- \frac{1}{2}$.

So $e_{11} = -\frac{1}{2}$ and $e_{12} = 1$.

You will find numerous examples of this on the net. Just ask for solving for eigenvectors.

Answer 2

If you end up with $0=0$ from solving those simultaneous equations, that means that there are infinitely many solutions. In order to get the solution set for this scenario, you need to let one of your variables, in this case, I'd choose $e_{11}$ to give a nicer set, equal $k$

This gives:

$2e_{11}+e_{12}=0$

$\therefore e_{12}=-2e_{11}$

Let $e_{11}=k$ where $k\in \mathbb{R}$

$\therefore e_{12}=-2k$

$\therefore \{e_{11}=k, e_{12}=-2k; k\in \mathbb{R}\}$