Q:

Calculating Principal Components - PCA - Calculating Eigenvectors as PCs

I've encountered difficulties while calculating (by hand) the principal components of the following covariance matrix:
$$\Sigma = \begin{bmatrix} 4 & -\sqrt{3}\\ -\sqrt{3} & 2\end{bmatrix}$$
I know how to calculate the eigenvalues ($\lambda_1 = 5$, $\lambda_2 = 1$) but I'm confused after that. I have the solution but I can't reproduce it or do not know what happens there.
So, here is the solution:

 

My questions are as follows. 

How is the solution for each $\lambda$ calculated? 
How do they come from the eigenvectors to the principal components? I thought the eigenvectors are already the principal components?

Would be nice if someone can explain me what is going on there, and maybe as detailed as possible and for non-mathematicians :) Cheers!

A:

So after writing down your EVD decomposition that is
\begin{equation}
 \Sigma = UDU^T
\end{equation}
Notice that $U^T = U^{-1}$ because $\Sigma$ is symmetric, so we have that (as you mention that eigenvalues are)
\begin{equation}
 D = \begin{bmatrix}
  1 & 0 \\
  0 & 5
 \end{bmatrix}
\end{equation}
and the second screenshot computes the eigenvectors (which i'm sure you know how to do:
\begin{equation}
 U = \begin{bmatrix}
  u_1 & u_2
 \end{bmatrix}
 =
 \frac{1}{2}
 \begin{bmatrix}
  -\sqrt{3} & 1 \\
  1 & \sqrt{3}
 \end{bmatrix}
\end{equation}
Now let any vector $x = \begin{bmatrix}
 x_1 & x_2
\end{bmatrix}^T$ be given so 
\begin{equation}
 y 
 =
 \begin{bmatrix}
  y_1\\y_2
 \end{bmatrix}
 =
 Ux
 =
 \frac{1}{2}
 \begin{bmatrix}
  -\sqrt{3} & 1 \\
  1 & \sqrt{3}
 \end{bmatrix}
 \begin{bmatrix}
  x_1 \\
  x_2
 \end{bmatrix}
 =
 \begin{bmatrix}
  -\frac{\sqrt{3}}{2}x_1 + \frac{1}{2}x_2\\
  \frac{1}{2}x_1 + \frac{\sqrt{3}}{2} x_2
 \end{bmatrix}
\end{equation}

