Q:

Math Explanations needed for algebraic steps

I am trying to understand the context below in the image. 
Somehow, I cannot follow some algebraic steps that I have underlined with red in the image. What stops me first is the first underlined equation. I cannot see How X is derived. If somebody knows, help me go through the steps for those underline in with read in the image. 

A:

We want the following:
$$\mathbf{x^`Ax}=c^2$$
They give us a formula for the left side:
$$\lambda_1(\mathbf{x^`e_1})^2+\lambda_2(\mathbf{x^`e_2})^2=c^2$$
Now, this is the key observation: if $x$ is on the axes, which means $a\mathbf{e_1}$ and $a\mathbf{e_2}$, then, assuming $\mathbf{e_1}$ and $\mathbf{e_2}$ are orthonormal eigenvectors, the $\mathbf{e_1}`\mathbf{e_2}=0$ will cancel out, leaving us with only one term, and the $\mathbf{e_i}^`\mathbf{e_i}=1$ can be dealt with easily, leaving us with only scalars. That might be a hard thing to notice, but if you just go out on a limb and try plugging in an eigenvector like $\mathbf{x}=a\mathbf{e_1}$, you will see the equation becomes easy to solve. In order to find these solutions, you just have to have the intuition to see that a lot of equations like these is easy to solve if $\mathbf{x}$ is an eigenvector. Thus, let's plug in $\mathbf{x}=a\mathbf{e_1}$:
$$\lambda_1((a\mathbf{e_1})^`\mathbf{e_1})^2+\lambda_2((a\mathbf{e_1})^`\mathbf{e_2})^2=c^2$$
Take $a$ out of the transpose operation:
$$\lambda_1(a\mathbf{e_1}^`\mathbf{e_1})^2+\lambda_2((a\mathbf{e_1}^`\mathbf{e_2})^2=c^2$$
Get rid of the second term because $\mathbf{e_1}^`\mathbf{e_2}=0$ and get rid of $\mathbf{e_1}^`\mathbf{e_1}=1$:
$$\lambda_1a^2=c^2$$
Divide by $\lambda_1$ and take the square root:
$$a=\frac{c}{\sqrt{\lambda_1}}$$
Thus, we get that $\mathbf{x}=\frac{c}{\sqrt{\lambda_1}}\mathbf{e_1}=c\lambda^{-1/2}\mathbf{e_1}$ is a solution to the equation.

For the red thing at the bottom, I think what they are saying is that if you substitute for all $1 \leq i \leq n$ where $i$ is an integer, $\mathbf{x}=\frac{c}{\sqrt{\lambda_i}}\mathbf{e_i}$ is a solution to  $c^2=\lambda_1(\mathbf{x}^`\mathbf{e_1})^2+...+\lambda_1(\mathbf{x}^`\mathbf{e_1})^2$, which can be shown by a similar process to above.

