I was watching justin solomon Numerical Analysis course and this is where I was stuck (link to youtube video)
Definition:
$$A^TA \overrightarrow{x_i} = \lambda_i\overrightarrow{x_i}$$
Taking another vector $\overrightarrow{y_i}$ as transformation of $\overrightarrow{x_i}$
$$\overrightarrow{y_i} = A \overrightarrow{x_i}\label{d}\tag{1}$$
So, $AA^T\overrightarrow{y_i}=\lambda_i\overrightarrow{y_i}$ $$\lambda_i\overrightarrow{y_i} = \lambda_iA\overrightarrow{x_i} = AA^TA\overrightarrow{x_i}=AA^T\overrightarrow{y_i}\label{c}\tag{2}$$ And, $||\overrightarrow{y_i}|| = \sqrt{\lambda_i}||\overrightarrow{x_i}||$ $$||\overrightarrow{y_i}|| = ||A\overrightarrow{x_i}||=\sqrt{(A\overrightarrow{x_i})^2}=\sqrt{\overrightarrow{x_i}^TA^TA\overrightarrow{x_i}}=\sqrt{\overrightarrow{x_i}^T\lambda_i\overrightarrow{x_i}} = \sqrt{\lambda_i}||\overrightarrow{x_i}||\label{b}\tag{3}$$ Let,
$\bar{U} \in \Re^{m\times k} $, matrix of $\overrightarrow{y_i}$
$\bar{V} \in \Re^{n\times k} $, matrix of $\overrightarrow{x_i}$
Derivation follows as:
$$ \begin{align*} \bar{U}^TA\bar{V}\overrightarrow{e_i} & = \bar{U}^TA\overrightarrow{x_i} \\ & = \frac{1}{\lambda_i} \bar{U}^TA\lambda_i\overrightarrow{x_i} \\ &= \frac{1}{\lambda_i} \bar{U}^TAA^TA\overrightarrow{x_i}\label{e}\tag{4} \\ &= \frac{1}{\sqrt{\lambda_i}} \bar{U}^TAA^T\overrightarrow{y_i}\label{a}\tag{5} \\ &= \frac{1}{\sqrt{\lambda_i}} \bar{U}^T\lambda_i\overrightarrow{y_i} \\ &=\sqrt{\lambda_i} \overrightarrow{e_i} \end{align*} $$
This is my question: From equation (4) to equation (5), $A\overrightarrow{x_i}$ must be equal to $\overrightarrow{y_i}$ from equation (1)
But, what he wrote is, $A\overrightarrow{x_i}$ equals $\sqrt{\lambda_i} \overrightarrow{y_i}$
He says something about the length, using equation (3) $A\overrightarrow{x_i}$ should equal to $\overrightarrow{y_i}\over{\sqrt{\lambda_i} }$, which is not the case in the derivation.
Can somebody explain me, why do $A\overrightarrow{x_i}$ equals $\sqrt{\lambda_i} \overrightarrow{y_i}$ ?
There is a step missing in your recounting of the computation. At some point one would have to demand that the $\vec x_i$ are unit vectors, $\|\vec x_i\|=1$. Then with the recognition that $\|\vec y_i\|=\sqrt{λ_i}$ for $\vec y_i=A\vec x_i$ one re-defines $$\vec y_i=\frac1{\sqrt{λ_i}}A\vec x_i$$ so that now also the $\vec y_i$ are unit vectors, $\|\vec y_i\|=1$.
One could also have that formulated less confusing by using $$ \vec u_i=\frac{\vec x_i}{\|\vec x_i\|}~\text{ and }~\vec v_i=\frac{\vec v_i}{\|\vec v_i\|} $$ and assemble $U$ and $V$ from these vectors as columns.