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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 $\sqrt{\lambda_i} \overrightarrow{y_i}$ ?