This is the proof in my textbook:
What I don't understand it why " $AA^{T}u = 0u $ means that u is an eigenvector.
Is this a theorem that I don't know? That if you multiply a matrix by a vector and it's 0, then that vector is an eigenvector?
Further, we are asked to prove that u is a unit eigenvector. The answer only proves that it is an eigenvector. How do we show that it is a unit eigenvector?
I'll write my SVD notation as $A=U \Sigma V^T$, assuming that $A \in \mathbb{R}^{m \times n}$, $U \in \mathbb{R}^{m \times m}$, $\Sigma \in \mathbb{R}^{m \times n}$, and $V \in \mathbb{R}^{n \times n}$. So we have a full SVD, $U^T U = I$, $V^T V = I$, and $\Sigma$ is diagonal with nonnegative entries. I mention this because everything is slightly different in the case of a reduced SVD.
Now if $u_i$ is the $i$th column of $U$, then $A^T u_i = V \Sigma^T U^T u_i$. $U^T u_i = e_i$ (check it). Then $\Sigma^T e_i$ is the $i$th column of $\Sigma^T$, which is $\sigma_i e_i$ (check it). Finally $V e_i$ is the $i$th column of $V$, so we have $A^T u_i = \sigma_i v_i$.
Doing the same thing for $A$ and $v_i$, we find that $A v_i = \sigma_i u_i$, so $A A^T u_i = A \left ( \sigma_i v_i \right ) = \sigma_i^2 u_i$.
Finally, $u_i$ are unit by the definition of the SVD, so there's no need to separately prove anything there.