Suppose $Rank(A)=r,A{\in}R^{m \times n}$, $v_1,v_2,...,v_r{\in}R^n$ forms an orthonormal basis of $C(A^T)$(i.e. row space of A) and are eigenvectors of $A^TA$.
Now, take $\sigma_1u_1=Av_1,\sigma_2u_2=Av_2,...,\sigma_ru_r=Av_r;\ \sigma_1,\sigma_2,...,\sigma_r$ are positive scaling factors and $u_1,u_2,...,u_r{\in}R^m$ are unit vectors in $C(A)$(column space of A).
Now,we have $A(v_1,v_2,...,v_r)=(u_1,u_2,...,u_r)\begin{bmatrix}\sigma_1& 0 &\cdots &0 \\ 0& \sigma_2 & \cdots &0 \\ \vdots &\vdots &\ddots &\vdots \\0&0&\cdots&\sigma_r\end{bmatrix}$. And denote this formula as $AV=U\Sigma,V{\in}R^{n\times r},U{\in}R^{m\times r},\Sigma$ is a r$\times$r diagonal matrix.
My problem is how to show $u_1,u_2,...,u_r$ is an orthonormal basis of $C(A)$? E.g. $u_1^Tu_2=0$ when take $\sigma_1u_1=Av_1,\sigma_2u_2=Av_2$.
Above thread comes from MIT's opencourse 18.06 linear algebra lecture 29
Assuming everything in the problem statement, ... \begin{align*} u_i^\mathrm{T} u_j &= \frac{1}{\sigma_i \sigma_j} \sigma_i \sigma_j u_i^\mathrm{T} u_j \\ &= \frac{1}{\sigma_i \sigma_j} \sigma_i u_i^\mathrm{T} \sigma_j u_j \\ &= \frac{1}{\sigma_i \sigma_j} (\sigma_i u_i)^\mathrm{T} (\sigma_j u_j) \\ &= \frac{1}{\sigma_i \sigma_j} (A v_i)^\mathrm{T} (A v_j) \\ &= \frac{1}{\sigma_i \sigma_j} v_i^\mathrm{T} (A^\mathrm{T} A) v_j \\ &= \frac{1}{\sigma_i \sigma_j} v_i^\mathrm{T} \varepsilon_j v_j \\ &= \frac{1}{\sigma_i \sigma_j} \varepsilon_j v_i^\mathrm{T} v_j \\ &= \frac{1}{\sigma_i \sigma_j} \varepsilon_j 0 \\ &= 0 \text{,} \end{align*} where $\varepsilon_j$ is the eigenvalue of $(A^\mathrm{T} A)$ associated to $v_j$.