# In SVD, why $u_i=Av_i/\sigma_i$?

In SVD, $$u_i=Av_i/\sigma_i$$
so we conclude that $A=U\Sigma V^{T}$.

I read this part several times but still hard to understand what they are trying to explain. Can anyone help me to understand this topic?

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The SVD of the matrix $A$ is given by $A = U\Sigma V^T$, which can be rewritten as $$A = \sum_j \sigma_j u_jv_j^T\ \ (1)$$ where $U$ and $V$ are Hermetian. Multiplying (1) from the right by some $v_i$, we obtain $$Av_i = {(\sum_j \sigma_j u_jv_j^T)}v_i$$ but because all the $(v_j)_i$ are orthonormal ($v_i^Tv_j = \delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta) we obtain $$Av_i = \sigma_i u_i$$