This question is about Singular Value Decomposition.

Given an arbitrary matrix $A$

$$A \in \mathbb{R}^{m\times n} $$

Its reduced SVD form is:-

$$A = U\Sigma V^T$$

Where as $\Sigma$ is the diagonal matrix containing scaling factors across its diagonal:-

$$ \Sigma = \left( \begin{array}{ccccc} \sigma_1 & \hfill & \hfill & \hfill & \hfill \\ \hfill & \sigma_2 & \hfill & \hfill & \hfill \\ \hfill & \hfill & \ddots &\hfill & \hfill \\ \hfill & \hfill & \hfill & \hfill & \sigma_n \\ \end{array} \right)$$

Now, it is said that the singular values across the diagonal are such that:-

$$\sigma_1 \geq \sigma_2 \geq \sigma_3 \geq \cdots \geq \sigma_n \geq 0 $$

Why do these values have to be necessarily in descending order across the diagonal? I read that it's a convention to write so, but there is more to the order than just convention. If this order is changed, we will get a totally different matrix instead of $A$. So the order is important.

Or is it so that there is always guaranteed to be one unique solution for SVD in which these singular values across $\Sigma$ are guaranteed to be in descending order?


1 Answer 1


It is just convention. If you change the order, and permute the columns of $U$ and $V$ correspondingly, you do get the same matrix $A$.


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