This is from my textbook
What I don't understand is, $V$ and $U$ are already square, why the textbook says "if we want to make them square"?
2026-03-29 21:52:53.1774821173
square eigenvectors for Singular Value Decomposition?
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No. Let us assume that $A$ is $n \times p$ in "portrait" format ($n>p$).
Then $V$ is $p \times r$ and $U$ is $p \times r$ ($r$ being the rank)