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In this answer the author writes:

PCA is very closely related to singular value decomposition (SVD), see Relationship between SVD and PCA. How to use SVD to perform PCA? for more details. If a $n\times p$ matrix $\mathbf X$ is SVD-ed as $\mathbf X = \mathbf {USV}^\top$ and one selects a $k$-dimensional vector $\mathbf z$ that represents the point in the "reduced" $U$-space of $k$ dimensions, then to map it back to $p$ dimensions one needs to multiply it with $\mathbf S^\phantom\top_{1:k,1:k}\mathbf V^\top_{:,1:k}$.

The linked answer in the quotation says that such a reconstruction has the same dimensions $n \times p$ but lower rank. What does this term of "rank" mean in the context of PCA?

a6623
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    The link in the quotation provides an extensive, explicit answer. What are you hoping we might add to it? – whuber Feb 20 '22 at 16:09
  • @whuber, sorry, I didn't read the linked answer entirely, now I see it. But I have a followup question, so I updated the original post now – a6623 Feb 20 '22 at 16:15
  • PCA is linear algebra, so the default understanding of the rank of a matrix is the dimension of its image. – whuber Feb 20 '22 at 19:27
  • @whuber so by the rank being $k$, does that mean there are $p - k$ linearly dependent columns, and this represents the data "lost" by the reconstruction of PCA? – a6623 Feb 20 '22 at 22:03
  • When the rank is $k,$ there exist $k$ linearly independent columns, but any set of $l\gt k$ columns will not be linearly independent. – whuber Feb 21 '22 at 13:19

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