I've looked at this and it doesn't help because I don't know anything about SVD. Can someone dumb it down for me please?
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3 Answers
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It is not exactly true that non-square matrices can have eigenvalues. Indeed, the definition of an eigenvalue is for square matrices. For non-square matrices, we can define singular values:
Definition: The singular values of a $m \times n$ matrix $A$ are the positive square roots of the nonzero eigenvalues of the corresponding matrix $A^{T}A$. The corresponding eigenvectors are called the singular vectors.
Of course, these have certain properties, that may or may not be useful for what you are trying to study.
answered Nov 27, 2013 at 23:28
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8$\begingroup$ Suggestion: change the first sentence for people who are looking for a quick answer and end up with inaccurate information. $\endgroup$– chrisMar 21, 2017 at 21:16
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$\begingroup$ The co-kurtosis matrix is non-square. does it have eigenvalues and eigenvectors, or singular values and singular vectors? $\endgroup$ Sep 9, 2020 at 1:53
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Eigenvalues and eigenvectors of a matrix, say $A$, help us find subspaces which are invariant under $A$ (when $A$ is seen as a linear transformation). If $A$ is non-square, then $A:\mathbb{R}^m\rightarrow \mathbb{R}^n$, where $m\neq n$. Hence $Av=\lambda v$ makes no sense, since $Av\notin\mathbb{R}^m$.
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1$\begingroup$ Can someone interpret the notation A: R^m => R^n. Does it mean all the elements in A are real numbers ? $\endgroup$ Jul 3, 2020 at 3:12
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$\begingroup$ Yes, it does. Of course we can consider matrices over $\mathbb{C}$ instead of the reals, etc $\endgroup$– AnyADJul 6, 2020 at 12:02
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$\begingroup$ @AmeyYadav If I'm not wrong, $A:\mathbb{R}^m\rightarrow \mathbb{R}^n$ represents $A$ as a transformation from an $m$-dimensional space to an $n$-dimensional space. I believe answerer here is assuming that $A$ contains only real numbers. $\endgroup$– koyaeJan 14, 2021 at 3:39
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Non-square matrices do not have eigenvalues. If the matrix X is a real matrix, the eigenvalues will either be all real, or else there will be complex conjugate pairs.
Source: Wikipedia.
answered Nov 27, 2013 at 23:22