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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?

Don Larynx
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3 Answers3

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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.

glS
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    Suggestion: change the first sentence for people who are looking for a quick answer and end up with inaccurate information. – chris Mar 21 '17 at 21:16
  • The co-kurtosis matrix is non-square. does it have eigenvalues and eigenvectors, or singular values and singular vectors? – develarist Sep 09 '20 at 01:53
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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 $Av\notin\mathbb{R}^m$.

AnyAD
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    Can someone interpret the notation A: R^m => R^n. Does it mean all the elements in A are real numbers ? – Amey Kumar Samala Jul 03 '20 at 03:12
  • Yes, it does. Of course we can consider matrices over $\mathbb{C}$ instead of the reals, etc – AnyAD Jul 06 '20 at 12:02
  • @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. – koyae Jan 14 '21 at 03:39
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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.