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?


3 Answers 3


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.

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    $\begingroup$ Suggestion: change the first sentence for people who are looking for a quick answer and end up with inaccurate information. $\endgroup$
    – chris
    Mar 21, 2017 at 21:16
  • $\begingroup$ The co-kurtosis matrix is non-square. does it have eigenvalues and eigenvectors, or singular values and singular vectors? $\endgroup$
    – develarist
    Sep 9, 2020 at 1:53

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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    $\begingroup$ 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
  • $\begingroup$ Yes, it does. Of course we can consider matrices over $\mathbb{C}$ instead of the reals, etc $\endgroup$
    – AnyAD
    Jul 6, 2020 at 12:02
  • $\begingroup$ @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. $\endgroup$
    – koyae
    Jan 14, 2021 at 3:39

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.


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