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In my opinion, both are the matrices that the number of the directions of eigenvector is smaller than their size. So, I guess singular matrices and degenerate matrices refer to the same thing. Is that right?

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    $\begingroup$ Singular matrices are matrices that have determinant zero; equivalently, they have a non-trivial kernel. Another term for a singular matrix is a deficient matrix which might be why you had a little bit of a mix-up with terminology. $\endgroup$ May 13 '14 at 3:00
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Yes.

A singular matrix, also known as a degenerate matrix, is a square matrix whose determinate is zero.

That is, they do not have an inverse.

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  • $\begingroup$ It is also worth mentioning that the two terms should not be confused with a defective matrix, which is a non-diagonalisable square matrix. $\endgroup$ Oct 28 '20 at 4:24
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No. Degenerate matrices are not the same as singular matrices. A matrix is degenerate if and only if the determinants of all its principal submatrices are non-zero (where the principal submatrices are the square matrices obtained by choosing the same subset of both the rows and the columns of the original matrix).

See e.g. "On the number of solutions to the complementarity problem and spanning properties of complementary cones", Murty (1972) http://www.sciencedirect.com/science/article/pii/0024379572900195

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    $\begingroup$ This does not sound like the most common definition for "Degenerate matrice". $\endgroup$
    – Ranc
    Oct 5 '15 at 17:31
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Because lately I am reading many articles that is including degenerate instead of non singular or non invert-able then I assume there is a difference. As far as I understood from Wikipedia the singularity theories in this link, it is mentioned that singular sometimes is stable (non degenerate) and sometimes is not stable (degenerate).

The determinant explanation is common for both, if ${\textbf{A}_{n\times n}}$ $det(\textbf{A}) \neq 0$, then $\textbf{A}$ is non singular, non degenerate and as a result is invertable.

A biological definition of degenerate is to become less specialized. I have seen that explanation several times when columns of a matrix are called flat columns (high correlation between columns), each column is becoming less specialized or less informative and as a result the matrix becomes degenerate. This is directly related to the condition number of a matrix. When Condition no. of a matrix is too high then it is an ill conditioned matrix and there is a high correlation between its columns.

Another explanation that I liked for singular matrix is that when you transform a matrix to another shape of less dimension (plane to line), it becomes singular and you will not be able to invert it back as information is lost. Again this is mentioned in Wikipedia link saying 2 points becomes 1. Also a good explanation is in that youtube video

I just found an answered question that proves it,

Simply degeneracy is related to correlation between columns or between rows. Condition of a matrix is decided based on the correlation of the matrix so there is a direct connection between degeneracy and correlation between columns. Adding that the relation between rank of matrix (full rank or not) and its size is another definition of degeneracy. ex. ${\bf{A}}_{I \times J}$ with rank $R$

if $R \le (I,J)$ then it is a non degenerate matrix

if $R > (I,J)$ then it is a degenerate matrix (under-determined system) more unknowns than givens, high information to be found represented by rank while available data is limited represented by size.

if $\sigma_1 \gg \sigma_{R} $ then the matrix is degenerate, where $\sigma_n$ is the singular values of a matrix and $R$ is the highest index where $\sigma$ > 0. ex. $\sigma_1 = 100$ while $\sigma_R = 0.0001$

if ${\bf{A}}$ is not full rank matrix then ${\bf{A}}$ is a degenerate matrix. Rank{$\bf{A}$} $\neq$ min{size($\bf{A}$)}

if ${\bf{A}}$ is a singular matrix then $cond\{ {\bf{A}} \} = \infty $, so is also a degenerate matrix.

high condition number is an ill condition matrix and low condition no. is well condition matrix.

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    $\begingroup$ Wrong. If ${\textbf{A}_{n\times n}}$ $det(\textbf{A}) \neq 0$, then $\textbf{A}$ is invert-able. $\endgroup$
    – user236182
    Jan 7 '18 at 20:03
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matrix is a degenerate if and only if it determinants are non-zero ( non-zero eigenvalues)

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