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Singular matrix is defined as a square matrix with determinant of zero. I am aware that linear dependency among columns or rows leads to determinant being equal to zero (e.g. one column is a linear composite of other columns).

I am interested whether other conditions exist, except for linear dependency, that can lead to determinant of 0 in Singular Matrix.

Excuse me if I am missing on something obvious, as I am somewhat of a beginner.

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  • $\begingroup$ What kind of answer are you looking for? Confirmation that singular matrices imply linear dependence of their columns? Or are you asking whether there are other tests for singularity besides computing the determinant or evaluating linear dependency? $\endgroup$ – whuber Oct 9 at 12:55
  • $\begingroup$ @whuber thank you for your reply. I am really looking to understand whether it is possible to have det(A) = 0 in a singular matrix, but no linear dependency between rows or columns. And if so, what situation could result in det(A) = 0 in a singular matrix, other than some sort of linear dependency. $\endgroup$ – PsychometStats Oct 9 at 13:14
  • $\begingroup$ The basic theorem of linear algebra involved here is that the determinant is zero if and only if the columns are linearly dependent. One way to understand that is afforded by the geometric interpretation of the determinant as the hypervolume of the parallelepiped formed by the origin and the column vectors. $\endgroup$ – whuber Oct 9 at 13:23
  • $\begingroup$ @whuber thank you so much! I greatly appreciate your explanation! $\endgroup$ – PsychometStats Oct 9 at 13:30
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I think you are operating on an incorrect understanding about the relationship between linear dependence and singularity. Many matrices are not square, and thus do not have a determinant, yet they can have columns that are linearly dependent or independent. In general, if the columns of the matrix $\mathbf{x}$ are linearly dependent then the determinant of the Gramian matrix of $\mathbf{x}$ is zero. That is, you have:

$$\det (\mathbf{x}^\text{T} \mathbf{x}) = 0 \quad \quad \iff \quad \quad \text{columns of matrix } \mathbf{x} \text{ are linearly dependent}.$$

This relationship holds for matrices of any dimension. However, in the special case where $\mathbf{x}$ is a square matrix, you then have $\det (\mathbf{x}^\text{T} \mathbf{x}) = (\det \mathbf{x})^2$, which means that:

$$\quad \quad \quad \quad \det (\mathbf{x}) = 0 \quad \quad \iff \quad \quad \text{columns of square matrix } \mathbf{x} \text{ are linearly dependent}.$$

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  • $\begingroup$ Thank you for your reply. I am aware that determinant can only be calculated for square matrices. Equally, I am aware that linear dependency can exist in both square and non-square matrices.I am really looking to understand whether it is possible to have det(A) = 0 in a singular matrix, but no linear dependency between rows or columns. And if so, what situation could result in det(A) = 0 in a singular matrix, other than some sort of linear dependency. $\endgroup$ – PsychometStats Oct 9 at 13:16
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    $\begingroup$ Okay, but I think you'll find that that is answered above. The $\iff$ symbol means "if and only if", so you can see that $\det \mathbf{x} = 0$ if and only if the columns of the square matrix $\mathbf{x}$ are linearly dependent. In other words, it is not possible to have a square matrix with zero determinant, but with linearly independent columns. $\endgroup$ – Reinstate Monica Oct 9 at 13:24
  • $\begingroup$ Thank you!!! I now understand this! $\endgroup$ – PsychometStats Oct 9 at 13:31

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