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I am a novice in the field of matrices, and I am doing some self-learning. I would like to ask what might be a very basic question, but I would be very appreciative of your help.

Singular matrix is defined as a square matrix with the determinant of zero. The determinant of zero occurs when matrix columns are linearly dependent (i.e. one of the columns can be defined as a linear combination of other columns).

However, some sources also note that the determinant can be zero when there is linear dependency not only among the columns but also among the rows of a square matrix. Therefore, I wanted to ask the following in the context of the singular matrices:

Does linear dependency among columns imply that there is automatically linear dependency among rows of data?

Or, does linear dependency result from only columns, or only rows, or both?

Note: I originally posted this question on Cross-validated, but I did not get an answer. So I thought that maybe this network is more appropriate for this type of question.

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Yes, it does! Row dependency implies column dependency and vice versa in square matrices i.e. the only matrices where you can calculate the determinant in the first place.

One way to prove this is by noticing that taking the transpose doesn't change the determinant of the matrix. So if your matrix was singular to begin with, so is its transpose, but this means there is dependency in the columns of the transpose, which are the rows of your initial matrix.

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  • $\begingroup$ That's an intuitive way to think about this problem indeed!! Thank you so much! $\endgroup$ Nov 11 '19 at 23:52
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For a square matrix the answer is yes as the rank of a matrix determines the dimensions of both the row and the column spaces. However for non-square matrices it is not true. For example, take $$A=\begin{bmatrix}1&2&3&4\\0&1&2&3\end{bmatrix}$$ then rows of $A$ are linearly independent but the columns are not.

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  • $\begingroup$ Thank you so much for the explanation and for the example with a non-square matrix! $\endgroup$ Nov 11 '19 at 23:53
  • $\begingroup$ Yet the column space has rank $2$, just like the row space. $\endgroup$
    – Bernard
    Nov 11 '19 at 23:53

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