Are all non-/sigular matrices linearly in/dependent?

Hi all, from the lecture, I learned that linearly dependent matrix implies that it is sigunlar, and linearly independent matrix implies that it is non-sigunlar.

I wonder if the other way around is true - are all sigular matrices linearly dependent, and all non-sigular matrices linearly independent?


Hi @chenhang

Welcome to the community!

A matrix M is singular if and only if det(M) = 0. So the logic goes in both directions! In the case det(M) = 0 the inverse M^{-1} is not defined resp. cannot be computed since at least one column vector is linearly dependent on at least one other column vector and the matrix does not have full rank. (The equation described by such a matrix might not have a unique solution.)

If the determinant is unequal zero, however this means the matrix has full rank (each column vector contributes to span the space) and can be inverted accordingly. Per definition it is non-singular since the vectors are not linearly dependent and an inverse can be calculated!

Please let me know if this answers your question.

Best regards

Thank you very much for your answer, Christian!

I think I might need to study more to understand your answer, as I am unfamiliar with terms such as inverse and rank. But from your answer, I can infer that all sigular matrices are linearly dependent, and all non-sigular matrices are linearly independent, correct?

Thank you!

1 Like

I think you cannot call a matrix just linear dependent. Or what should it be dependent on?
I think you mean the vectors or columns inside that matrix with respect to linear dependency. But then your statement would be fine.

Therefore, I would phrase it:

  • for all singular matrices it applies: the belonging vectors are not linearly independent (= at least one vector is linearly dependent on at least one other vector)
  • for all non-singular matrices it applies that belonging vectors are linearly independent

Hope that answers your question, @chenhang!

Best regards