Linearly Independent Columns of a Matrix

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If the determinant of a matrix P is not equal to zero, then the columns of P are linearly independent, indicating that P is invertible and has maximal rank. The reasoning is that a matrix can be transformed into upper diagonal form through row operations, and a non-zero determinant implies that the final matrix is diagonal with non-zero entries. This confirms that the columns of the original matrix are independent. Understanding this concept is crucial for linear algebra, especially in contexts like exams. The discussion highlights the importance of grasping the relationship between determinants and linear independence.
sajama
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Hi wondering if can anyone help me... I've gotten so bogged down in all the rules and stuff for singular/non-singular matrices I've completely confused myself!

Can anyone tell me is it true to say that if I have a matrix P, det(P) is NOT EQUAL to 0, then the vectors that would form the columns of P are linearly independent?

Cheers
 
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Yes because then P is invertible, and thus have maximal rank.
 
Brilliant - cheers! :)
 
do you know why this is true? i.e. do you know what it really means that the determinant is not zero?

recall that any matrix can be rendered into upper diagonal form by repeatedly performing row operations, hence also by repeatedly multiplying by special invertible matrices.

then the determinant is non zero iff the final matrix is actually diagonal and has non zero entries on the diagonal. these can then be made 1's.

hence it has been inverted by matrix multiplication, and the columns are visibly independent, hence were also originally.
this is just one of many ways to see it.
 
Thank you - it does make sense now - I had a Linear Maths exam this afternoon and even though at some stage I had understood the reasoning behind what made a matrix singular, my mind seemed to be blanking on me in the hours leading up to the exam!

Thanks again for the help :)
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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