Non-square matrix and linear independence

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A non-square matrix can have linearly independent columns if it has more rows than columns. For example, a 3x2 matrix can have independent columns. However, if a matrix has more columns than rows, the columns cannot be independent due to dimensional limitations. Specifically, if there are m columns and n rows with m greater than n, the columns will not be independent. Thus, the independence of columns in a non-square matrix depends on its dimensions.
torquerotates
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Can a non square matrix have linearly independent columns? I can't take the determinant so I can't tell.
 
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torquerotates said:
Can a non square matrix have linearly independent columns? I can't take the determinant so I can't tell.

Yes. For instance,
\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right]
Of course it will have to have more rows than columns.
 
If, on the other hand, the matrix has more columns than rows, the columns cannot be independent. If there are say, m columns and n rows, with m> n, then the columns are n dimensional vectors and a set of m vectors of n dimensions, with m> n, cannot be independent.
 
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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