Can the determinant (assuming it's non-zero) be used to determine that the vectors given are linearly independent, span the subspace and are a basis of that subspace?

12 you can take the vectors to form a matrix and check its determinant. If the determinant is non zero, then the vectors are linearly independent. Otherwise, they are linearly dependent.

I have a very hard time remembering which is which between linear independence and linear dependence... that is, if I am asked to specify whether a set of vectors are linearly dependent …

Aug 4, 2021 · I have attached a screenshot in which a variable is defined for an object somehow that it linearly decreases from 500 micrometers at the top of the object to 50 micrometers at …

Jan 24, 2016 · Old thread, but in fact putting the vectors in as columns and then computing reduced row echelon form gives you more insight about linear dependence than if you put …

None of the columns are multiples of the others, but the columns do form a linearly dependent set. You know this without any real work, since $3$ vectors in $\mathbb {R}^2$ cannot form a …

Jul 20, 2013 · Another alternative for testing is to check for the determinant for each matrices (this may look tedious for a complicated matrix system), If the determinant is non zero, It is said to …

Feb 23, 2017 · Any set of linearly independent vectors can be said to span a space. If you have linearly dependent vectors, then there is at least one redundant vector in the mix.

How to prove that eigenvectors from different eigenvalues are linearly independent [duplicate] Ask Question Asked 14 years, 8 months ago Modified 3 years, 11 months ago

A set of vectors is linearly dependent when there are an infinite amount of solutions to the system of equations. This is non-trivial? Where does no solution come in? I understand that if there is …