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 …

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?

Jan 24, 2011 · If you are considering lists, and repetitions are allowed (and make a list linearly dependent), then Andres's answer is completely correct. If you are considering sets, and …

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 …

Oct 4, 2017 · Well i'm reading in a book that the rank of a matrix is equal to the maximum number of linearly independent rows or, equivalently, the dimension of the row space. So does that …

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.

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 …

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 …

However, every orthonormal set is linearly independent by the above theorem, as every orthonormal set is an orthogonal set consisting of nonzero vectors.

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 …