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I'm having problems with the best way to work out linearly independent sets of matrices.

When the set can be made into a square matrix, such as $ \begin{bmatrix} 1 \\ 0\end{bmatrix}, \begin{bmatrix} 1 \\ 3\end{bmatrix} $, the determinate $ det(\begin{bmatrix} 1 & 1 \\ 0 & 3\end{bmatrix}) $ can be used, where a non-zero number means linearly independent, otherwise it is dependent.

However, I'm not sure how to work it out with a set that does not generate a square matrix and hence cannot use the determinate such as $ \begin{bmatrix} 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ 3 \end{bmatrix} $ or $ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} $?

Another rule I could see that may apply here is that if $ c_1 \mathbf v_1 + c_2 \mathbf v_2 + ... + c_n \mathbf v_n = 0 $ then the vectors are said to be linearly dependent, however applying this rule to the example of $ \begin{bmatrix} 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ 3 \end{bmatrix} $ produces two equations with three unknowns.

What is the proper way of finding if a set of matrices are linearly dependent or independent where the sets don't produce a square matrix?

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You only need to worry about matrices with more rows than columns. If it has more columns than rows, it's always linearly dependent, because every set with more than $n$ elements is linearly dependent in a vector space with dimension $n$. – fgp Oct 16 '12 at 0:38

Some of the general properties of square matrices still follow to non-square matrices.

If you row reduce the matrix to reduced row echelon form, then the non-zero rows will still form a basis for the rowspace. Likewise, the columns corresponding to the pivot columns will still form a basis for the columnspace.

All of these results are actually extraneous since you seem to only be interested in whether the vectors are linearly independent or not. In that case, you only need to calculate the rank of the matrix (i.e. count the number of pivots), which will be the size of the maximal linearly independent set that your vectors can form.

If you are feeling particularly masochistic, you can even calculate $A^\mathrm{T}A$ which will be square with the same rank.

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Simply compute the rank of the matrix formed by the vectors.

If you insist using the determinant, you can compute the Gram determinant, which is zero for linear dependent vectors.

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