Standard Basis of a Subspace S of R^n

If S is a subspace of R^n spanned by the vectors x1, x2, . . . xn, form the matrix
A=[x1 x2 x3 . . . xn], that is the columns of A are the vectors in the spanning set. Find the GJSF of Transpose[A], call it M. The nonzero columns of Transpose[M] form the standard basis for S.

Why does this work? Finding the GJSF uses row operations, so we look at Transpose[A] so the original columns become rows in Transpose[A]. The row operations eliminate unnecessary rows (rows that are linear combinations of other rows, therefore not needed in the basis), and leave the remaining rows in simplified form. Taking the transpose back gives us columns, like we started with.

Example

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