Basis theorem: for an n-dimensional vector space any linearly independent set with n elements is a basis, as is any spanning set with n elements; dimension of the column space of a matrix equals the number of pivot columns of the matrix; dimension of the null space of a matrix equals the number of free variables of the matrix
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- August 25th, 2017
- 7 years ago
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Any linearly independent set can be expanded to a basis for the (sub)space, arbitrary vector space. math.la.t.vsp.linindep.basis.arb
If a vector space has dimension n, then any subset of n vectors that is linearly independent must be a basis, arbitrary vector space. math.la.t.vsp.dim.linindep.arb
If a vector space has dimension n, then any subset set of n vectors that spans the space must be a basis, arbitrary vector space. math.la.t.vsp.dim.span.arb
The dimension of a subspace is less than or equal to the dimension of the whole space, arbitrary vector space. math.la.t.vsp.subspace.dim.arb
The pivot columns of a matrix are a basis for the column space. math.la.t.mat.col_space.pivot
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August 25th, 2017 7 years ago
Submitted by Petra Taylor
