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Dimension and rank 07/04

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Dimension and rank 07/04

Summary (in reverse order to the lecture!)

Given a matrix A ∈ Rm,n, define its kernel or null space KerA = {X ∈ Rn,1 :AX = 0},

consisting of the solutions to the homogeneous system.

The Gauss-Jordan method will give us one basis element for each free parameter. The dimension of a subspace is the number of elements in a basis, so dim(Ker A) = n − r where r is the rank of A.

If R1, . . . , Rm are the rows of A, then its row space RowA = L {R1, . . . , Rm} ⊆ R1,n

is the subspace generated by the rows. A basis consists of the non-zero rows in any step-reduced version of A, so dim(Row A) = r . This gives us a way of finding a basis of any subspace generated by a set of vectors, better than

‘trashing dependents’.

Elements of KerA are columns, and those of Row A rows, but both can be regarded as subspaces of Rn. The methods we have developed can be used to prove the

Theorem. Given two matrices A, A0 of the same size, A ∼ A0 a KerA = Ker A0

a RowA = Row A0.

Any subspace can be defined as RowA for some A, and one can use this to prove that the dimension of a subspace of Rn is at most n and independent of the choice of basis.

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