(a) Prove or disprove: Let A be mxn matrix, and let it be right invertible, then rank(A) = m (b)Show that an nxn matrix A with a unique right inverse A-R must be invertible and A-R = A-1
Please provide proofs.
a) Let A be m x n right invertible matrix. Then there exists n x m matrix such that AB = I where I is m x m matrix .
Thus Range of AB is m dimensional subspace and rank(AB) = m
Then Null space of AB = { x ; AB(x)= 0 } contains only the zero vector.
B is a linear map from Rm Rn
Null space of B , N(B) = { x ; Bx = 0 } is always a subset of Null space of AB. This can be seen as follows.
Let x N(B) . Then Bx = 0 .
AB(x) =A(Bx) = 0 implies x Null space of AB
Thus {0} N(B) Null space of AB = {0}
Therefore Null space of B , N(B) = {0} , that is nullity (B) = 0
By rank nullity theorem rank(B) + nullity(B) = m
Hence rank(B) = m or Range (B) is m dimensional subspace of Rn
Then m n
A is a linear map from Rn to Rm
Rank (A) min { m , n }= m ---(1)
Also m = rank(AB) rank(A) ---(2) { note rank(AB) min { rank(A), rank(B) }
From (1) and (2) rank(A) = m
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