Linear Algebra
Prove that dim(nullA) = dim(null(AV)) A is a m x n matrix. V is a n x n matrix and it is invertible.
so you need to prove that dim(null(AV) is a subset of nullA?
This is what I did:
Therefore d = dimA = dim(AV) and let nullA= span{X1,..Xd} and null(AV) = span{ X1,..Xd}.
Null(AV) = span{X1,..Xd}
Null(A) = span{V-1X1,.., V-1Xd}
If a1(V-1X1 ) + …+a2(V-1Xd) = 0
So 0 = VV-1(t1X1 +… + t1Xd)
0 = t1X1 +… + t1Xd all of ti = 0 meaning it is linearly independent, and also span {V-1X1,.., V-1Xd} is a basis of nullA
AX = 0 so AVX=O then
V0 = AVX which is in null(AV)
This also means V0 = t1X1 +… + t1Xd
which also means 0 = t1V-1X1 +…+ tdV-1Xd which spans nullA.
Is this correct?
Prove that dim(nullA) = dim(null(AV)) A is a m x n matrix. V is a n x n matrix and it is invertible.
so you need to prove that dim(null(AV) is a subset of nullA?
This is what I did:
Therefore d = dimA = dim(AV) and let nullA= span{X1,..Xd} and null(AV) = span{ X1,..Xd}.
Null(AV) = span{X1,..Xd}
Null(A) = span{V-1X1,.., V-1Xd}
If a1(V-1X1 ) + …+a2(V-1Xd) = 0
So 0 = VV-1(t1X1 +… + t1Xd)
0 = t1X1 +… + t1Xd all of ti = 0 meaning it is linearly independent, and also span {V-1X1,.., V-1Xd} is a basis of nullA
AX = 0 so AVX=O then
V0 = AVX which is in null(AV)
This also means V0 = t1X1 +… + t1Xd
which also means 0 = t1V-1X1 +…+ tdV-1Xd which spans nullA.
Is this correct?