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## Homework Statement

Prove that dim(nullA) = dim(null(AV))

(A is a m x n matrix, V is a n x n matrix and is invertible

## Homework Equations

AX=0 and AVX = 0

Null(AV) = span{X1,..Xd}

Null(A) = span{V-1X1,.., V-1Xd}

## The Attempt at a Solution

so you need to prove that dim(null(AV) is a subset of nullA?

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} which 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?

Thank you, in advance :)