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Instead of showing it for this particular case, it is actually less work to prove that the null space and row space are orthogonal complements (perpendicular) for any matrix `A.` To do this, suppose that `x in N(A),` so that `Ax=0.` Also, the row space of `A` is the same as the column space of `A^T.` Thus `y in R(A)` if and only if `A^T bary=y` for some vector `bary.` We must show that `x*y=0,` and we can use the fact that `x*y=x^Ty.` We get
`x^Ty=x^T(A^Tbary)=(x^TA^T)bary=(Ax)^Tbary=0^Tbary=0,` ``which is what we needed to show. See the link for a slightly different explanation.
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