# dim(nul(A)) = dim(nul(A^T))?

I am currently trying to prove that the union of an orthogonal subspace $$W$$ and its orthogonal complement $$W^\perp$$ span $$\Bbb R^n$$. In order to do this, I am trying to use the Rank-Nullity theorem.

If $$A$$ is a matrix with its columns being the basis of $$W$$, then I know $$ColA = W$$. I also know $$(colA)^\perp = W^\perp = nul(A^T)$$.

The only thing I'm stuck on is $$\dim(nul(A)) = \dim(nul(A^T))$$? If this is the case, then by the Rank-Nullity theorem $$\dim(nul(A)) + \dim(col(A)) = \dim(\Bbb R^n) \iff \dim(W^T) + \dim(W) = \dim(R^n)$$.

Is it right in general to assume $$\dim(nul(A)) = \dim(nul(A^T))$$? If not, how should I go about solving this problem?

• Yes. Row rank equals column rank, i.e. $\dim Col A^T = \dim Col A$. Now apply the rank-nullity theorem. Commented May 30, 2020 at 3:02

Not generally. The rank of $$A$$ equals the rank of $$A^T$$. If $$A$$ is $$m\times n$$, then the rank-nullity theorem applied to $$A$$ gives $$n=\dim\operatorname{nul}(A)+\operatorname{rk}(A)$$ and applied to $$A^T$$ it gives $$m=\dim\operatorname{nul}(A^T)+\operatorname{rk}(A^T)$$ By subtraction, in general you can say that $$n+\dim\operatorname{nul}(A^T)=m+\dim\operatorname{nul}(A)$$ and, when $$m=n$$, which is your case, $$\dim\operatorname{nul}(A)=\dim\operatorname{nul}(A^T)$$
Well... $$\dim(Nul(A^T))=\dim (colA)^\perp=n-\dim(col(A))=\dim(nul(A))$$.