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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?

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

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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, $$ \dim\operatorname{nul}(A)=\dim\operatorname{nul}(A^T) $$

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Well... $\dim(Nul(A^T))=\dim (colA)^\perp=n-\dim(col(A))=\dim(nul(A))$.

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  • $\begingroup$ Ahhhhhhh right on. Thanks! $\endgroup$ Commented May 21, 2020 at 17:27

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