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Prove that if $A$ ∈ R$^n$$^x$$^m$ then rank($A$) = rank($A^T$).

I'm trying to make a very mathematical proof. I've tried everything and I have no clue what to do. I have so far been able to use rank-nullity theorem to show that null($A$) ⊆ null($A^TA)$, but I don't know where to go from there.

J. W. Tanner
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prismarine
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2 Answers2

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Let $A$ be a matrix (not necessarily square). Reduce $A$ to its reduced row echelon form $A'$ using row transformations. Then the row rank of $A$ is the same as the row rank of $A'$, the same as the number of pivots is $A'$, and the same as the column rank of $A'$. But when we do a row transformation of a matrix, its column rank stays the same (easy to check because row operations preserve linear dependencies of columns). Hence the column rank of $A'$ is the same as the column rank of $A$ and the same as the row rank of $A$. But the row rank of $A$ is $\mathrm{rank}(A)$ while the column rank of $A$ is $\mathrm{rank}(A^T)$. QED

markvs
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  • I don't really understand. How does this show that rank(A) = rank(A$^T$)? – prismarine Sep 29 '20 at 03:54
  • The first rank is the row rank of $A$ while the second rank is the column rank of $A$, and the answer proves that the ranks are the same because both are equal to the rank of $A'$. – markvs Sep 29 '20 at 04:01
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Here's another strategy, although I like JCAA's explanation better.

Notice $[\text{Col}(A)]^{\perp}=N(A^T)$ and $\text{Col}(A) \oplus [\text{Col}(A)]^{\perp}=\mathbb{R}^n$. From this we immediately see $\text{nullity}(A^T)=n-\text{rank}(A)$ and so with rank nullity

$$n=\text{rank}(A^T)+\text{nullity}(A^T)=\text{rank}(A^T)+n-\text{rank}(A)$$ which boils down to $\text{rank}(A)=\text{rank}(A^T)$. Also, if you're looking to prove that $N(A^TA)\subseteq N(A)$ note $\vec{x}\in N(A^TA)$ implies $$0=\vec{x}^TA^TA\vec{x}=(A\vec{x})^T(A\vec{x})=||A\vec{x}||^2$$ and so $A\vec{x}=0$ i.e. $\vec{x}\in N(A)$.

  • This really helps! Only question is it wasn't clear why nullity()=−rank(). Would you mind explaining this a bit more? – prismarine Sep 29 '20 at 05:26