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If $A$ is a $n\times m$ matrix, is the formula $(\text{ker }A)^\perp=\text{im }A^T$ necessarily true?

I'm thinking that rank-nullity would be the simplest and easiest way to prove this, but would applying the definition of linear independence to the components of $(A^T)$ be more illustrative? I'm a bit hung up on this "necessity" issue, and not sure if it has a particular logical meaning in linear algebra.

Travis Willse
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Benjamin Loya
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    Rank-nullity might enable you to prove that the two spaces you mention have the same dimension, but without extra information, this cannot possibly prove that they are the same space. – David Mar 25 '15 at 03:14
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    "Necessarily" here just emphasizes that the question is whether the identity holds for all matrices $A$. – Travis Willse Mar 25 '15 at 03:23
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    by definition, any vector in $\ker A$ is orthogonal to every row of $A.$ that means $\ker A \subset (im A^T)^\perp$ or equivalently $im A^T \subset (\ker A)^\perp$ – abel Mar 25 '15 at 03:41
  • @abel but is this sufficient to show $(\text{ker }A)^\perp=\text{im }A^T$? Would a contradiction showing a vector $\vec{v}\in \text{im }(A^T), \notin(\text{ker }A)^\perp$ even be possible? My intuition says yes but the definitions seem to leave no exception. – Benjamin Loya Mar 25 '15 at 04:01
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    i believe you need the rank nullity theorem for that. i am not sure you can prove that without it. – abel Mar 25 '15 at 04:02
  • And also this more recent question – Arnaud D. Aug 22 '18 at 09:43

1 Answers1

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I assume that the underlying field is $\mathbb R$. As the result does not hold in infinite-dimensional spaces (see closed range theorem), understandably the proof uses a dimensionality argument.

Let $y \in im A^T$ and $x\in \ker A$. Then there is $u$ such that $y = A^Tu$. This gives $$ x^T y = x^T(A^Tu) = (Ax)^Tu=0. $$ Hence $im A^T \subset (\ker A)^\perp$.

Let now $x\in (im A^T)^\perp$. Then $x^T (A^T u)=0$ for all $u$, which implies $(Ax)^Tu=0$ for all $u$. Setting $u:=Ax$ yields $x\in \ker A$, this proves $(im A^T)^\perp \subset \ker A$, and $\ker A^\perp \subset (im A^T)^{\perp\perp}$.

Finally, using $U^{\perp\perp}=U$, which holds for all subspaces of finite-dimensional inner product spaces, we conclude $$ im A^T \subset (\ker A)^\perp \subset (im A^T)^{\perp\perp}=im A^T, $$ which is the claim. The proof $U^{\perp\perp}=U$ uses a dimension argument for subspaces (no rank-nullity). It is easy to see that $U\subset U^{\perp\perp}$ and $\dim U = \dim U^{\perp\perp}$.

daw
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