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I am really struggling with understanding how to do this proof directly.

Let A be a linear transformation of a vector space V and let A' be its adjoint. Show directly that $\mathrm{nullspace}(A)^o=\mathrm{range}(A')$.

This is where I am at...

  1. Let $h \in \text{Im } A^{*}$. Then there exists $f \in V^{*}$ such that $f(A(v)) = (A^{*}f)(v) = h(v)$ for all $v \in V$. But then $h \in (\ker A )^{\circ} \iff h(v) =0$ for all $v \in \ker A$, but $v \in \ker A \implies h(v) = f(A(v)) = f(0) = 0$. Hence, $\text{Im }A^{*} \subset (\ker A)^{\circ}$.
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PrincessEev
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MathIsHard
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  • Related thread: http://math.stackexchange.com/questions/21144/intuitive-explanation-of-the-fundamental-theorem-of-linear-algebra – littleO Apr 08 '16 at 02:31

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  1. Let $f \in (\text{Im } A)^{\circ}$. Then $f \in V^{*}$ annihilates the image so that for all $v \in V$, we have $0 = f(A(v)) = (A^{*}(f))(v) \implies A^{*}f =0 \implies f \in \ker A^{*}$. We have $(\text{Im } A)^{\circ} \subset \ker A^{*}$.

  2. Let $g \in \ker A^{*}$. Then for all $v \in V$, $0= (A^{*}g)(v) =g(A(v) \implies g \in (\text{Im } A)^{\circ}$. We have $\ker A^{*} \subset (\text{Im } A)^{\circ}$.

  3. Let $h \in \text{Im } A^{*}$. Then there exists $f \in V^{*}$ such that $f(A(v)) = (A^{*}f)(v) = h(v)$ for all $v \in V$. But then $h \in (\ker A )^{\circ} \iff h(v) =0$ for all $v \in \ker A$, but $v \in \ker A \implies h(v) = f(A(v)) = f(0) = 0$. Hence, $\text{Im }A^{*} \subset (\ker A)^{\circ}$.

  4. Can you use 1, 2, and 3 and anything else you know about rank-nullity theorem and how dimensions of all these objects are related to show that $\dim \text{Im } A^{*} = \dim (\ker A)^{\circ}$ and then conclude the equality?

Daniel Akech Thiong
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  • Thank you so much for the response. I will definitely take the time to go through your proof. I appreciate it. – MathIsHard Apr 08 '16 at 02:24
  • I need a bit more help if the values starts in N(A)^o – MathIsHard Apr 08 '16 at 21:18
  • Can you use 1, 2, and 3 and anything else you know about rank-nullity theorem and how dimensions of all these objects are related to show that $\dim \text{Im } A^{*} = \dim (\ker A)^{\circ}$ and then conclude the equality? – Daniel Akech Thiong Apr 08 '16 at 22:07
  • Oh so I need to prove the dimensions are the same to show that $kernal(A)^o$ is a subspace of $image(A)^*$? – MathIsHard Apr 08 '16 at 22:18
  • You do not have to [perhaps you do not know how to --- you cannot ignore using Rank-Nullity theorem because it is quite powerful a tool], but it is a standard way of cutting down your work in finite dimensional linear algebra [as opposed to infinite dimensional linear algebra a.k functional analysis] that if you know that two subspaces have the same dimension and one contains the other then they are in fact equal. – Daniel Akech Thiong Apr 08 '16 at 22:22
  • I have been looking at this for a while now. I said to let x exist in $N(A)^o$ then x exists in V' annihilates N(A) so that for all w in N(A) we have 0=x(w) but I am not sure what to do now... I would really appreciate any help you could give. Thanks. – MathIsHard Apr 09 '16 at 04:41
  • Can you show that the two subspaces in question have the same dimension? By the way, there are usually reasons as to why proceeding directly may be harder and this is why tools in your disposal do not only consist of definitions but also of any results proven in the text or lecture notes prior to your questions. – Daniel Akech Thiong Apr 09 '16 at 10:56