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Let $A:E \rightarrow F$ be a linear mapping and E,F two finite-dimension inner vector spaces. I want to prove that 1) $N(A^*)=Im(A)^\bot$ 2) $Im(A^*)=N(A)^\bot$ 3) $N(A)=Im(A^*)^\bot$ 4) $Im(A)=N(A^*)^\bot$

Where $A^*$ is the adjoint of $A$ and $\bot$ denotes the orthogonal complement, i.e., $Im(A)^\bot$, for example, is the set whose elements are orthogonal to all vectors of $Im(A)$. I managed to prove (1) this way:

1) $v\in N(A^*) \Rightarrow A^*v=0 \Rightarrow <u,A^*v>=0, \forall u\in E \Rightarrow <Au,v>=0, \forall u\in E \Rightarrow v \in Im(A)^\bot$.

I want some help to prove (2),(3),(4). Thank you.

PinkyWay
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Walter r
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1 Answers1

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For your prove of (1) note that you have to replace the $\Rightarrow$ by $\iff $, because as written, you only proved $\def\N{\mathop{\rm N}}\N(A^*) \subseteq \def\Im{\mathop{\rm Im}}\Im(A)^\bot$.

For (2) just apply (1) to $A^*$, using $A^{**} = A$, giving $\N(A) = \Im(A^*)^\bot$, now take orthogonal complements and use that your spaces are finite-dimensional.

(3) and (4) are (1) and (2) applied to $A^*$.

Addendum. If you want to prove (2) as you did with (1), say, you can do this as follows: Let $v \in \Im(A^*)$, say $v = A^*w$, for any $x \in \N(A)$ we have $\def\sp#1{\left<#1\right>}$ $$ \sp{v, x} = \sp{A^*w, x} = \sp{w, Ax} = \sp{w, 0} = 0 $$ so $v \in \N(A)^\bot$. So we have $\Im(A^*) \subseteq \N(A)^\bot$. Counting dimensions, we have \begin{align*} \dim\N(A)^\bot &= \dim E - \dim \N(A)\\ &= \dim E - \bigl(\dim E - \dim \Im(A)\bigr) \\ &= \dim\Im(A)\\ &= \dim\Im(A^*) \end{align*} So we must have $\Im(A^*) = \N(A)^\bot$.

martini
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  • Good morning, Martini. Ok, but I search a direct proof of (2), (3) and (4). – Walter r Oct 16 '13 at 13:33
  • In what way is the given proof "indirect"? – martini Oct 16 '13 at 13:35
  • Added something for (2). (3) and (4) are exactly the same as (1) and (2), just exchange $A$ and $A^*$. You can do this for yourself, I'm sure. – martini Oct 16 '13 at 13:53
  • Thank you, Martini. Your explanation in "adendum" was very clear! – Walter r Oct 16 '13 at 14:29
  • I think this is another way to prove the second part of (2): $v \in N(A)^\bot \rightarrow <v,x>=0, \forall x \in N(A)$.But $x \in N(A) \rightarrow Ax=0. Then, \forall w \in F, <Ax,w>=<x,A^w>=0$. This implies $<v,x>=<x,A^w> \Rightarrow v=A^w \Rightarrow v \in Im(A^)$ – Walter r Oct 16 '13 at 14:57
  • But $\def\sp#1{\left<#1\right>}$$\sp{v,x} = \sp{A^w, x}$ for all $x \in {\rm N}(A)$ does not* imply $v = A^w$, but only $v- A^w \in {\rm N}(A)^\bot$. – martini Oct 16 '13 at 15:01