Proof Verification: the orthogonal complement of the column space is the left nullspace












1












$begingroup$


Can someone please check my proof and my definitions.



Let $A in mathbb{R}^{n times m}$ be my matrix.



The left null space of $A$ is written as,



$mathcal{N}(A^top) = {x in mathbb{R}^n| A^top x = 0}$



The orthogonal complement of the column space $mathcal{C}(A)$ is written as,



$mathcal{C}(A)^perp = {x in mathbb{R}^n | x^top y = 0, forall y in mathcal{C}(A)}$




We want to show that $mathcal{N}(A^top) = mathcal{C}(A)^perp $




First, we show, $mathcal{N}(A^top) subseteq mathcal{C}(A)^perp$



Let $x in mathcal{N}(A^top)$, then $A^top x = 0 implies x^top A = 0^top implies x^top Av= 0^top v, forall v in mathcal{C}(A) implies x^top y = 0 , y = Av$, $implies x in C(A)^perp$.



Next, we show, $mathcal{N}(A^top) supseteq mathcal{C}(A)^perp$



Let $x in C(A)^perp$, then $x^top y = 0$, forall $y in C(A)$. But $y = Av, forall v in mathbb{R}^n$. Hence, $x^top y = x^top Av = v^top A^top x.$ For all $v neq 0, A^top x = 0$, hence $x in mathcal{N}(A^top)$.




I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.






linear-algebra matrices proof-verification vector-spaces linear-transformations







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$endgroup$

















    1












    $begingroup$


    Can someone please check my proof and my definitions.



    Let $A in mathbb{R}^{n times m}$ be my matrix.



    The left null space of $A$ is written as,



    $mathcal{N}(A^top) = {x in mathbb{R}^n| A^top x = 0}$



    The orthogonal complement of the column space $mathcal{C}(A)$ is written as,



    $mathcal{C}(A)^perp = {x in mathbb{R}^n | x^top y = 0, forall y in mathcal{C}(A)}$




    We want to show that $mathcal{N}(A^top) = mathcal{C}(A)^perp $




    First, we show, $mathcal{N}(A^top) subseteq mathcal{C}(A)^perp$



    Let $x in mathcal{N}(A^top)$, then $A^top x = 0 implies x^top A = 0^top implies x^top Av= 0^top v, forall v in mathcal{C}(A) implies x^top y = 0 , y = Av$, $implies x in C(A)^perp$.



    Next, we show, $mathcal{N}(A^top) supseteq mathcal{C}(A)^perp$



    Let $x in C(A)^perp$, then $x^top y = 0$, forall $y in C(A)$. But $y = Av, forall v in mathbb{R}^n$. Hence, $x^top y = x^top Av = v^top A^top x.$ For all $v neq 0, A^top x = 0$, hence $x in mathcal{N}(A^top)$.




    I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.






    linear-algebra matrices proof-verification vector-spaces linear-transformations







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Can someone please check my proof and my definitions.



      Let $A in mathbb{R}^{n times m}$ be my matrix.



      The left null space of $A$ is written as,



      $mathcal{N}(A^top) = {x in mathbb{R}^n| A^top x = 0}$



      The orthogonal complement of the column space $mathcal{C}(A)$ is written as,



      $mathcal{C}(A)^perp = {x in mathbb{R}^n | x^top y = 0, forall y in mathcal{C}(A)}$




      We want to show that $mathcal{N}(A^top) = mathcal{C}(A)^perp $




      First, we show, $mathcal{N}(A^top) subseteq mathcal{C}(A)^perp$



      Let $x in mathcal{N}(A^top)$, then $A^top x = 0 implies x^top A = 0^top implies x^top Av= 0^top v, forall v in mathcal{C}(A) implies x^top y = 0 , y = Av$, $implies x in C(A)^perp$.



      Next, we show, $mathcal{N}(A^top) supseteq mathcal{C}(A)^perp$



      Let $x in C(A)^perp$, then $x^top y = 0$, forall $y in C(A)$. But $y = Av, forall v in mathbb{R}^n$. Hence, $x^top y = x^top Av = v^top A^top x.$ For all $v neq 0, A^top x = 0$, hence $x in mathcal{N}(A^top)$.




      I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.






      linear-algebra matrices proof-verification vector-spaces linear-transformations







      share|cite|improve this question











      $endgroup$




      Can someone please check my proof and my definitions.



      Let $A in mathbb{R}^{n times m}$ be my matrix.



      The left null space of $A$ is written as,



      $mathcal{N}(A^top) = {x in mathbb{R}^n| A^top x = 0}$



      The orthogonal complement of the column space $mathcal{C}(A)$ is written as,



      $mathcal{C}(A)^perp = {x in mathbb{R}^n | x^top y = 0, forall y in mathcal{C}(A)}$




      We want to show that $mathcal{N}(A^top) = mathcal{C}(A)^perp $




      First, we show, $mathcal{N}(A^top) subseteq mathcal{C}(A)^perp$



      Let $x in mathcal{N}(A^top)$, then $A^top x = 0 implies x^top A = 0^top implies x^top Av= 0^top v, forall v in mathcal{C}(A) implies x^top y = 0 , y = Av$, $implies x in C(A)^perp$.



      Next, we show, $mathcal{N}(A^top) supseteq mathcal{C}(A)^perp$



      Let $x in C(A)^perp$, then $x^top y = 0$, forall $y in C(A)$. But $y = Av, forall v in mathbb{R}^n$. Hence, $x^top y = x^top Av = v^top A^top x.$ For all $v neq 0, A^top x = 0$, hence $x in mathcal{N}(A^top)$.




      I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.






      linear-algebra matrices proof-verification vector-spaces linear-transformations




      linear-algebra matrices proof-verification vector-spaces linear-transformations






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      edited Jan 25 at 16:56







      Aåkon

















      asked Jan 25 at 16:13









      AåkonAåkon

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          $begingroup$

          $y in C(A)$ means that there exists (at least one) $v$ of appropriate dimension such that $y = Av$.



          So we can say: For $x in C(A)^{perp}$, then $x^T y = 0$ for every $y in C(A)$. For every $y in C(A)$, we can express $y = Av$ for some (nonzero) $v$. So we can always express $x^T y$ as $x^T Av$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $v neq 0$, so we must have $A^T x = 0$, i.e., $x in N(A^T)$.






          share|cite|improve this answer











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            $begingroup$

            $y in C(A)$ means that there exists (at least one) $v$ of appropriate dimension such that $y = Av$.



            So we can say: For $x in C(A)^{perp}$, then $x^T y = 0$ for every $y in C(A)$. For every $y in C(A)$, we can express $y = Av$ for some (nonzero) $v$. So we can always express $x^T y$ as $x^T Av$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $v neq 0$, so we must have $A^T x = 0$, i.e., $x in N(A^T)$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              $y in C(A)$ means that there exists (at least one) $v$ of appropriate dimension such that $y = Av$.



              So we can say: For $x in C(A)^{perp}$, then $x^T y = 0$ for every $y in C(A)$. For every $y in C(A)$, we can express $y = Av$ for some (nonzero) $v$. So we can always express $x^T y$ as $x^T Av$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $v neq 0$, so we must have $A^T x = 0$, i.e., $x in N(A^T)$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                $y in C(A)$ means that there exists (at least one) $v$ of appropriate dimension such that $y = Av$.



                So we can say: For $x in C(A)^{perp}$, then $x^T y = 0$ for every $y in C(A)$. For every $y in C(A)$, we can express $y = Av$ for some (nonzero) $v$. So we can always express $x^T y$ as $x^T Av$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $v neq 0$, so we must have $A^T x = 0$, i.e., $x in N(A^T)$.






                share|cite|improve this answer











                $endgroup$



                $y in C(A)$ means that there exists (at least one) $v$ of appropriate dimension such that $y = Av$.



                So we can say: For $x in C(A)^{perp}$, then $x^T y = 0$ for every $y in C(A)$. For every $y in C(A)$, we can express $y = Av$ for some (nonzero) $v$. So we can always express $x^T y$ as $x^T Av$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $v neq 0$, so we must have $A^T x = 0$, i.e., $x in N(A^T)$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 25 at 17:11

























                answered Jan 25 at 16:51









                jjjjjjjjjjjj

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