Linear Algebra Invertible Matrix Theorem Proof












4












$begingroup$


Part of the proof for this theorem asks you to show that if $A$ is an $n times n$-matrix and there exists an $n times n$-matrix $D$ such that $AD = I$ (the $n times n$-identity matrix), then the equation $Ax = b$ has at least one solution for each $b$ in $mathbb{R}^n$.



In order to prove this, I simply started with the hypothesis that $AD = I$ and right-multiplied each side of the equation by the $mathbb{R}^n$ vector $b$ to get:



$AD(b) = I(b) = b.$



By the associative property for matrix multiplication, this leads to $A(Db) = b$, which means that for every $b$ in $mathbb{R}^n$, the vector $Db$ is a solution so that $Ax = b$ has at least that one solution for every $b$ in $mathbb{R}^n$.



I'm pretty sure that's one way of proving it, but can anyone give me any hints about a different or possibly more insightful way of proving this step? It's definitely one of the least straight-forward inferences in the theorem.






linear-algebra matrices linear-transformations







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$endgroup$












  • $begingroup$
    Your proof is correct although (depending on your professor) it can be written much more succinctly. For example: for all $binmathbb{R}^n$, if we let $x=Dbinmathbb{R}^n$, then $Ax=A(Db)=(AD)b=Ib=b$.
    $endgroup$
    – yurnero
    Oct 19 '15 at 0:10








  • 3




    $begingroup$
    @Meta = welcome to math.stackexchange! This is correct and it is probably the most straight forward way to prove it. I also like the good pace (no gaps, no jumps, no stalling) in your argument.
    $endgroup$
    – Hans Engler
    Oct 19 '15 at 0:11
















4












$begingroup$


Part of the proof for this theorem asks you to show that if $A$ is an $n times n$-matrix and there exists an $n times n$-matrix $D$ such that $AD = I$ (the $n times n$-identity matrix), then the equation $Ax = b$ has at least one solution for each $b$ in $mathbb{R}^n$.



In order to prove this, I simply started with the hypothesis that $AD = I$ and right-multiplied each side of the equation by the $mathbb{R}^n$ vector $b$ to get:



$AD(b) = I(b) = b.$



By the associative property for matrix multiplication, this leads to $A(Db) = b$, which means that for every $b$ in $mathbb{R}^n$, the vector $Db$ is a solution so that $Ax = b$ has at least that one solution for every $b$ in $mathbb{R}^n$.



I'm pretty sure that's one way of proving it, but can anyone give me any hints about a different or possibly more insightful way of proving this step? It's definitely one of the least straight-forward inferences in the theorem.






linear-algebra matrices linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your proof is correct although (depending on your professor) it can be written much more succinctly. For example: for all $binmathbb{R}^n$, if we let $x=Dbinmathbb{R}^n$, then $Ax=A(Db)=(AD)b=Ib=b$.
    $endgroup$
    – yurnero
    Oct 19 '15 at 0:10








  • 3




    $begingroup$
    @Meta = welcome to math.stackexchange! This is correct and it is probably the most straight forward way to prove it. I also like the good pace (no gaps, no jumps, no stalling) in your argument.
    $endgroup$
    – Hans Engler
    Oct 19 '15 at 0:11














4












4








4





$begingroup$


Part of the proof for this theorem asks you to show that if $A$ is an $n times n$-matrix and there exists an $n times n$-matrix $D$ such that $AD = I$ (the $n times n$-identity matrix), then the equation $Ax = b$ has at least one solution for each $b$ in $mathbb{R}^n$.



In order to prove this, I simply started with the hypothesis that $AD = I$ and right-multiplied each side of the equation by the $mathbb{R}^n$ vector $b$ to get:



$AD(b) = I(b) = b.$



By the associative property for matrix multiplication, this leads to $A(Db) = b$, which means that for every $b$ in $mathbb{R}^n$, the vector $Db$ is a solution so that $Ax = b$ has at least that one solution for every $b$ in $mathbb{R}^n$.



I'm pretty sure that's one way of proving it, but can anyone give me any hints about a different or possibly more insightful way of proving this step? It's definitely one of the least straight-forward inferences in the theorem.






linear-algebra matrices linear-transformations







share|cite|improve this question











$endgroup$




Part of the proof for this theorem asks you to show that if $A$ is an $n times n$-matrix and there exists an $n times n$-matrix $D$ such that $AD = I$ (the $n times n$-identity matrix), then the equation $Ax = b$ has at least one solution for each $b$ in $mathbb{R}^n$.



In order to prove this, I simply started with the hypothesis that $AD = I$ and right-multiplied each side of the equation by the $mathbb{R}^n$ vector $b$ to get:



$AD(b) = I(b) = b.$



By the associative property for matrix multiplication, this leads to $A(Db) = b$, which means that for every $b$ in $mathbb{R}^n$, the vector $Db$ is a solution so that $Ax = b$ has at least that one solution for every $b$ in $mathbb{R}^n$.



I'm pretty sure that's one way of proving it, but can anyone give me any hints about a different or possibly more insightful way of proving this step? It's definitely one of the least straight-forward inferences in the theorem.






linear-algebra matrices linear-transformations




linear-algebra matrices linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 19 '15 at 0:58









Martin Argerami

127k1182183




127k1182183










asked Oct 19 '15 at 0:06









MetaMeta

334




334












  • $begingroup$
    Your proof is correct although (depending on your professor) it can be written much more succinctly. For example: for all $binmathbb{R}^n$, if we let $x=Dbinmathbb{R}^n$, then $Ax=A(Db)=(AD)b=Ib=b$.
    $endgroup$
    – yurnero
    Oct 19 '15 at 0:10








  • 3




    $begingroup$
    @Meta = welcome to math.stackexchange! This is correct and it is probably the most straight forward way to prove it. I also like the good pace (no gaps, no jumps, no stalling) in your argument.
    $endgroup$
    – Hans Engler
    Oct 19 '15 at 0:11


















  • $begingroup$
    Your proof is correct although (depending on your professor) it can be written much more succinctly. For example: for all $binmathbb{R}^n$, if we let $x=Dbinmathbb{R}^n$, then $Ax=A(Db)=(AD)b=Ib=b$.
    $endgroup$
    – yurnero
    Oct 19 '15 at 0:10








  • 3




    $begingroup$
    @Meta = welcome to math.stackexchange! This is correct and it is probably the most straight forward way to prove it. I also like the good pace (no gaps, no jumps, no stalling) in your argument.
    $endgroup$
    – Hans Engler
    Oct 19 '15 at 0:11
















$begingroup$
Your proof is correct although (depending on your professor) it can be written much more succinctly. For example: for all $binmathbb{R}^n$, if we let $x=Dbinmathbb{R}^n$, then $Ax=A(Db)=(AD)b=Ib=b$.
$endgroup$
– yurnero
Oct 19 '15 at 0:10






$begingroup$
Your proof is correct although (depending on your professor) it can be written much more succinctly. For example: for all $binmathbb{R}^n$, if we let $x=Dbinmathbb{R}^n$, then $Ax=A(Db)=(AD)b=Ib=b$.
$endgroup$
– yurnero
Oct 19 '15 at 0:10






3




3




$begingroup$
@Meta = welcome to math.stackexchange! This is correct and it is probably the most straight forward way to prove it. I also like the good pace (no gaps, no jumps, no stalling) in your argument.
$endgroup$
– Hans Engler
Oct 19 '15 at 0:11




$begingroup$
@Meta = welcome to math.stackexchange! This is correct and it is probably the most straight forward way to prove it. I also like the good pace (no gaps, no jumps, no stalling) in your argument.
$endgroup$
– Hans Engler
Oct 19 '15 at 0:11










2 Answers
2






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$begingroup$

The OP thought that his proof/argument was somehow not straightforward. Actually, in the theory of sets it is well known that $f circ g = 1_{id} text{ iff } f text{ is surjective and } g text{ is injective}$. His argument actually does not use linear algebra and is used in the more general context.



So, in passing, I felt obliged to mention that the argument is indeed straightforward, albeit abstract.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Here's a hint to another approach: There exists a $D$ such that $AD = I$ if and only if the column space of $A$ is all of $mathbb{R}^n$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
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      $begingroup$

      The OP thought that his proof/argument was somehow not straightforward. Actually, in the theory of sets it is well known that $f circ g = 1_{id} text{ iff } f text{ is surjective and } g text{ is injective}$. His argument actually does not use linear algebra and is used in the more general context.



      So, in passing, I felt obliged to mention that the argument is indeed straightforward, albeit abstract.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The OP thought that his proof/argument was somehow not straightforward. Actually, in the theory of sets it is well known that $f circ g = 1_{id} text{ iff } f text{ is surjective and } g text{ is injective}$. His argument actually does not use linear algebra and is used in the more general context.



        So, in passing, I felt obliged to mention that the argument is indeed straightforward, albeit abstract.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The OP thought that his proof/argument was somehow not straightforward. Actually, in the theory of sets it is well known that $f circ g = 1_{id} text{ iff } f text{ is surjective and } g text{ is injective}$. His argument actually does not use linear algebra and is used in the more general context.



          So, in passing, I felt obliged to mention that the argument is indeed straightforward, albeit abstract.






          share|cite|improve this answer









          $endgroup$



          The OP thought that his proof/argument was somehow not straightforward. Actually, in the theory of sets it is well known that $f circ g = 1_{id} text{ iff } f text{ is surjective and } g text{ is injective}$. His argument actually does not use linear algebra and is used in the more general context.



          So, in passing, I felt obliged to mention that the argument is indeed straightforward, albeit abstract.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 '18 at 2:29









          CopyPasteItCopyPasteIt

          4,2031628




          4,2031628























              0












              $begingroup$

              Here's a hint to another approach: There exists a $D$ such that $AD = I$ if and only if the column space of $A$ is all of $mathbb{R}^n$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Here's a hint to another approach: There exists a $D$ such that $AD = I$ if and only if the column space of $A$ is all of $mathbb{R}^n$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Here's a hint to another approach: There exists a $D$ such that $AD = I$ if and only if the column space of $A$ is all of $mathbb{R}^n$.






                  share|cite|improve this answer









                  $endgroup$



                  Here's a hint to another approach: There exists a $D$ such that $AD = I$ if and only if the column space of $A$ is all of $mathbb{R}^n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 19 '15 at 0:17









                  Michael BiroMichael Biro

                  10.8k21831




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