Linear Algebra Invertible Matrix Theorem Proof
$begingroup$
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
edited Oct 19 '15 at 0:58
Martin Argerami
127k1182183
asked Oct 19 '15 at 0:06
MetaMeta
334
$endgroup$
add a comment |
$begingroup$
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
edited Oct 19 '15 at 0:58
Martin Argerami
127k1182183
asked Oct 19 '15 at 0:06
MetaMeta
334
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Oct 19 '15 at 0:58
Martin Argerami
127k1182183
asked Oct 19 '15 at 0:06
MetaMeta
334
$endgroup$
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
edited Oct 19 '15 at 0:58
Martin Argerami
127k1182183
asked Oct 19 '15 at 0:06
MetaMeta
334
edited Oct 19 '15 at 0:58
Martin Argerami
127k1182183
asked Oct 19 '15 at 0:06
MetaMeta
334
edited Oct 19 '15 at 0:58
Martin Argerami
127k1182183
edited Oct 19 '15 at 0:58
Martin Argerami
127k1182183
edited Oct 19 '15 at 0:58
Martin Argerami
127k1182183
127k1182183
asked Oct 19 '15 at 0:06
MetaMeta
334
asked Oct 19 '15 at 0:06
MetaMeta
334
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Jan 5 '18 at 2:29
CopyPasteItCopyPasteIt
4,2031628
$endgroup$
add a comment |
$begingroup$
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 $mathbb{R}^n$.
answered Oct 19 '15 at 0:17
Michael BiroMichael Biro
10.8k21831
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add a comment |
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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.
answered Jan 5 '18 at 2:29
CopyPasteItCopyPasteIt
4,2031628
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 '18 at 2:29
CopyPasteItCopyPasteIt
4,2031628
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 '18 at 2:29
CopyPasteItCopyPasteIt
4,2031628
$endgroup$
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.
answered Jan 5 '18 at 2:29
CopyPasteItCopyPasteIt
4,2031628
answered Jan 5 '18 at 2:29
CopyPasteItCopyPasteIt
4,2031628
answered Jan 5 '18 at 2:29
CopyPasteItCopyPasteIt
4,2031628
answered Jan 5 '18 at 2:29
CopyPasteItCopyPasteIt
4,2031628
4,2031628
add a comment |
add a comment |
$begingroup$
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 $mathbb{R}^n$.
answered Oct 19 '15 at 0:17
Michael BiroMichael Biro
10.8k21831
$endgroup$
add a comment |
$begingroup$
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 $mathbb{R}^n$.
answered Oct 19 '15 at 0:17
Michael BiroMichael Biro
10.8k21831
$endgroup$
add a comment |
$begingroup$
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 $mathbb{R}^n$.
answered Oct 19 '15 at 0:17
Michael BiroMichael Biro
10.8k21831
$endgroup$
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 $mathbb{R}^n$.
answered Oct 19 '15 at 0:17
Michael BiroMichael Biro
10.8k21831
answered Oct 19 '15 at 0:17
Michael BiroMichael Biro
10.8k21831
answered Oct 19 '15 at 0:17
Michael BiroMichael Biro
10.8k21831
answered Oct 19 '15 at 0:17
Michael BiroMichael Biro
10.8k21831
10.8k21831
add a comment |
add a comment |
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$begingroup$
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