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How to prove that $A^T y=0implies y=0$ if and only if $A$ defines a surjective linear transformation?
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I know that in vector spaces, the cokernel of a linear transformation $f$ is isomorphic to the set of all $y$ such that $f^T(y)=0.$ (This is the transpose of $f$.) Then, how can I prove that $text{coker} f={0}$ if and only if $f$ is surjective?
I would apreciate a proof without any heavy machinery (that is, without something that is not taught in a linear algebra course).
Thank you
linear-algebra abstract-algebra category-theory
asked Jan 10 at 11:33
Gabriel RibeiroGabriel Ribeiro
1,441522
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I know that in vector spaces, the cokernel of a linear transformation $f$ is isomorphic to the set of all $y$ such that $f^T(y)=0.$ (This is the transpose of $f$.) Then, how can I prove that $text{coker} f={0}$ if and only if $f$ is surjective?
I would apreciate a proof without any heavy machinery (that is, without something that is not taught in a linear algebra course).
Thank you
linear-algebra abstract-algebra category-theory
asked Jan 10 at 11:33
Gabriel RibeiroGabriel Ribeiro
1,441522
$endgroup$
add a comment |
$begingroup$
I know that in vector spaces, the cokernel of a linear transformation $f$ is isomorphic to the set of all $y$ such that $f^T(y)=0.$ (This is the transpose of $f$.) Then, how can I prove that $text{coker} f={0}$ if and only if $f$ is surjective?
I would apreciate a proof without any heavy machinery (that is, without something that is not taught in a linear algebra course).
Thank you
linear-algebra abstract-algebra category-theory
asked Jan 10 at 11:33
Gabriel RibeiroGabriel Ribeiro
1,441522
$endgroup$
I know that in vector spaces, the cokernel of a linear transformation $f$ is isomorphic to the set of all $y$ such that $f^T(y)=0.$ (This is the transpose of $f$.) Then, how can I prove that $text{coker} f={0}$ if and only if $f$ is surjective?
I would apreciate a proof without any heavy machinery (that is, without something that is not taught in a linear algebra course).
Thank you
linear-algebra abstract-algebra category-theory
linear-algebra abstract-algebra category-theory
asked Jan 10 at 11:33
Gabriel RibeiroGabriel Ribeiro
1,441522
asked Jan 10 at 11:33
Gabriel RibeiroGabriel Ribeiro
1,441522
asked Jan 10 at 11:33
Gabriel RibeiroGabriel Ribeiro
1,441522
asked Jan 10 at 11:33
Gabriel RibeiroGabriel Ribeiro
1,441522
asked Jan 10 at 11:33
Gabriel RibeiroGabriel Ribeiro
1,441522
1,441522
add a comment |
add a comment |
3 Answers
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The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m times n$.
So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.
answered Jan 10 at 11:39
MindlackMindlack
3,53717
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Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.
answered Jan 10 at 16:05
André 3000André 3000
12.6k22243
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So, we have a function $f: Xrightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.
In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $rightarrow$ $A$ has a left inverse $rightarrow$ $A^T$ has a right inverse $rightarrow$ $A$ is surjective.
answered Jan 10 at 22:23
AcccumulationAcccumulation
6,9042618
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
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votes
$begingroup$
The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m times n$.
So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.
answered Jan 10 at 11:39
MindlackMindlack
3,53717
$endgroup$
add a comment |
$begingroup$
The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m times n$.
So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.
answered Jan 10 at 11:39
MindlackMindlack
3,53717
$endgroup$
add a comment |
$begingroup$
The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m times n$.
So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.
answered Jan 10 at 11:39
MindlackMindlack
3,53717
$endgroup$
The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m times n$.
So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.
answered Jan 10 at 11:39
MindlackMindlack
3,53717
answered Jan 10 at 11:39
MindlackMindlack
3,53717
answered Jan 10 at 11:39
MindlackMindlack
3,53717
answered Jan 10 at 11:39
MindlackMindlack
3,53717
3,53717
add a comment |
add a comment |
$begingroup$
Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.
answered Jan 10 at 16:05
André 3000André 3000
12.6k22243
$endgroup$
add a comment |
$begingroup$
Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.
answered Jan 10 at 16:05
André 3000André 3000
12.6k22243
$endgroup$
add a comment |
$begingroup$
Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.
answered Jan 10 at 16:05
André 3000André 3000
12.6k22243
$endgroup$
Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.
answered Jan 10 at 16:05
André 3000André 3000
12.6k22243
answered Jan 10 at 16:05
André 3000André 3000
12.6k22243
answered Jan 10 at 16:05
André 3000André 3000
12.6k22243
answered Jan 10 at 16:05
André 3000André 3000
12.6k22243
12.6k22243
add a comment |
add a comment |
$begingroup$
So, we have a function $f: Xrightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.
In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $rightarrow$ $A$ has a left inverse $rightarrow$ $A^T$ has a right inverse $rightarrow$ $A$ is surjective.
answered Jan 10 at 22:23
AcccumulationAcccumulation
6,9042618
$endgroup$
add a comment |
$begingroup$
So, we have a function $f: Xrightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.
In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $rightarrow$ $A$ has a left inverse $rightarrow$ $A^T$ has a right inverse $rightarrow$ $A$ is surjective.
answered Jan 10 at 22:23
AcccumulationAcccumulation
6,9042618
$endgroup$
add a comment |
$begingroup$
So, we have a function $f: Xrightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.
In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $rightarrow$ $A$ has a left inverse $rightarrow$ $A^T$ has a right inverse $rightarrow$ $A$ is surjective.
answered Jan 10 at 22:23
AcccumulationAcccumulation
6,9042618
$endgroup$
So, we have a function $f: Xrightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.
In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $rightarrow$ $A$ has a left inverse $rightarrow$ $A^T$ has a right inverse $rightarrow$ $A$ is surjective.
answered Jan 10 at 22:23
AcccumulationAcccumulation
6,9042618
answered Jan 10 at 22:23
AcccumulationAcccumulation
6,9042618
answered Jan 10 at 22:23
AcccumulationAcccumulation
6,9042618
answered Jan 10 at 22:23
AcccumulationAcccumulation
6,9042618
6,9042618
add a comment |
add a comment |
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