Proof of Thm: For a linear map $f:Vto W$ with $Xsubset V$, $Ysubset W$ with $Xcap f^{-1}[Y]={0}$, we have a...
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I'm reading a chapter in my Linear Algebra book about nilpotent endomorphisms and it contains the following theorem: Let $f:Vto W$ be linear map and $Xsubset V$, $Ysubset W$ such that $Xcap f^{-1}[Y]={0}$. Then $f$ restricted to $X$ is injective and $f[X]cap Y={0}$. The first rule of the proof is: The function $f|_X$ satisfies $ker(f|_X)=Xcapker(f)subset Xcap f^{-1}[Y]={0}$. However I have no clue whatsoever why this inclusion holds. May be there is something I'm missing and I hope someone can help.
linear-algebra
asked Dec 22 '18 at 15:28
Sander KortewegSander Korteweg
205
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I'm reading a chapter in my Linear Algebra book about nilpotent endomorphisms and it contains the following theorem: Let $f:Vto W$ be linear map and $Xsubset V$, $Ysubset W$ such that $Xcap f^{-1}[Y]={0}$. Then $f$ restricted to $X$ is injective and $f[X]cap Y={0}$. The first rule of the proof is: The function $f|_X$ satisfies $ker(f|_X)=Xcapker(f)subset Xcap f^{-1}[Y]={0}$. However I have no clue whatsoever why this inclusion holds. May be there is something I'm missing and I hope someone can help.
linear-algebra
asked Dec 22 '18 at 15:28
Sander KortewegSander Korteweg
205
$endgroup$
add a comment |
$begingroup$
I'm reading a chapter in my Linear Algebra book about nilpotent endomorphisms and it contains the following theorem: Let $f:Vto W$ be linear map and $Xsubset V$, $Ysubset W$ such that $Xcap f^{-1}[Y]={0}$. Then $f$ restricted to $X$ is injective and $f[X]cap Y={0}$. The first rule of the proof is: The function $f|_X$ satisfies $ker(f|_X)=Xcapker(f)subset Xcap f^{-1}[Y]={0}$. However I have no clue whatsoever why this inclusion holds. May be there is something I'm missing and I hope someone can help.
linear-algebra
asked Dec 22 '18 at 15:28
Sander KortewegSander Korteweg
205
$endgroup$
I'm reading a chapter in my Linear Algebra book about nilpotent endomorphisms and it contains the following theorem: Let $f:Vto W$ be linear map and $Xsubset V$, $Ysubset W$ such that $Xcap f^{-1}[Y]={0}$. Then $f$ restricted to $X$ is injective and $f[X]cap Y={0}$. The first rule of the proof is: The function $f|_X$ satisfies $ker(f|_X)=Xcapker(f)subset Xcap f^{-1}[Y]={0}$. However I have no clue whatsoever why this inclusion holds. May be there is something I'm missing and I hope someone can help.
linear-algebra
linear-algebra
asked Dec 22 '18 at 15:28
Sander KortewegSander Korteweg
205
asked Dec 22 '18 at 15:28
Sander KortewegSander Korteweg
205
edited Dec 22 '18 at 15:35
Sander Korteweg
asked Dec 22 '18 at 15:28
Sander KortewegSander Korteweg
205
asked Dec 22 '18 at 15:28
Sander KortewegSander Korteweg
205
asked Dec 22 '18 at 15:28
Sander KortewegSander Korteweg
205
205
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3 Answers
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Let $x in X capker(f)$. Assume that $x ne 0$. Then $x notin f^{-1}(Y)$ since $X cap f^{-1}(Y) = { 0 }$. Hence $f(x) notin Y$ and therefore $f(x) ne 0$. Thus $x notinker(f)$, a contradiction.
Edited: Although probably $X, Y$ are assumed to be linear subspaces of $V, W$, the proof works for arbitrary subsets $X, Y$. The key is that $f(x) notin Y$ implies $f(x) ne 0$. This follows from $X cap f^{-1}(Y) = { 0 }$: We have $0 in f^{-1}(Y)$ and therefore $0 = f(0) in Y$.
answered Dec 22 '18 at 15:45
Paul FrostPaul Frost
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$Xcap f^{-1}[Y]={0_V}implies 0_Vin f^{-1}[Y]implies f(0_V)=0_Win Y$
Since $0_W$ belongs to $Y$, the entire null-space of $f,ker(f)subseteq f^{-1}[Y]$.$$therefore Xcapker(f)subseteq Xcap f^{-1}[Y]$$Since $Xcap f^{-1}[Y]$ contains only $0_V$, no other element of $ker(f)$ is contained in $X.$
$$thereforeker(fBig|_X)={0_V}$$
answered Dec 22 '18 at 15:45
Shubham JohriShubham Johri
5,082717
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The kernel is $ker(f)=f^{-1}[{0}]$ and $f^{-1}[{0}]subseteq f^{-1}[Y]$. More generally, if $S$ and $T$ are subsets of $W$, then $Ssubseteq T$ implies $f^{-1}[S]subseteq f^{-1}[T]$; indeed, if $vin f^{-1}[S]$, then $f(v)in S$, so $f(v)in T$, hence $vin f^{-1}[T]$.
Therefore
$$
Xcapker(f)=Xcap f^{-1}[{0}]subseteq Xcap f^{-1}[Y]
$$
The fact that $ker(f|_X)=Xcapker(f)$ follows immediately from the definition of kernel and of restriction map.
Why is $f[X]cap Y={0}$? Well, if $vin X$, then $f(v)in Y$ implies $vin f^{-1}[Y]$; therefore $vin Xcap f^{-1}[Y]$ and therefore $v=0$, so $f(v)=0$.
answered Dec 22 '18 at 17:31
egregegreg
181k1485203
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3 Answers
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3 Answers
3
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Let $x in X capker(f)$. Assume that $x ne 0$. Then $x notin f^{-1}(Y)$ since $X cap f^{-1}(Y) = { 0 }$. Hence $f(x) notin Y$ and therefore $f(x) ne 0$. Thus $x notinker(f)$, a contradiction.
Edited: Although probably $X, Y$ are assumed to be linear subspaces of $V, W$, the proof works for arbitrary subsets $X, Y$. The key is that $f(x) notin Y$ implies $f(x) ne 0$. This follows from $X cap f^{-1}(Y) = { 0 }$: We have $0 in f^{-1}(Y)$ and therefore $0 = f(0) in Y$.
answered Dec 22 '18 at 15:45
Paul FrostPaul Frost
10.3k3933
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add a comment |
$begingroup$
Let $x in X capker(f)$. Assume that $x ne 0$. Then $x notin f^{-1}(Y)$ since $X cap f^{-1}(Y) = { 0 }$. Hence $f(x) notin Y$ and therefore $f(x) ne 0$. Thus $x notinker(f)$, a contradiction.
Edited: Although probably $X, Y$ are assumed to be linear subspaces of $V, W$, the proof works for arbitrary subsets $X, Y$. The key is that $f(x) notin Y$ implies $f(x) ne 0$. This follows from $X cap f^{-1}(Y) = { 0 }$: We have $0 in f^{-1}(Y)$ and therefore $0 = f(0) in Y$.
answered Dec 22 '18 at 15:45
Paul FrostPaul Frost
10.3k3933
$endgroup$
add a comment |
$begingroup$
Let $x in X capker(f)$. Assume that $x ne 0$. Then $x notin f^{-1}(Y)$ since $X cap f^{-1}(Y) = { 0 }$. Hence $f(x) notin Y$ and therefore $f(x) ne 0$. Thus $x notinker(f)$, a contradiction.
Edited: Although probably $X, Y$ are assumed to be linear subspaces of $V, W$, the proof works for arbitrary subsets $X, Y$. The key is that $f(x) notin Y$ implies $f(x) ne 0$. This follows from $X cap f^{-1}(Y) = { 0 }$: We have $0 in f^{-1}(Y)$ and therefore $0 = f(0) in Y$.
answered Dec 22 '18 at 15:45
Paul FrostPaul Frost
10.3k3933
$endgroup$
Let $x in X capker(f)$. Assume that $x ne 0$. Then $x notin f^{-1}(Y)$ since $X cap f^{-1}(Y) = { 0 }$. Hence $f(x) notin Y$ and therefore $f(x) ne 0$. Thus $x notinker(f)$, a contradiction.
Edited: Although probably $X, Y$ are assumed to be linear subspaces of $V, W$, the proof works for arbitrary subsets $X, Y$. The key is that $f(x) notin Y$ implies $f(x) ne 0$. This follows from $X cap f^{-1}(Y) = { 0 }$: We have $0 in f^{-1}(Y)$ and therefore $0 = f(0) in Y$.
answered Dec 22 '18 at 15:45
Paul FrostPaul Frost
10.3k3933
edited Dec 22 '18 at 16:02
answered Dec 22 '18 at 15:45
Paul FrostPaul Frost
10.3k3933
answered Dec 22 '18 at 15:45
Paul FrostPaul Frost
10.3k3933
answered Dec 22 '18 at 15:45
Paul FrostPaul Frost
10.3k3933
10.3k3933
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$Xcap f^{-1}[Y]={0_V}implies 0_Vin f^{-1}[Y]implies f(0_V)=0_Win Y$
Since $0_W$ belongs to $Y$, the entire null-space of $f,ker(f)subseteq f^{-1}[Y]$.$$therefore Xcapker(f)subseteq Xcap f^{-1}[Y]$$Since $Xcap f^{-1}[Y]$ contains only $0_V$, no other element of $ker(f)$ is contained in $X.$
$$thereforeker(fBig|_X)={0_V}$$
answered Dec 22 '18 at 15:45
Shubham JohriShubham Johri
5,082717
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$begingroup$
$Xcap f^{-1}[Y]={0_V}implies 0_Vin f^{-1}[Y]implies f(0_V)=0_Win Y$
Since $0_W$ belongs to $Y$, the entire null-space of $f,ker(f)subseteq f^{-1}[Y]$.$$therefore Xcapker(f)subseteq Xcap f^{-1}[Y]$$Since $Xcap f^{-1}[Y]$ contains only $0_V$, no other element of $ker(f)$ is contained in $X.$
$$thereforeker(fBig|_X)={0_V}$$
answered Dec 22 '18 at 15:45
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
$Xcap f^{-1}[Y]={0_V}implies 0_Vin f^{-1}[Y]implies f(0_V)=0_Win Y$
Since $0_W$ belongs to $Y$, the entire null-space of $f,ker(f)subseteq f^{-1}[Y]$.$$therefore Xcapker(f)subseteq Xcap f^{-1}[Y]$$Since $Xcap f^{-1}[Y]$ contains only $0_V$, no other element of $ker(f)$ is contained in $X.$
$$thereforeker(fBig|_X)={0_V}$$
answered Dec 22 '18 at 15:45
Shubham JohriShubham Johri
5,082717
$endgroup$
$Xcap f^{-1}[Y]={0_V}implies 0_Vin f^{-1}[Y]implies f(0_V)=0_Win Y$
Since $0_W$ belongs to $Y$, the entire null-space of $f,ker(f)subseteq f^{-1}[Y]$.$$therefore Xcapker(f)subseteq Xcap f^{-1}[Y]$$Since $Xcap f^{-1}[Y]$ contains only $0_V$, no other element of $ker(f)$ is contained in $X.$
$$thereforeker(fBig|_X)={0_V}$$
answered Dec 22 '18 at 15:45
Shubham JohriShubham Johri
5,082717
answered Dec 22 '18 at 15:45
Shubham JohriShubham Johri
5,082717
answered Dec 22 '18 at 15:45
Shubham JohriShubham Johri
5,082717
answered Dec 22 '18 at 15:45
Shubham JohriShubham Johri
5,082717
5,082717
add a comment |
add a comment |
$begingroup$
The kernel is $ker(f)=f^{-1}[{0}]$ and $f^{-1}[{0}]subseteq f^{-1}[Y]$. More generally, if $S$ and $T$ are subsets of $W$, then $Ssubseteq T$ implies $f^{-1}[S]subseteq f^{-1}[T]$; indeed, if $vin f^{-1}[S]$, then $f(v)in S$, so $f(v)in T$, hence $vin f^{-1}[T]$.
Therefore
$$
Xcapker(f)=Xcap f^{-1}[{0}]subseteq Xcap f^{-1}[Y]
$$
The fact that $ker(f|_X)=Xcapker(f)$ follows immediately from the definition of kernel and of restriction map.
Why is $f[X]cap Y={0}$? Well, if $vin X$, then $f(v)in Y$ implies $vin f^{-1}[Y]$; therefore $vin Xcap f^{-1}[Y]$ and therefore $v=0$, so $f(v)=0$.
answered Dec 22 '18 at 17:31
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
The kernel is $ker(f)=f^{-1}[{0}]$ and $f^{-1}[{0}]subseteq f^{-1}[Y]$. More generally, if $S$ and $T$ are subsets of $W$, then $Ssubseteq T$ implies $f^{-1}[S]subseteq f^{-1}[T]$; indeed, if $vin f^{-1}[S]$, then $f(v)in S$, so $f(v)in T$, hence $vin f^{-1}[T]$.
Therefore
$$
Xcapker(f)=Xcap f^{-1}[{0}]subseteq Xcap f^{-1}[Y]
$$
The fact that $ker(f|_X)=Xcapker(f)$ follows immediately from the definition of kernel and of restriction map.
Why is $f[X]cap Y={0}$? Well, if $vin X$, then $f(v)in Y$ implies $vin f^{-1}[Y]$; therefore $vin Xcap f^{-1}[Y]$ and therefore $v=0$, so $f(v)=0$.
answered Dec 22 '18 at 17:31
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
The kernel is $ker(f)=f^{-1}[{0}]$ and $f^{-1}[{0}]subseteq f^{-1}[Y]$. More generally, if $S$ and $T$ are subsets of $W$, then $Ssubseteq T$ implies $f^{-1}[S]subseteq f^{-1}[T]$; indeed, if $vin f^{-1}[S]$, then $f(v)in S$, so $f(v)in T$, hence $vin f^{-1}[T]$.
Therefore
$$
Xcapker(f)=Xcap f^{-1}[{0}]subseteq Xcap f^{-1}[Y]
$$
The fact that $ker(f|_X)=Xcapker(f)$ follows immediately from the definition of kernel and of restriction map.
Why is $f[X]cap Y={0}$? Well, if $vin X$, then $f(v)in Y$ implies $vin f^{-1}[Y]$; therefore $vin Xcap f^{-1}[Y]$ and therefore $v=0$, so $f(v)=0$.
answered Dec 22 '18 at 17:31
egregegreg
181k1485203
$endgroup$
The kernel is $ker(f)=f^{-1}[{0}]$ and $f^{-1}[{0}]subseteq f^{-1}[Y]$. More generally, if $S$ and $T$ are subsets of $W$, then $Ssubseteq T$ implies $f^{-1}[S]subseteq f^{-1}[T]$; indeed, if $vin f^{-1}[S]$, then $f(v)in S$, so $f(v)in T$, hence $vin f^{-1}[T]$.
Therefore
$$
Xcapker(f)=Xcap f^{-1}[{0}]subseteq Xcap f^{-1}[Y]
$$
The fact that $ker(f|_X)=Xcapker(f)$ follows immediately from the definition of kernel and of restriction map.
Why is $f[X]cap Y={0}$? Well, if $vin X$, then $f(v)in Y$ implies $vin f^{-1}[Y]$; therefore $vin Xcap f^{-1}[Y]$ and therefore $v=0$, so $f(v)=0$.
answered Dec 22 '18 at 17:31
egregegreg
181k1485203
answered Dec 22 '18 at 17:31
egregegreg
181k1485203
answered Dec 22 '18 at 17:31
egregegreg
181k1485203
answered Dec 22 '18 at 17:31
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
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