how can I prove that $operatornameim(K circ L) subseteq ker(K)$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why is $ker(rho-sigma)subseteqker(rho)oplusoperatornameim(sigma)$?How to prove a property of ranks: $operatornamerank(AB)= operatornamerank(B)- dim(operatornameIm B cap ker A)$$ker(T) subseteqker(S)$ implies the exist some $H$ s.t. $Hcirc T=S$Problem proving: $V = ker T oplus operatornameimT$Prove that $textrank T = operatornamerank T^2 iff operatornameImT cap ker T = vec 0$Prove that $operatornamerank (f) + operatornamerank (g) -dim Wleq operatornamerank(gcirc f)$.given A linear map $T:Vlongrightarrow V$, let $dim V = n$, prove that for every $kgeq n, operatornameImT^kcap ker T^k=0$.show that $operatornamerank(gcirc f) leq operatornamerank(f)+operatornamerank(g)-n$How to show that $dimker(AB) le dim ker A + dim ker B $?How to prove that if $f^2=f$ then $V= ker f + operatornameIm f$.
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how can I prove that $operatornameim(K circ L) subseteq ker(K)$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why is $ker(rho-sigma)subseteqker(rho)oplusoperatornameim(sigma)$?How to prove a property of ranks: $operatornamerank(AB)= operatornamerank(B)- dim(operatornameIm B cap ker A)$$ker(T) subseteqker(S)$ implies the exist some $H$ s.t. $Hcirc T=S$Problem proving: $V = ker T oplus operatornameimT$Prove that $textrank T = operatornamerank T^2 iff operatornameImT cap ker T = vec 0$Prove that $operatornamerank (f) + operatornamerank (g) -dim Wleq operatornamerank(gcirc f)$.given A linear map $T:Vlongrightarrow V$, let $dim V = n$, prove that for every $kgeq n, operatornameImT^kcap ker T^k=0$.show that $operatornamerank(gcirc f) leq operatornamerank(f)+operatornamerank(g)-n$How to show that $dimker(AB) le dim ker A + dim ker B $?How to prove that if $f^2=f$ then $V= ker f + operatornameIm f$.
$begingroup$
I am struggling with an algebra problem here is what we got :
Assume we have :
$$L : E_0 to E_1 quadtextandquad K : E_1 to E_2 $$
We have to show that :
$$dim(ker(K circ L)) leq dim(ker(L)) + dim(ker(K)).$$
I define $H$ as :
$$H : ker(K circ L) to E_2, qquad H(v) = L(v) quad textfor quad v ∈ ker(K circ L)$$
Now I have to prove that :
$$ ker(L) = ker(H) $$
But how can I show that :
$$Rightarrow qquad ker(L) subseteq ker(K circ L) quad textand quad ker(K circ L) subseteq ker(L) quad ?$$
I got an idea for this one : $L(v) = 0$ implies $K(L(v)) = 0$ ... But I stuck with this one.
$$ Rightarrow qquad operatornameim(H) subseteq ker(K) quad ? $$
How can I prove? I only know and prove that $operatornameim(Kcirc L) subseteq operatornameim(K)$.
Thanks.
linear-algebra
edited Mar 18 at 0:45
Sangchul Lee
96.6k12173283
asked Mar 17 at 23:57
JoshuaKJoshuaK
607
$endgroup$
add a comment |
$begingroup$
I am struggling with an algebra problem here is what we got :
Assume we have :
$$L : E_0 to E_1 quadtextandquad K : E_1 to E_2 $$
We have to show that :
$$dim(ker(K circ L)) leq dim(ker(L)) + dim(ker(K)).$$
I define $H$ as :
$$H : ker(K circ L) to E_2, qquad H(v) = L(v) quad textfor quad v ∈ ker(K circ L)$$
Now I have to prove that :
$$ ker(L) = ker(H) $$
But how can I show that :
$$Rightarrow qquad ker(L) subseteq ker(K circ L) quad textand quad ker(K circ L) subseteq ker(L) quad ?$$
I got an idea for this one : $L(v) = 0$ implies $K(L(v)) = 0$ ... But I stuck with this one.
$$ Rightarrow qquad operatornameim(H) subseteq ker(K) quad ? $$
How can I prove? I only know and prove that $operatornameim(Kcirc L) subseteq operatornameim(K)$.
Thanks.
linear-algebra
edited Mar 18 at 0:45
Sangchul Lee
96.6k12173283
asked Mar 17 at 23:57
JoshuaKJoshuaK
607
$endgroup$
add a comment |
$begingroup$
I am struggling with an algebra problem here is what we got :
Assume we have :
$$L : E_0 to E_1 quadtextandquad K : E_1 to E_2 $$
We have to show that :
$$dim(ker(K circ L)) leq dim(ker(L)) + dim(ker(K)).$$
I define $H$ as :
$$H : ker(K circ L) to E_2, qquad H(v) = L(v) quad textfor quad v ∈ ker(K circ L)$$
Now I have to prove that :
$$ ker(L) = ker(H) $$
But how can I show that :
$$Rightarrow qquad ker(L) subseteq ker(K circ L) quad textand quad ker(K circ L) subseteq ker(L) quad ?$$
I got an idea for this one : $L(v) = 0$ implies $K(L(v)) = 0$ ... But I stuck with this one.
$$ Rightarrow qquad operatornameim(H) subseteq ker(K) quad ? $$
How can I prove? I only know and prove that $operatornameim(Kcirc L) subseteq operatornameim(K)$.
Thanks.
linear-algebra
edited Mar 18 at 0:45
Sangchul Lee
96.6k12173283
asked Mar 17 at 23:57
JoshuaKJoshuaK
607
$endgroup$
I am struggling with an algebra problem here is what we got :
Assume we have :
$$L : E_0 to E_1 quadtextandquad K : E_1 to E_2 $$
We have to show that :
$$dim(ker(K circ L)) leq dim(ker(L)) + dim(ker(K)).$$
I define $H$ as :
$$H : ker(K circ L) to E_2, qquad H(v) = L(v) quad textfor quad v ∈ ker(K circ L)$$
Now I have to prove that :
$$ ker(L) = ker(H) $$
But how can I show that :
$$Rightarrow qquad ker(L) subseteq ker(K circ L) quad textand quad ker(K circ L) subseteq ker(L) quad ?$$
I got an idea for this one : $L(v) = 0$ implies $K(L(v)) = 0$ ... But I stuck with this one.
$$ Rightarrow qquad operatornameim(H) subseteq ker(K) quad ? $$
How can I prove? I only know and prove that $operatornameim(Kcirc L) subseteq operatornameim(K)$.
Thanks.
linear-algebra
linear-algebra
edited Mar 18 at 0:45
Sangchul Lee
96.6k12173283
asked Mar 17 at 23:57
JoshuaKJoshuaK
607
edited Mar 18 at 0:45
Sangchul Lee
96.6k12173283
asked Mar 17 at 23:57
JoshuaKJoshuaK
607
edited Mar 18 at 0:45
Sangchul Lee
96.6k12173283
edited Mar 18 at 0:45
Sangchul Lee
96.6k12173283
edited Mar 18 at 0:45
Sangchul Lee
96.6k12173283
96.6k12173283
asked Mar 17 at 23:57
JoshuaKJoshuaK
607
asked Mar 17 at 23:57
JoshuaKJoshuaK
607
asked Mar 17 at 23:57
JoshuaKJoshuaK
607
607
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Generally you cannot show that $ker (mathcal KL) = ker mathcal L$: if $mathcal Kv =0$ for all $v in E_1$, where $E_1 neq 0$, and $mathcal L$ surjective, then $ker(mathcal KL) = E_0$ but $ker mathcal Lsubsetneq E_0$ [otherwise $mathcal L$ would not be surjective: some $v in E_1 setminus 0$ cannot be mapped by $mathcal L$].
Assume $dim(E_0) = n <+infty$. By the Rank-Nullity Theorem, $ dim(ker mathcal L)+ dim(mathrm im, mathcal L) = n$, also $
dim(ker (mathcal KL)) + dim(mathrm im (mathcal KL)) =n
$.
Then
beginalign*
dim(ker (mathcal KL)) &= n - dim (mathrm im (mathcal KL))\
&= dim (ker mathcal L) + dim(mathrm im (mathcal L )) \
&quad - dim (mathrm im(mathcal KL)).
endalign*
Note that the mapping $mathcal Kvert _mathrm im,mathcal L colon mathrm im, mathcal L to E_2$ is a linear mapping, and we could know that $mathrm im(mathcal Kvert _mathrm im, mathcal L) = mathrm im (mathcal KL)$: $z in mathrm im (mathcal Kvert _mathrm im ,mathcal L) iff exists y in mathrm im, mathcal L, mathcal Ky = z iff exists x in E_0, y = mathcal Lx, z = mathcal Ky iff exists x in E_0, z = mathcal KLx$. So again by the Rank-Nullity Theorem,
$$
dim (mathrm im, mathcal L) = dim(ker (mathcal Kvert _mathrm im,mathcal L)) + dim (mathrm im(mathcal KL)).
$$
Thus
$$
dim (mathrm im, mathcal L) - dim (mathrm im(mathcal KL)) =dim(ker (mathcal Kvert _mathrm im,mathcal L)) leqslant dim (ker mathcal K),
$$
the $leqslant$ holds because $mathrm im ,mathcal L subset E_1$, and possibly there is some $y in (E_1setminus mathrm im, mathcal L)$ that $mathcal Ky = 0$.
Hence the inequality we want to prove.
answered Mar 18 at 0:37
xbhxbh
6,3701522
$endgroup$
add a comment |
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$begingroup$
Generally you cannot show that $ker (mathcal KL) = ker mathcal L$: if $mathcal Kv =0$ for all $v in E_1$, where $E_1 neq 0$, and $mathcal L$ surjective, then $ker(mathcal KL) = E_0$ but $ker mathcal Lsubsetneq E_0$ [otherwise $mathcal L$ would not be surjective: some $v in E_1 setminus 0$ cannot be mapped by $mathcal L$].
Assume $dim(E_0) = n <+infty$. By the Rank-Nullity Theorem, $ dim(ker mathcal L)+ dim(mathrm im, mathcal L) = n$, also $
dim(ker (mathcal KL)) + dim(mathrm im (mathcal KL)) =n
$.
Then
beginalign*
dim(ker (mathcal KL)) &= n - dim (mathrm im (mathcal KL))\
&= dim (ker mathcal L) + dim(mathrm im (mathcal L )) \
&quad - dim (mathrm im(mathcal KL)).
endalign*
Note that the mapping $mathcal Kvert _mathrm im,mathcal L colon mathrm im, mathcal L to E_2$ is a linear mapping, and we could know that $mathrm im(mathcal Kvert _mathrm im, mathcal L) = mathrm im (mathcal KL)$: $z in mathrm im (mathcal Kvert _mathrm im ,mathcal L) iff exists y in mathrm im, mathcal L, mathcal Ky = z iff exists x in E_0, y = mathcal Lx, z = mathcal Ky iff exists x in E_0, z = mathcal KLx$. So again by the Rank-Nullity Theorem,
$$
dim (mathrm im, mathcal L) = dim(ker (mathcal Kvert _mathrm im,mathcal L)) + dim (mathrm im(mathcal KL)).
$$
Thus
$$
dim (mathrm im, mathcal L) - dim (mathrm im(mathcal KL)) =dim(ker (mathcal Kvert _mathrm im,mathcal L)) leqslant dim (ker mathcal K),
$$
the $leqslant$ holds because $mathrm im ,mathcal L subset E_1$, and possibly there is some $y in (E_1setminus mathrm im, mathcal L)$ that $mathcal Ky = 0$.
Hence the inequality we want to prove.
answered Mar 18 at 0:37
xbhxbh
6,3701522
$endgroup$
add a comment |
$begingroup$
Generally you cannot show that $ker (mathcal KL) = ker mathcal L$: if $mathcal Kv =0$ for all $v in E_1$, where $E_1 neq 0$, and $mathcal L$ surjective, then $ker(mathcal KL) = E_0$ but $ker mathcal Lsubsetneq E_0$ [otherwise $mathcal L$ would not be surjective: some $v in E_1 setminus 0$ cannot be mapped by $mathcal L$].
Assume $dim(E_0) = n <+infty$. By the Rank-Nullity Theorem, $ dim(ker mathcal L)+ dim(mathrm im, mathcal L) = n$, also $
dim(ker (mathcal KL)) + dim(mathrm im (mathcal KL)) =n
$.
Then
beginalign*
dim(ker (mathcal KL)) &= n - dim (mathrm im (mathcal KL))\
&= dim (ker mathcal L) + dim(mathrm im (mathcal L )) \
&quad - dim (mathrm im(mathcal KL)).
endalign*
Note that the mapping $mathcal Kvert _mathrm im,mathcal L colon mathrm im, mathcal L to E_2$ is a linear mapping, and we could know that $mathrm im(mathcal Kvert _mathrm im, mathcal L) = mathrm im (mathcal KL)$: $z in mathrm im (mathcal Kvert _mathrm im ,mathcal L) iff exists y in mathrm im, mathcal L, mathcal Ky = z iff exists x in E_0, y = mathcal Lx, z = mathcal Ky iff exists x in E_0, z = mathcal KLx$. So again by the Rank-Nullity Theorem,
$$
dim (mathrm im, mathcal L) = dim(ker (mathcal Kvert _mathrm im,mathcal L)) + dim (mathrm im(mathcal KL)).
$$
Thus
$$
dim (mathrm im, mathcal L) - dim (mathrm im(mathcal KL)) =dim(ker (mathcal Kvert _mathrm im,mathcal L)) leqslant dim (ker mathcal K),
$$
the $leqslant$ holds because $mathrm im ,mathcal L subset E_1$, and possibly there is some $y in (E_1setminus mathrm im, mathcal L)$ that $mathcal Ky = 0$.
Hence the inequality we want to prove.
answered Mar 18 at 0:37
xbhxbh
6,3701522
$endgroup$
add a comment |
$begingroup$
Generally you cannot show that $ker (mathcal KL) = ker mathcal L$: if $mathcal Kv =0$ for all $v in E_1$, where $E_1 neq 0$, and $mathcal L$ surjective, then $ker(mathcal KL) = E_0$ but $ker mathcal Lsubsetneq E_0$ [otherwise $mathcal L$ would not be surjective: some $v in E_1 setminus 0$ cannot be mapped by $mathcal L$].
Assume $dim(E_0) = n <+infty$. By the Rank-Nullity Theorem, $ dim(ker mathcal L)+ dim(mathrm im, mathcal L) = n$, also $
dim(ker (mathcal KL)) + dim(mathrm im (mathcal KL)) =n
$.
Then
beginalign*
dim(ker (mathcal KL)) &= n - dim (mathrm im (mathcal KL))\
&= dim (ker mathcal L) + dim(mathrm im (mathcal L )) \
&quad - dim (mathrm im(mathcal KL)).
endalign*
Note that the mapping $mathcal Kvert _mathrm im,mathcal L colon mathrm im, mathcal L to E_2$ is a linear mapping, and we could know that $mathrm im(mathcal Kvert _mathrm im, mathcal L) = mathrm im (mathcal KL)$: $z in mathrm im (mathcal Kvert _mathrm im ,mathcal L) iff exists y in mathrm im, mathcal L, mathcal Ky = z iff exists x in E_0, y = mathcal Lx, z = mathcal Ky iff exists x in E_0, z = mathcal KLx$. So again by the Rank-Nullity Theorem,
$$
dim (mathrm im, mathcal L) = dim(ker (mathcal Kvert _mathrm im,mathcal L)) + dim (mathrm im(mathcal KL)).
$$
Thus
$$
dim (mathrm im, mathcal L) - dim (mathrm im(mathcal KL)) =dim(ker (mathcal Kvert _mathrm im,mathcal L)) leqslant dim (ker mathcal K),
$$
the $leqslant$ holds because $mathrm im ,mathcal L subset E_1$, and possibly there is some $y in (E_1setminus mathrm im, mathcal L)$ that $mathcal Ky = 0$.
Hence the inequality we want to prove.
answered Mar 18 at 0:37
xbhxbh
6,3701522
$endgroup$
Generally you cannot show that $ker (mathcal KL) = ker mathcal L$: if $mathcal Kv =0$ for all $v in E_1$, where $E_1 neq 0$, and $mathcal L$ surjective, then $ker(mathcal KL) = E_0$ but $ker mathcal Lsubsetneq E_0$ [otherwise $mathcal L$ would not be surjective: some $v in E_1 setminus 0$ cannot be mapped by $mathcal L$].
Assume $dim(E_0) = n <+infty$. By the Rank-Nullity Theorem, $ dim(ker mathcal L)+ dim(mathrm im, mathcal L) = n$, also $
dim(ker (mathcal KL)) + dim(mathrm im (mathcal KL)) =n
$.
Then
beginalign*
dim(ker (mathcal KL)) &= n - dim (mathrm im (mathcal KL))\
&= dim (ker mathcal L) + dim(mathrm im (mathcal L )) \
&quad - dim (mathrm im(mathcal KL)).
endalign*
Note that the mapping $mathcal Kvert _mathrm im,mathcal L colon mathrm im, mathcal L to E_2$ is a linear mapping, and we could know that $mathrm im(mathcal Kvert _mathrm im, mathcal L) = mathrm im (mathcal KL)$: $z in mathrm im (mathcal Kvert _mathrm im ,mathcal L) iff exists y in mathrm im, mathcal L, mathcal Ky = z iff exists x in E_0, y = mathcal Lx, z = mathcal Ky iff exists x in E_0, z = mathcal KLx$. So again by the Rank-Nullity Theorem,
$$
dim (mathrm im, mathcal L) = dim(ker (mathcal Kvert _mathrm im,mathcal L)) + dim (mathrm im(mathcal KL)).
$$
Thus
$$
dim (mathrm im, mathcal L) - dim (mathrm im(mathcal KL)) =dim(ker (mathcal Kvert _mathrm im,mathcal L)) leqslant dim (ker mathcal K),
$$
the $leqslant$ holds because $mathrm im ,mathcal L subset E_1$, and possibly there is some $y in (E_1setminus mathrm im, mathcal L)$ that $mathcal Ky = 0$.
Hence the inequality we want to prove.
answered Mar 18 at 0:37
xbhxbh
6,3701522
edited Mar 18 at 0:54
answered Mar 18 at 0:37
xbhxbh
6,3701522
answered Mar 18 at 0:37
xbhxbh
6,3701522
answered Mar 18 at 0:37
xbhxbh
6,3701522
6,3701522
add a comment |
add a comment |
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