Mixing
How to prove that if $f^2=f$ then $V= ker f + operatorname{Im} f$.
$begingroup$
I need help to solve this algebra problem.
Having $f$, an endomorphism, and knowing that $f(f(v))$ is equal to $f(v)$, I need to prove that $V= operatorname{Ker}f + operatorname{Im}f$, being complementary.
I also need to show that if we reduce $f$ to $ operatorname{Im}f$, What we have is the identity application.
I have already proved that if $dim V=n$ and $dim operatorname{Ker}f=s$ , $dim operatorname{Im}f=n-s$, but I don´t know how to justify that $0_V$ is the only vector in the intersection.
It would be amazing if you helped me.
Thanks!
linear-algebra
edited Jan 24 at 9:47
Bernard
123k741116
asked Jan 24 at 8:45
AneAne
12
$endgroup$
add a comment |
$begingroup$
I need help to solve this algebra problem.
Having $f$, an endomorphism, and knowing that $f(f(v))$ is equal to $f(v)$, I need to prove that $V= operatorname{Ker}f + operatorname{Im}f$, being complementary.
I also need to show that if we reduce $f$ to $ operatorname{Im}f$, What we have is the identity application.
I have already proved that if $dim V=n$ and $dim operatorname{Ker}f=s$ , $dim operatorname{Im}f=n-s$, but I don´t know how to justify that $0_V$ is the only vector in the intersection.
It would be amazing if you helped me.
Thanks!
linear-algebra
edited Jan 24 at 9:47
Bernard
123k741116
asked Jan 24 at 8:45
AneAne
12
$endgroup$
$begingroup$
Hint: assume there is a vector other than $0v$ in the intersection and see what happens.
$endgroup$
– P.Diddy
Jan 24 at 8:50
add a comment |
$begingroup$
I need help to solve this algebra problem.
Having $f$, an endomorphism, and knowing that $f(f(v))$ is equal to $f(v)$, I need to prove that $V= operatorname{Ker}f + operatorname{Im}f$, being complementary.
I also need to show that if we reduce $f$ to $ operatorname{Im}f$, What we have is the identity application.
I have already proved that if $dim V=n$ and $dim operatorname{Ker}f=s$ , $dim operatorname{Im}f=n-s$, but I don´t know how to justify that $0_V$ is the only vector in the intersection.
It would be amazing if you helped me.
Thanks!
linear-algebra
edited Jan 24 at 9:47
Bernard
123k741116
asked Jan 24 at 8:45
AneAne
12
$endgroup$
I need help to solve this algebra problem.
Having $f$, an endomorphism, and knowing that $f(f(v))$ is equal to $f(v)$, I need to prove that $V= operatorname{Ker}f + operatorname{Im}f$, being complementary.
I also need to show that if we reduce $f$ to $ operatorname{Im}f$, What we have is the identity application.
I have already proved that if $dim V=n$ and $dim operatorname{Ker}f=s$ , $dim operatorname{Im}f=n-s$, but I don´t know how to justify that $0_V$ is the only vector in the intersection.
It would be amazing if you helped me.
Thanks!
linear-algebra
linear-algebra
edited Jan 24 at 9:47
Bernard
123k741116
asked Jan 24 at 8:45
AneAne
12
edited Jan 24 at 9:47
Bernard
123k741116
asked Jan 24 at 8:45
AneAne
12
edited Jan 24 at 9:47
Bernard
123k741116
edited Jan 24 at 9:47
Bernard
123k741116
edited Jan 24 at 9:47
Bernard
123k741116
123k741116
asked Jan 24 at 8:45
AneAne
12
asked Jan 24 at 8:45
AneAne
12
asked Jan 24 at 8:45
AneAne
12
12
$begingroup$
Hint: assume there is a vector other than $0v$ in the intersection and see what happens.
$endgroup$
– P.Diddy
Jan 24 at 8:50
add a comment |
$begingroup$
Hint: assume there is a vector other than $0v$ in the intersection and see what happens.
$endgroup$
– P.Diddy
Jan 24 at 8:50
$begingroup$
Hint: assume there is a vector other than $0v$ in the intersection and see what happens.
$endgroup$
– P.Diddy
Jan 24 at 8:50
$begingroup$
Hint: assume there is a vector other than $0v$ in the intersection and see what happens.
$endgroup$
– P.Diddy
Jan 24 at 8:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For all $x in V$ we have $x=f(x)+(x-f(x))$. Since $f^2=f$, we have $x-f(x) in ker(f)$, hence $x=f(x)+(x-f(x) in Im(f)+ker(f).$
answered Jan 24 at 8:58
FredFred
48.3k1849
$endgroup$
$begingroup$
The question also asks to show they are complementary spaces. That would mean showing that $Im(f)$ is closed.
$endgroup$
– Chrystomath
Jan 24 at 10:40
add a comment |
$begingroup$
If $v in V$ then $v=f(v) +(v-f(v))$. Since the first term belongs to image of $f$ and $f(v-f(v))=f(v)-f(f(v))=f(v)-f(v)=0$ we see that the second term belongs to kernel of $f$. Hence $V =ker (f) +Im (f)$. If $v in ker (f) cap Im(f)$ then $v=f(w)$ for some $w$. Hence $f(v)=f(f(w))=f(w)=v$ but $f(v)=0$. Hence $v=0$. It follows that $ker (f) cap Im(f)={0}$. Finally, if $v in Im(f)$ then $v=f(w)$ for some $w$. Hence $f(v)=f(f(w))=f(w)=v$ so $f$ is the identity map on $Im (f)$.
answered Jan 24 at 8:50
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
$begingroup$
Why is $v= f(v) + (v-(f(v))$? I don´t understand it
$endgroup$
– Ane
Jan 24 at 10:59
$begingroup$
$f(v)+(v-f(v))=f(v)+v-f(v)=v.$
$endgroup$
– Fred
Jan 24 at 11:24
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For all $x in V$ we have $x=f(x)+(x-f(x))$. Since $f^2=f$, we have $x-f(x) in ker(f)$, hence $x=f(x)+(x-f(x) in Im(f)+ker(f).$
answered Jan 24 at 8:58
FredFred
48.3k1849
$endgroup$
$begingroup$
The question also asks to show they are complementary spaces. That would mean showing that $Im(f)$ is closed.
$endgroup$
– Chrystomath
Jan 24 at 10:40
add a comment |
$begingroup$
For all $x in V$ we have $x=f(x)+(x-f(x))$. Since $f^2=f$, we have $x-f(x) in ker(f)$, hence $x=f(x)+(x-f(x) in Im(f)+ker(f).$
answered Jan 24 at 8:58
FredFred
48.3k1849
$endgroup$
$begingroup$
The question also asks to show they are complementary spaces. That would mean showing that $Im(f)$ is closed.
$endgroup$
– Chrystomath
Jan 24 at 10:40
add a comment |
$begingroup$
For all $x in V$ we have $x=f(x)+(x-f(x))$. Since $f^2=f$, we have $x-f(x) in ker(f)$, hence $x=f(x)+(x-f(x) in Im(f)+ker(f).$
answered Jan 24 at 8:58
FredFred
48.3k1849
$endgroup$
For all $x in V$ we have $x=f(x)+(x-f(x))$. Since $f^2=f$, we have $x-f(x) in ker(f)$, hence $x=f(x)+(x-f(x) in Im(f)+ker(f).$
answered Jan 24 at 8:58
FredFred
48.3k1849
answered Jan 24 at 8:58
FredFred
48.3k1849
answered Jan 24 at 8:58
FredFred
48.3k1849
answered Jan 24 at 8:58
FredFred
48.3k1849
48.3k1849
$begingroup$
The question also asks to show they are complementary spaces. That would mean showing that $Im(f)$ is closed.
$endgroup$
– Chrystomath
Jan 24 at 10:40
add a comment |
$begingroup$
The question also asks to show they are complementary spaces. That would mean showing that $Im(f)$ is closed.
$endgroup$
– Chrystomath
Jan 24 at 10:40
$begingroup$
The question also asks to show they are complementary spaces. That would mean showing that $Im(f)$ is closed.
$endgroup$
– Chrystomath
Jan 24 at 10:40
$begingroup$
The question also asks to show they are complementary spaces. That would mean showing that $Im(f)$ is closed.
$endgroup$
– Chrystomath
Jan 24 at 10:40
add a comment |
$begingroup$
If $v in V$ then $v=f(v) +(v-f(v))$. Since the first term belongs to image of $f$ and $f(v-f(v))=f(v)-f(f(v))=f(v)-f(v)=0$ we see that the second term belongs to kernel of $f$. Hence $V =ker (f) +Im (f)$. If $v in ker (f) cap Im(f)$ then $v=f(w)$ for some $w$. Hence $f(v)=f(f(w))=f(w)=v$ but $f(v)=0$. Hence $v=0$. It follows that $ker (f) cap Im(f)={0}$. Finally, if $v in Im(f)$ then $v=f(w)$ for some $w$. Hence $f(v)=f(f(w))=f(w)=v$ so $f$ is the identity map on $Im (f)$.
answered Jan 24 at 8:50
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
$begingroup$
Why is $v= f(v) + (v-(f(v))$? I don´t understand it
$endgroup$
– Ane
Jan 24 at 10:59
$begingroup$
$f(v)+(v-f(v))=f(v)+v-f(v)=v.$
$endgroup$
– Fred
Jan 24 at 11:24
add a comment |
$begingroup$
If $v in V$ then $v=f(v) +(v-f(v))$. Since the first term belongs to image of $f$ and $f(v-f(v))=f(v)-f(f(v))=f(v)-f(v)=0$ we see that the second term belongs to kernel of $f$. Hence $V =ker (f) +Im (f)$. If $v in ker (f) cap Im(f)$ then $v=f(w)$ for some $w$. Hence $f(v)=f(f(w))=f(w)=v$ but $f(v)=0$. Hence $v=0$. It follows that $ker (f) cap Im(f)={0}$. Finally, if $v in Im(f)$ then $v=f(w)$ for some $w$. Hence $f(v)=f(f(w))=f(w)=v$ so $f$ is the identity map on $Im (f)$.
answered Jan 24 at 8:50
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
$begingroup$
Why is $v= f(v) + (v-(f(v))$? I don´t understand it
$endgroup$
– Ane
Jan 24 at 10:59
$begingroup$
$f(v)+(v-f(v))=f(v)+v-f(v)=v.$
$endgroup$
– Fred
Jan 24 at 11:24
add a comment |
$begingroup$
If $v in V$ then $v=f(v) +(v-f(v))$. Since the first term belongs to image of $f$ and $f(v-f(v))=f(v)-f(f(v))=f(v)-f(v)=0$ we see that the second term belongs to kernel of $f$. Hence $V =ker (f) +Im (f)$. If $v in ker (f) cap Im(f)$ then $v=f(w)$ for some $w$. Hence $f(v)=f(f(w))=f(w)=v$ but $f(v)=0$. Hence $v=0$. It follows that $ker (f) cap Im(f)={0}$. Finally, if $v in Im(f)$ then $v=f(w)$ for some $w$. Hence $f(v)=f(f(w))=f(w)=v$ so $f$ is the identity map on $Im (f)$.
answered Jan 24 at 8:50
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
If $v in V$ then $v=f(v) +(v-f(v))$. Since the first term belongs to image of $f$ and $f(v-f(v))=f(v)-f(f(v))=f(v)-f(v)=0$ we see that the second term belongs to kernel of $f$. Hence $V =ker (f) +Im (f)$. If $v in ker (f) cap Im(f)$ then $v=f(w)$ for some $w$. Hence $f(v)=f(f(w))=f(w)=v$ but $f(v)=0$. Hence $v=0$. It follows that $ker (f) cap Im(f)={0}$. Finally, if $v in Im(f)$ then $v=f(w)$ for some $w$. Hence $f(v)=f(f(w))=f(w)=v$ so $f$ is the identity map on $Im (f)$.
answered Jan 24 at 8:50
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Jan 24 at 8:50
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Jan 24 at 8:50
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Jan 24 at 8:50
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
67.6k53067
$begingroup$
Why is $v= f(v) + (v-(f(v))$? I don´t understand it
$endgroup$
– Ane
Jan 24 at 10:59
$begingroup$
$f(v)+(v-f(v))=f(v)+v-f(v)=v.$
$endgroup$
– Fred
Jan 24 at 11:24
add a comment |
$begingroup$
Why is $v= f(v) + (v-(f(v))$? I don´t understand it
$endgroup$
– Ane
Jan 24 at 10:59
$begingroup$
$f(v)+(v-f(v))=f(v)+v-f(v)=v.$
$endgroup$
– Fred
Jan 24 at 11:24
$begingroup$
Why is $v= f(v) + (v-(f(v))$? I don´t understand it
$endgroup$
– Ane
Jan 24 at 10:59
$begingroup$
Why is $v= f(v) + (v-(f(v))$? I don´t understand it
$endgroup$
– Ane
Jan 24 at 10:59
$begingroup$
$f(v)+(v-f(v))=f(v)+v-f(v)=v.$
$endgroup$
– Fred
Jan 24 at 11:24
$begingroup$
$f(v)+(v-f(v))=f(v)+v-f(v)=v.$
$endgroup$
– Fred
Jan 24 at 11:24
add a comment |
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$begingroup$
Hint: assume there is a vector other than $0v$ in the intersection and see what happens.
$endgroup$
– P.Diddy
Jan 24 at 8:50