If $f(T)g(T)=0$ and $V_1=g(T)(V)$, $V_2=f(T)(V)$ proof that $V=V_1oplus V_2$

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$Bbb K$ is a field, $V$ and let $T:Vlongrightarrow V$ be a linear map.
$Bbb K[x]$ is the vector space of the polynomials over $Bbb K$.

Suppose that

$f,gin Bbb K[x]$, $f(T)g(T)=0$, $V_1=g(T)(V)$ $ $ and $V_2=f(T)(V) space$; then prove that $$V=V_1oplus V_2$$ and that $V_1$ and $V_2 $are $T-$invariant.

This is what i got so far:

if $$uin V_1cap V_2 Rightarrow u=f(T)(v_1)=g(T)(v_2)$$
$$f(T)(u)=f(T)f(T)(v_1)=f(T)g(T)(v_2)=0$$

I'm trying to proof that the intersection is equal to 0, and than i would proof $forall v in V ;v=v_1+v_2$ with $v_1in V_1 wedge v_2in V_2$

but i wasn't able to do more than that, so if you have any idea, i would be glad to hear it.

linear-algebra polynomials vector-spaces linear-transformations invariance

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edited Jul 16 at 17:21

Legoman

4,65421033

asked Jul 16 at 15:46

Jonatas Teo

31

add a comment | 

up vote
0
down vote

favorite

$Bbb K$ is a field, $V$ and let $T:Vlongrightarrow V$ be a linear map.
$Bbb K[x]$ is the vector space of the polynomials over $Bbb K$.

Suppose that

$f,gin Bbb K[x]$, $f(T)g(T)=0$, $V_1=g(T)(V)$ $ $ and $V_2=f(T)(V) space$; then prove that $$V=V_1oplus V_2$$ and that $V_1$ and $V_2 $are $T-$invariant.

This is what i got so far:

if $$uin V_1cap V_2 Rightarrow u=f(T)(v_1)=g(T)(v_2)$$
$$f(T)(u)=f(T)f(T)(v_1)=f(T)g(T)(v_2)=0$$

I'm trying to proof that the intersection is equal to 0, and than i would proof $forall v in V ;v=v_1+v_2$ with $v_1in V_1 wedge v_2in V_2$

but i wasn't able to do more than that, so if you have any idea, i would be glad to hear it.

linear-algebra polynomials vector-spaces linear-transformations invariance

share|cite|improve this question

edited Jul 16 at 17:21

Legoman

4,65421033

asked Jul 16 at 15:46

Jonatas Teo

31

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

$Bbb K$ is a field, $V$ and let $T:Vlongrightarrow V$ be a linear map.
$Bbb K[x]$ is the vector space of the polynomials over $Bbb K$.

Suppose that

$f,gin Bbb K[x]$, $f(T)g(T)=0$, $V_1=g(T)(V)$ $ $ and $V_2=f(T)(V) space$; then prove that $$V=V_1oplus V_2$$ and that $V_1$ and $V_2 $are $T-$invariant.

This is what i got so far:

if $$uin V_1cap V_2 Rightarrow u=f(T)(v_1)=g(T)(v_2)$$
$$f(T)(u)=f(T)f(T)(v_1)=f(T)g(T)(v_2)=0$$

I'm trying to proof that the intersection is equal to 0, and than i would proof $forall v in V ;v=v_1+v_2$ with $v_1in V_1 wedge v_2in V_2$

but i wasn't able to do more than that, so if you have any idea, i would be glad to hear it.

linear-algebra polynomials vector-spaces linear-transformations invariance

share|cite|improve this question

edited Jul 16 at 17:21

Legoman

4,65421033

asked Jul 16 at 15:46

Jonatas Teo

31

$Bbb K$ is a field, $V$ and let $T:Vlongrightarrow V$ be a linear map.
$Bbb K[x]$ is the vector space of the polynomials over $Bbb K$.

Suppose that

$f,gin Bbb K[x]$, $f(T)g(T)=0$, $V_1=g(T)(V)$ $ $ and $V_2=f(T)(V) space$; then prove that $$V=V_1oplus V_2$$ and that $V_1$ and $V_2 $are $T-$invariant.

This is what i got so far:

if $$uin V_1cap V_2 Rightarrow u=f(T)(v_1)=g(T)(v_2)$$
$$f(T)(u)=f(T)f(T)(v_1)=f(T)g(T)(v_2)=0$$

I'm trying to proof that the intersection is equal to 0, and than i would proof $forall v in V ;v=v_1+v_2$ with $v_1in V_1 wedge v_2in V_2$

but i wasn't able to do more than that, so if you have any idea, i would be glad to hear it.

linear-algebra polynomials vector-spaces linear-transformations invariance

share|cite|improve this question

edited Jul 16 at 17:21

Legoman

4,65421033

asked Jul 16 at 15:46

Jonatas Teo

31

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 17:21

Legoman

4,65421033

edited Jul 16 at 17:21

Legoman

4,65421033

edited Jul 16 at 17:21

Legoman

4,65421033

4,65421033

asked Jul 16 at 15:46

Jonatas Teo

31

asked Jul 16 at 15:46

Jonatas Teo

31

asked Jul 16 at 15:46

Jonatas Teo

31

31

add a comment | 

add a comment | 

1 Answer
1

active

oldest

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0
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accepted

This is not true as stated. For example, if $f=g$ then $V_1=V_2$, in which case we certainly cannot have $V=V_1oplus V_2$ (unless $V=0$, in which case everything equals $0$...)

It is true if $f$ and $g$ have no common factor. In that case, since the polynomials over a field form a PID there exist polynomials $p$ and $q$ with $$f(t)p(t)+g(t)q(t)=1.$$So $f(T)p(T)+g(T)q(T)=I$, hence for every $xin V$ we have $$x=f(T)(p(T)x)=g(T)(q(T)x)in V_1+V_2.$$Now suppose $xin V_1cap V_2$, so $x=p(T)v=g(T)w$. Then $$x=f(T)p(T)g(T)v+g(T)q(T)f(T)w=f(T)g(T)(p(T)v+q(T)w)=0.$$

(Note here that if $r$ and $s$ are polynomials it's clear that $r(T)s(T)=s(T)r(T)$.)

share|cite|improve this answer

edited Jul 16 at 17:51

answered Jul 16 at 16:07

David C. Ullrich

54.3k33583

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

This is not true as stated. For example, if $f=g$ then $V_1=V_2$, in which case we certainly cannot have $V=V_1oplus V_2$ (unless $V=0$, in which case everything equals $0$...)

It is true if $f$ and $g$ have no common factor. In that case, since the polynomials over a field form a PID there exist polynomials $p$ and $q$ with $$f(t)p(t)+g(t)q(t)=1.$$So $f(T)p(T)+g(T)q(T)=I$, hence for every $xin V$ we have $$x=f(T)(p(T)x)=g(T)(q(T)x)in V_1+V_2.$$Now suppose $xin V_1cap V_2$, so $x=p(T)v=g(T)w$. Then $$x=f(T)p(T)g(T)v+g(T)q(T)f(T)w=f(T)g(T)(p(T)v+q(T)w)=0.$$

(Note here that if $r$ and $s$ are polynomials it's clear that $r(T)s(T)=s(T)r(T)$.)

share|cite|improve this answer

edited Jul 16 at 17:51

answered Jul 16 at 16:07

David C. Ullrich

54.3k33583

add a comment | 

up vote
0
down vote



accepted

This is not true as stated. For example, if $f=g$ then $V_1=V_2$, in which case we certainly cannot have $V=V_1oplus V_2$ (unless $V=0$, in which case everything equals $0$...)

It is true if $f$ and $g$ have no common factor. In that case, since the polynomials over a field form a PID there exist polynomials $p$ and $q$ with $$f(t)p(t)+g(t)q(t)=1.$$So $f(T)p(T)+g(T)q(T)=I$, hence for every $xin V$ we have $$x=f(T)(p(T)x)=g(T)(q(T)x)in V_1+V_2.$$Now suppose $xin V_1cap V_2$, so $x=p(T)v=g(T)w$. Then $$x=f(T)p(T)g(T)v+g(T)q(T)f(T)w=f(T)g(T)(p(T)v+q(T)w)=0.$$

(Note here that if $r$ and $s$ are polynomials it's clear that $r(T)s(T)=s(T)r(T)$.)

share|cite|improve this answer

edited Jul 16 at 17:51

answered Jul 16 at 16:07

David C. Ullrich

54.3k33583

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

This is not true as stated. For example, if $f=g$ then $V_1=V_2$, in which case we certainly cannot have $V=V_1oplus V_2$ (unless $V=0$, in which case everything equals $0$...)

It is true if $f$ and $g$ have no common factor. In that case, since the polynomials over a field form a PID there exist polynomials $p$ and $q$ with $$f(t)p(t)+g(t)q(t)=1.$$So $f(T)p(T)+g(T)q(T)=I$, hence for every $xin V$ we have $$x=f(T)(p(T)x)=g(T)(q(T)x)in V_1+V_2.$$Now suppose $xin V_1cap V_2$, so $x=p(T)v=g(T)w$. Then $$x=f(T)p(T)g(T)v+g(T)q(T)f(T)w=f(T)g(T)(p(T)v+q(T)w)=0.$$

(Note here that if $r$ and $s$ are polynomials it's clear that $r(T)s(T)=s(T)r(T)$.)

share|cite|improve this answer

edited Jul 16 at 17:51

answered Jul 16 at 16:07

David C. Ullrich

54.3k33583

This is not true as stated. For example, if $f=g$ then $V_1=V_2$, in which case we certainly cannot have $V=V_1oplus V_2$ (unless $V=0$, in which case everything equals $0$...)

It is true if $f$ and $g$ have no common factor. In that case, since the polynomials over a field form a PID there exist polynomials $p$ and $q$ with $$f(t)p(t)+g(t)q(t)=1.$$So $f(T)p(T)+g(T)q(T)=I$, hence for every $xin V$ we have $$x=f(T)(p(T)x)=g(T)(q(T)x)in V_1+V_2.$$Now suppose $xin V_1cap V_2$, so $x=p(T)v=g(T)w$. Then $$x=f(T)p(T)g(T)v+g(T)q(T)f(T)w=f(T)g(T)(p(T)v+q(T)w)=0.$$

(Note here that if $r$ and $s$ are polynomials it's clear that $r(T)s(T)=s(T)r(T)$.)

share|cite|improve this answer

edited Jul 16 at 17:51

answered Jul 16 at 16:07

David C. Ullrich

54.3k33583

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 17:51

edited Jul 16 at 17:51

edited Jul 16 at 17:51

answered Jul 16 at 16:07

David C. Ullrich

54.3k33583

answered Jul 16 at 16:07

David C. Ullrich

54.3k33583

answered Jul 16 at 16:07

David C. Ullrich

54.3k33583

54.3k33583

add a comment | 

add a comment | 

 

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