Subspace, Direct Sum, Polynomials, Basis
$begingroup$
Let $U = {p in mathcal{P}_4(mathbb{F}) ;colon; p''(6) = 0}$.
a. Find a basis for $U$.
b. Extend the basis in part (a) to a basis for $mathcal{P}_4(mathbb{F})$.
c. Find a subspace $W$ of $mathcal{P}_4(mathbb{F})$ such that
$mathcal{P}_4(mathbb{F}) = U oplus W$.
If I take the basis as $1$, $x$, $x^3 -18x^2$, and $x^4-12x^3$.
Now $x^2$ can't be produced by the basis elements so adding $x^2$ to the previous basis I get the basis of $mathcal{P}_4(mathbb{F})$.
Please let me know if I am correct.
Moreover I am stuck with the part (c).
linear-algebra
edited May 1 '15 at 21:56
Mike Pierce
11.5k103584
asked May 1 '15 at 21:48
jyoti prokash royjyoti prokash roy
459510
$endgroup$
add a comment |
$begingroup$
Let $U = {p in mathcal{P}_4(mathbb{F}) ;colon; p''(6) = 0}$.
a. Find a basis for $U$.
b. Extend the basis in part (a) to a basis for $mathcal{P}_4(mathbb{F})$.
c. Find a subspace $W$ of $mathcal{P}_4(mathbb{F})$ such that
$mathcal{P}_4(mathbb{F}) = U oplus W$.
If I take the basis as $1$, $x$, $x^3 -18x^2$, and $x^4-12x^3$.
Now $x^2$ can't be produced by the basis elements so adding $x^2$ to the previous basis I get the basis of $mathcal{P}_4(mathbb{F})$.
Please let me know if I am correct.
Moreover I am stuck with the part (c).
linear-algebra
edited May 1 '15 at 21:56
Mike Pierce
11.5k103584
asked May 1 '15 at 21:48
jyoti prokash royjyoti prokash roy
459510
$endgroup$
1
$begingroup$
Well, you already have a $W$: it's the subspace generated by $x^2$.
$endgroup$
– Bernard
May 1 '15 at 21:59
$begingroup$
Did I do it correctly?
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:01
$begingroup$
Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
$endgroup$
– Bernard
May 1 '15 at 22:05
$begingroup$
thanks for pointing it out.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
add a comment |
$begingroup$
Let $U = {p in mathcal{P}_4(mathbb{F}) ;colon; p''(6) = 0}$.
a. Find a basis for $U$.
b. Extend the basis in part (a) to a basis for $mathcal{P}_4(mathbb{F})$.
c. Find a subspace $W$ of $mathcal{P}_4(mathbb{F})$ such that
$mathcal{P}_4(mathbb{F}) = U oplus W$.
If I take the basis as $1$, $x$, $x^3 -18x^2$, and $x^4-12x^3$.
Now $x^2$ can't be produced by the basis elements so adding $x^2$ to the previous basis I get the basis of $mathcal{P}_4(mathbb{F})$.
Please let me know if I am correct.
Moreover I am stuck with the part (c).
linear-algebra
edited May 1 '15 at 21:56
Mike Pierce
11.5k103584
asked May 1 '15 at 21:48
jyoti prokash royjyoti prokash roy
459510
$endgroup$
Let $U = {p in mathcal{P}_4(mathbb{F}) ;colon; p''(6) = 0}$.
a. Find a basis for $U$.
b. Extend the basis in part (a) to a basis for $mathcal{P}_4(mathbb{F})$.
c. Find a subspace $W$ of $mathcal{P}_4(mathbb{F})$ such that
$mathcal{P}_4(mathbb{F}) = U oplus W$.
If I take the basis as $1$, $x$, $x^3 -18x^2$, and $x^4-12x^3$.
Now $x^2$ can't be produced by the basis elements so adding $x^2$ to the previous basis I get the basis of $mathcal{P}_4(mathbb{F})$.
Please let me know if I am correct.
Moreover I am stuck with the part (c).
linear-algebra
linear-algebra
edited May 1 '15 at 21:56
Mike Pierce
11.5k103584
asked May 1 '15 at 21:48
jyoti prokash royjyoti prokash roy
459510
edited May 1 '15 at 21:56
Mike Pierce
11.5k103584
asked May 1 '15 at 21:48
jyoti prokash royjyoti prokash roy
459510
edited May 1 '15 at 21:56
Mike Pierce
11.5k103584
edited May 1 '15 at 21:56
Mike Pierce
11.5k103584
edited May 1 '15 at 21:56
Mike Pierce
11.5k103584
11.5k103584
asked May 1 '15 at 21:48
jyoti prokash royjyoti prokash roy
459510
asked May 1 '15 at 21:48
jyoti prokash royjyoti prokash roy
459510
asked May 1 '15 at 21:48
jyoti prokash royjyoti prokash roy
459510
459510
1
$begingroup$
Well, you already have a $W$: it's the subspace generated by $x^2$.
$endgroup$
– Bernard
May 1 '15 at 21:59
$begingroup$
Did I do it correctly?
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:01
$begingroup$
Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
$endgroup$
– Bernard
May 1 '15 at 22:05
$begingroup$
thanks for pointing it out.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
add a comment |
1
$begingroup$
Well, you already have a $W$: it's the subspace generated by $x^2$.
$endgroup$
– Bernard
May 1 '15 at 21:59
$begingroup$
Did I do it correctly?
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:01
$begingroup$
Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
$endgroup$
– Bernard
May 1 '15 at 22:05
$begingroup$
thanks for pointing it out.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
1
1
$begingroup$
Well, you already have a $W$: it's the subspace generated by $x^2$.
$endgroup$
– Bernard
May 1 '15 at 21:59
$begingroup$
Well, you already have a $W$: it's the subspace generated by $x^2$.
$endgroup$
– Bernard
May 1 '15 at 21:59
$begingroup$
Did I do it correctly?
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:01
$begingroup$
Did I do it correctly?
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:01
$begingroup$
Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
$endgroup$
– Bernard
May 1 '15 at 22:05
$begingroup$
Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
$endgroup$
– Bernard
May 1 '15 at 22:05
$begingroup$
thanks for pointing it out.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
$begingroup$
thanks for pointing it out.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
begin{align}
p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\
p''(x)&=2a_2+6a_3x+12a_4x^2
end{align}
so
$$
p''(6)=2a_2+36a_3+432a_4
$$
and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by
$$
{1,x,-18x^2+x^3,-216x^2+x^4}
$$
Now you have just to find a polynomial $q$ such that $qnotin U$, for instance $q(x)=x^2$.
Add this to the above basis; of course, you can also take $W=operatorname{Span}(x^2)$.
So you're correct. I just found a different basis, with a more systematic approach.
answered May 1 '15 at 22:04
egregegreg
183k1486204
$endgroup$
$begingroup$
thanks a lot,really appreciate your help.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
add a comment |
$begingroup$
Let us assume $mathbb F=mathbb R$. (Or $mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)
You could use the fact that very polynomial in $mathcal P_4$ can be expressed (uniquely) as
$$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
You have
begin{align*}
p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\
p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\
p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
end{align*}
and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
So you have
$$U={p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3inmathbb F}.$$
So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.
The only difference is if $mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=mathcal P_4$.
answered Dec 30 '18 at 9:23
Martin SleziakMartin Sleziak
44.7k10119272
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
begin{align}
p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\
p''(x)&=2a_2+6a_3x+12a_4x^2
end{align}
so
$$
p''(6)=2a_2+36a_3+432a_4
$$
and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by
$$
{1,x,-18x^2+x^3,-216x^2+x^4}
$$
Now you have just to find a polynomial $q$ such that $qnotin U$, for instance $q(x)=x^2$.
Add this to the above basis; of course, you can also take $W=operatorname{Span}(x^2)$.
So you're correct. I just found a different basis, with a more systematic approach.
answered May 1 '15 at 22:04
egregegreg
183k1486204
$endgroup$
$begingroup$
thanks a lot,really appreciate your help.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
add a comment |
$begingroup$
If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
begin{align}
p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\
p''(x)&=2a_2+6a_3x+12a_4x^2
end{align}
so
$$
p''(6)=2a_2+36a_3+432a_4
$$
and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by
$$
{1,x,-18x^2+x^3,-216x^2+x^4}
$$
Now you have just to find a polynomial $q$ such that $qnotin U$, for instance $q(x)=x^2$.
Add this to the above basis; of course, you can also take $W=operatorname{Span}(x^2)$.
So you're correct. I just found a different basis, with a more systematic approach.
answered May 1 '15 at 22:04
egregegreg
183k1486204
$endgroup$
$begingroup$
thanks a lot,really appreciate your help.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
add a comment |
$begingroup$
If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
begin{align}
p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\
p''(x)&=2a_2+6a_3x+12a_4x^2
end{align}
so
$$
p''(6)=2a_2+36a_3+432a_4
$$
and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by
$$
{1,x,-18x^2+x^3,-216x^2+x^4}
$$
Now you have just to find a polynomial $q$ such that $qnotin U$, for instance $q(x)=x^2$.
Add this to the above basis; of course, you can also take $W=operatorname{Span}(x^2)$.
So you're correct. I just found a different basis, with a more systematic approach.
answered May 1 '15 at 22:04
egregegreg
183k1486204
$endgroup$
If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
begin{align}
p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\
p''(x)&=2a_2+6a_3x+12a_4x^2
end{align}
so
$$
p''(6)=2a_2+36a_3+432a_4
$$
and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by
$$
{1,x,-18x^2+x^3,-216x^2+x^4}
$$
Now you have just to find a polynomial $q$ such that $qnotin U$, for instance $q(x)=x^2$.
Add this to the above basis; of course, you can also take $W=operatorname{Span}(x^2)$.
So you're correct. I just found a different basis, with a more systematic approach.
answered May 1 '15 at 22:04
egregegreg
183k1486204
answered May 1 '15 at 22:04
egregegreg
183k1486204
answered May 1 '15 at 22:04
egregegreg
183k1486204
answered May 1 '15 at 22:04
egregegreg
183k1486204
183k1486204
$begingroup$
thanks a lot,really appreciate your help.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
add a comment |
$begingroup$
thanks a lot,really appreciate your help.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
$begingroup$
thanks a lot,really appreciate your help.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
$begingroup$
thanks a lot,really appreciate your help.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09
add a comment |
$begingroup$
Let us assume $mathbb F=mathbb R$. (Or $mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)
You could use the fact that very polynomial in $mathcal P_4$ can be expressed (uniquely) as
$$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
You have
begin{align*}
p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\
p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\
p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
end{align*}
and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
So you have
$$U={p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3inmathbb F}.$$
So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.
The only difference is if $mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=mathcal P_4$.
answered Dec 30 '18 at 9:23
Martin SleziakMartin Sleziak
44.7k10119272
$endgroup$
add a comment |
$begingroup$
Let us assume $mathbb F=mathbb R$. (Or $mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)
You could use the fact that very polynomial in $mathcal P_4$ can be expressed (uniquely) as
$$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
You have
begin{align*}
p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\
p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\
p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
end{align*}
and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
So you have
$$U={p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3inmathbb F}.$$
So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.
The only difference is if $mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=mathcal P_4$.
answered Dec 30 '18 at 9:23
Martin SleziakMartin Sleziak
44.7k10119272
$endgroup$
add a comment |
$begingroup$
Let us assume $mathbb F=mathbb R$. (Or $mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)
You could use the fact that very polynomial in $mathcal P_4$ can be expressed (uniquely) as
$$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
You have
begin{align*}
p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\
p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\
p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
end{align*}
and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
So you have
$$U={p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3inmathbb F}.$$
So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.
The only difference is if $mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=mathcal P_4$.
answered Dec 30 '18 at 9:23
Martin SleziakMartin Sleziak
44.7k10119272
$endgroup$
Let us assume $mathbb F=mathbb R$. (Or $mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)
You could use the fact that very polynomial in $mathcal P_4$ can be expressed (uniquely) as
$$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
You have
begin{align*}
p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\
p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\
p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
end{align*}
and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
So you have
$$U={p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3inmathbb F}.$$
So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.
The only difference is if $mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=mathcal P_4$.
answered Dec 30 '18 at 9:23
Martin SleziakMartin Sleziak
44.7k10119272
answered Dec 30 '18 at 9:23
Martin SleziakMartin Sleziak
44.7k10119272
answered Dec 30 '18 at 9:23
Martin SleziakMartin Sleziak
44.7k10119272
answered Dec 30 '18 at 9:23
Martin SleziakMartin Sleziak
44.7k10119272
44.7k10119272
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1
$begingroup$
Well, you already have a $W$: it's the subspace generated by $x^2$.
$endgroup$
– Bernard
May 1 '15 at 21:59
$begingroup$
Did I do it correctly?
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:01
$begingroup$
Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
$endgroup$
– Bernard
May 1 '15 at 22:05
$begingroup$
thanks for pointing it out.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09