homomorphism question Announcing the arrival of Valued Associate #679: Cesar Manara ...
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homomorphism question
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Question on Linear TransformationDetermining kernel and image of linear mapKernel of a linear map.Every subspace is the kernel of a linear mapIs every jordan homomorphism a ring homomorphismKernel of the differential of a projectionHow to find a linear map given the image/kernelA homomorphism that is an $R$-module homomorphism but is not a ring homomorphism.Determining the dimension of the kernel and the dimension of the image of the linear map $T:mathbb{R}^3rightarrowmathbb{R}^4$Algebra - Homomorphism
$begingroup$
Check $p(x) + p(-x) ∈ P^e$ for every $p(x) ∈ mathbb{R}[x]$. Check that the map $Ψ:mathbb{R}[x]to P^e$ given by $Ψ(p(x))=(p(x)+p(-x))/2$ is a linear map. Further check that $Ψ^2=Ψ$. Determine the kernel of $Ψ$.
linear-algebra ring-homomorphism
edited Mar 23 at 11:40
egreg
186k1486208
asked Mar 23 at 11:30
nitz visionnitz vision
31
$endgroup$
add a comment |
$begingroup$
Check $p(x) + p(-x) ∈ P^e$ for every $p(x) ∈ mathbb{R}[x]$. Check that the map $Ψ:mathbb{R}[x]to P^e$ given by $Ψ(p(x))=(p(x)+p(-x))/2$ is a linear map. Further check that $Ψ^2=Ψ$. Determine the kernel of $Ψ$.
linear-algebra ring-homomorphism
edited Mar 23 at 11:40
egreg
186k1486208
asked Mar 23 at 11:30
nitz visionnitz vision
31
$endgroup$
1
$begingroup$
What is $P^e$? ...
$endgroup$
– Maria Mazur
Mar 23 at 11:33
$begingroup$
Is it $Ψ$ correctly defined?
$endgroup$
– Maria Mazur
Mar 23 at 11:34
1
$begingroup$
Surely you can answer some of these questions right?
$endgroup$
– Kavi Rama Murthy
Mar 23 at 11:35
$begingroup$
$p(x)in R(x)$, don't you mean $p(x)in R[x]$ ?
$endgroup$
– Jakobian
Mar 23 at 11:37
$begingroup$
i need its answer guys ...its urgent
$endgroup$
– nitz vision
Mar 23 at 11:57
add a comment |
$begingroup$
Check $p(x) + p(-x) ∈ P^e$ for every $p(x) ∈ mathbb{R}[x]$. Check that the map $Ψ:mathbb{R}[x]to P^e$ given by $Ψ(p(x))=(p(x)+p(-x))/2$ is a linear map. Further check that $Ψ^2=Ψ$. Determine the kernel of $Ψ$.
linear-algebra ring-homomorphism
edited Mar 23 at 11:40
egreg
186k1486208
asked Mar 23 at 11:30
nitz visionnitz vision
31
$endgroup$
Check $p(x) + p(-x) ∈ P^e$ for every $p(x) ∈ mathbb{R}[x]$. Check that the map $Ψ:mathbb{R}[x]to P^e$ given by $Ψ(p(x))=(p(x)+p(-x))/2$ is a linear map. Further check that $Ψ^2=Ψ$. Determine the kernel of $Ψ$.
linear-algebra ring-homomorphism
linear-algebra ring-homomorphism
edited Mar 23 at 11:40
egreg
186k1486208
asked Mar 23 at 11:30
nitz visionnitz vision
31
edited Mar 23 at 11:40
egreg
186k1486208
asked Mar 23 at 11:30
nitz visionnitz vision
31
edited Mar 23 at 11:40
egreg
186k1486208
edited Mar 23 at 11:40
egreg
186k1486208
edited Mar 23 at 11:40
egreg
186k1486208
186k1486208
asked Mar 23 at 11:30
nitz visionnitz vision
31
asked Mar 23 at 11:30
nitz visionnitz vision
31
asked Mar 23 at 11:30
nitz visionnitz vision
31
31
1
$begingroup$
What is $P^e$? ...
$endgroup$
– Maria Mazur
Mar 23 at 11:33
$begingroup$
Is it $Ψ$ correctly defined?
$endgroup$
– Maria Mazur
Mar 23 at 11:34
1
$begingroup$
Surely you can answer some of these questions right?
$endgroup$
– Kavi Rama Murthy
Mar 23 at 11:35
$begingroup$
$p(x)in R(x)$, don't you mean $p(x)in R[x]$ ?
$endgroup$
– Jakobian
Mar 23 at 11:37
$begingroup$
i need its answer guys ...its urgent
$endgroup$
– nitz vision
Mar 23 at 11:57
add a comment |
1
$begingroup$
What is $P^e$? ...
$endgroup$
– Maria Mazur
Mar 23 at 11:33
$begingroup$
Is it $Ψ$ correctly defined?
$endgroup$
– Maria Mazur
Mar 23 at 11:34
1
$begingroup$
Surely you can answer some of these questions right?
$endgroup$
– Kavi Rama Murthy
Mar 23 at 11:35
$begingroup$
$p(x)in R(x)$, don't you mean $p(x)in R[x]$ ?
$endgroup$
– Jakobian
Mar 23 at 11:37
$begingroup$
i need its answer guys ...its urgent
$endgroup$
– nitz vision
Mar 23 at 11:57
1
1
$begingroup$
What is $P^e$? ...
$endgroup$
– Maria Mazur
Mar 23 at 11:33
$begingroup$
What is $P^e$? ...
$endgroup$
– Maria Mazur
Mar 23 at 11:33
$begingroup$
Is it $Ψ$ correctly defined?
$endgroup$
– Maria Mazur
Mar 23 at 11:34
$begingroup$
Is it $Ψ$ correctly defined?
$endgroup$
– Maria Mazur
Mar 23 at 11:34
1
1
$begingroup$
Surely you can answer some of these questions right?
$endgroup$
– Kavi Rama Murthy
Mar 23 at 11:35
$begingroup$
Surely you can answer some of these questions right?
$endgroup$
– Kavi Rama Murthy
Mar 23 at 11:35
$begingroup$
$p(x)in R(x)$, don't you mean $p(x)in R[x]$ ?
$endgroup$
– Jakobian
Mar 23 at 11:37
$begingroup$
$p(x)in R(x)$, don't you mean $p(x)in R[x]$ ?
$endgroup$
– Jakobian
Mar 23 at 11:37
$begingroup$
i need its answer guys ...its urgent
$endgroup$
– nitz vision
Mar 23 at 11:57
$begingroup$
i need its answer guys ...its urgent
$endgroup$
– nitz vision
Mar 23 at 11:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I assume that
$$
P^e = left{ p(x) in mathbb{R}[X] middle| p(x)=p(-x) right}
$$
is the set of even polynomials.
For a $p(x) in mathbb{R}[X]$, define $f_p(x)=p(x)+p(-x)$. We calculate
$$
f_p(-x) = p(-x)+p(x) = p(x)+p(-x) = f_p(x)
$$
and thus $f_p(x) in P^e$ for every $p(x)$. This implies that $Psi colon mathbb{R}[X] to P^e$ is well defined.
We check linearity as follows: Let $p(x),q(x) in mathbb{R}[X]$, then
begin{align*}
Psi (p(x)+q(x)) &= frac{p(x)+q(x)+p(-x)+q(-x)}{2} \
&= frac{p(x)+p(-x)}{2} + frac{q(x)+q(-x)}{2} = Psi (p(x)) + Psi (q(x))
end{align*}
and for $lambda in mathbb{R}$,
$$
Psi(lambda p(x)) = frac{lambda p(x) + lambda p(-x) }{2} = lambda frac{p(x)+p(-x)}{2} = lambda Psi (p(x))
$$
which concludes the proof of the linearity.
We calculate
$$
Psi (Psi (p(x))) = Psi left( frac{p(x)+p(-x)}{2}right) = frac{frac{p(x)+p(-x)}{2} + frac{p(-x)+p(x)}{2}}{2} = frac{p(x)+p(-x)}{2} = Psi(p(x)),
$$
i.e. $Psi ^2 = Psi$.
Finally $Psi (p(x)) = frac{p(x)+p(-x)}{2} = 0$, if and only if $p(x)+p(-x) =0$, so
$$
ker Psi = left{ p(x) in mathbb{R}[X] middle| p(-x)=-p(x) right},
$$
i.e. the kernel of $Psi$ is the set of odd polynomials.
answered Mar 23 at 13:21
StrichcoderStrichcoder
5047
$endgroup$
$begingroup$
thank you very much :)
$endgroup$
– nitz vision
Mar 23 at 13:45
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
I assume that
$$
P^e = left{ p(x) in mathbb{R}[X] middle| p(x)=p(-x) right}
$$
is the set of even polynomials.
For a $p(x) in mathbb{R}[X]$, define $f_p(x)=p(x)+p(-x)$. We calculate
$$
f_p(-x) = p(-x)+p(x) = p(x)+p(-x) = f_p(x)
$$
and thus $f_p(x) in P^e$ for every $p(x)$. This implies that $Psi colon mathbb{R}[X] to P^e$ is well defined.
We check linearity as follows: Let $p(x),q(x) in mathbb{R}[X]$, then
begin{align*}
Psi (p(x)+q(x)) &= frac{p(x)+q(x)+p(-x)+q(-x)}{2} \
&= frac{p(x)+p(-x)}{2} + frac{q(x)+q(-x)}{2} = Psi (p(x)) + Psi (q(x))
end{align*}
and for $lambda in mathbb{R}$,
$$
Psi(lambda p(x)) = frac{lambda p(x) + lambda p(-x) }{2} = lambda frac{p(x)+p(-x)}{2} = lambda Psi (p(x))
$$
which concludes the proof of the linearity.
We calculate
$$
Psi (Psi (p(x))) = Psi left( frac{p(x)+p(-x)}{2}right) = frac{frac{p(x)+p(-x)}{2} + frac{p(-x)+p(x)}{2}}{2} = frac{p(x)+p(-x)}{2} = Psi(p(x)),
$$
i.e. $Psi ^2 = Psi$.
Finally $Psi (p(x)) = frac{p(x)+p(-x)}{2} = 0$, if and only if $p(x)+p(-x) =0$, so
$$
ker Psi = left{ p(x) in mathbb{R}[X] middle| p(-x)=-p(x) right},
$$
i.e. the kernel of $Psi$ is the set of odd polynomials.
answered Mar 23 at 13:21
StrichcoderStrichcoder
5047
$endgroup$
$begingroup$
thank you very much :)
$endgroup$
– nitz vision
Mar 23 at 13:45
add a comment |
$begingroup$
I assume that
$$
P^e = left{ p(x) in mathbb{R}[X] middle| p(x)=p(-x) right}
$$
is the set of even polynomials.
For a $p(x) in mathbb{R}[X]$, define $f_p(x)=p(x)+p(-x)$. We calculate
$$
f_p(-x) = p(-x)+p(x) = p(x)+p(-x) = f_p(x)
$$
and thus $f_p(x) in P^e$ for every $p(x)$. This implies that $Psi colon mathbb{R}[X] to P^e$ is well defined.
We check linearity as follows: Let $p(x),q(x) in mathbb{R}[X]$, then
begin{align*}
Psi (p(x)+q(x)) &= frac{p(x)+q(x)+p(-x)+q(-x)}{2} \
&= frac{p(x)+p(-x)}{2} + frac{q(x)+q(-x)}{2} = Psi (p(x)) + Psi (q(x))
end{align*}
and for $lambda in mathbb{R}$,
$$
Psi(lambda p(x)) = frac{lambda p(x) + lambda p(-x) }{2} = lambda frac{p(x)+p(-x)}{2} = lambda Psi (p(x))
$$
which concludes the proof of the linearity.
We calculate
$$
Psi (Psi (p(x))) = Psi left( frac{p(x)+p(-x)}{2}right) = frac{frac{p(x)+p(-x)}{2} + frac{p(-x)+p(x)}{2}}{2} = frac{p(x)+p(-x)}{2} = Psi(p(x)),
$$
i.e. $Psi ^2 = Psi$.
Finally $Psi (p(x)) = frac{p(x)+p(-x)}{2} = 0$, if and only if $p(x)+p(-x) =0$, so
$$
ker Psi = left{ p(x) in mathbb{R}[X] middle| p(-x)=-p(x) right},
$$
i.e. the kernel of $Psi$ is the set of odd polynomials.
answered Mar 23 at 13:21
StrichcoderStrichcoder
5047
$endgroup$
$begingroup$
thank you very much :)
$endgroup$
– nitz vision
Mar 23 at 13:45
add a comment |
$begingroup$
I assume that
$$
P^e = left{ p(x) in mathbb{R}[X] middle| p(x)=p(-x) right}
$$
is the set of even polynomials.
For a $p(x) in mathbb{R}[X]$, define $f_p(x)=p(x)+p(-x)$. We calculate
$$
f_p(-x) = p(-x)+p(x) = p(x)+p(-x) = f_p(x)
$$
and thus $f_p(x) in P^e$ for every $p(x)$. This implies that $Psi colon mathbb{R}[X] to P^e$ is well defined.
We check linearity as follows: Let $p(x),q(x) in mathbb{R}[X]$, then
begin{align*}
Psi (p(x)+q(x)) &= frac{p(x)+q(x)+p(-x)+q(-x)}{2} \
&= frac{p(x)+p(-x)}{2} + frac{q(x)+q(-x)}{2} = Psi (p(x)) + Psi (q(x))
end{align*}
and for $lambda in mathbb{R}$,
$$
Psi(lambda p(x)) = frac{lambda p(x) + lambda p(-x) }{2} = lambda frac{p(x)+p(-x)}{2} = lambda Psi (p(x))
$$
which concludes the proof of the linearity.
We calculate
$$
Psi (Psi (p(x))) = Psi left( frac{p(x)+p(-x)}{2}right) = frac{frac{p(x)+p(-x)}{2} + frac{p(-x)+p(x)}{2}}{2} = frac{p(x)+p(-x)}{2} = Psi(p(x)),
$$
i.e. $Psi ^2 = Psi$.
Finally $Psi (p(x)) = frac{p(x)+p(-x)}{2} = 0$, if and only if $p(x)+p(-x) =0$, so
$$
ker Psi = left{ p(x) in mathbb{R}[X] middle| p(-x)=-p(x) right},
$$
i.e. the kernel of $Psi$ is the set of odd polynomials.
answered Mar 23 at 13:21
StrichcoderStrichcoder
5047
$endgroup$
I assume that
$$
P^e = left{ p(x) in mathbb{R}[X] middle| p(x)=p(-x) right}
$$
is the set of even polynomials.
For a $p(x) in mathbb{R}[X]$, define $f_p(x)=p(x)+p(-x)$. We calculate
$$
f_p(-x) = p(-x)+p(x) = p(x)+p(-x) = f_p(x)
$$
and thus $f_p(x) in P^e$ for every $p(x)$. This implies that $Psi colon mathbb{R}[X] to P^e$ is well defined.
We check linearity as follows: Let $p(x),q(x) in mathbb{R}[X]$, then
begin{align*}
Psi (p(x)+q(x)) &= frac{p(x)+q(x)+p(-x)+q(-x)}{2} \
&= frac{p(x)+p(-x)}{2} + frac{q(x)+q(-x)}{2} = Psi (p(x)) + Psi (q(x))
end{align*}
and for $lambda in mathbb{R}$,
$$
Psi(lambda p(x)) = frac{lambda p(x) + lambda p(-x) }{2} = lambda frac{p(x)+p(-x)}{2} = lambda Psi (p(x))
$$
which concludes the proof of the linearity.
We calculate
$$
Psi (Psi (p(x))) = Psi left( frac{p(x)+p(-x)}{2}right) = frac{frac{p(x)+p(-x)}{2} + frac{p(-x)+p(x)}{2}}{2} = frac{p(x)+p(-x)}{2} = Psi(p(x)),
$$
i.e. $Psi ^2 = Psi$.
Finally $Psi (p(x)) = frac{p(x)+p(-x)}{2} = 0$, if and only if $p(x)+p(-x) =0$, so
$$
ker Psi = left{ p(x) in mathbb{R}[X] middle| p(-x)=-p(x) right},
$$
i.e. the kernel of $Psi$ is the set of odd polynomials.
answered Mar 23 at 13:21
StrichcoderStrichcoder
5047
answered Mar 23 at 13:21
StrichcoderStrichcoder
5047
answered Mar 23 at 13:21
StrichcoderStrichcoder
5047
answered Mar 23 at 13:21
StrichcoderStrichcoder
5047
5047
$begingroup$
thank you very much :)
$endgroup$
– nitz vision
Mar 23 at 13:45
add a comment |
$begingroup$
thank you very much :)
$endgroup$
– nitz vision
Mar 23 at 13:45
$begingroup$
thank you very much :)
$endgroup$
– nitz vision
Mar 23 at 13:45
$begingroup$
thank you very much :)
$endgroup$
– nitz vision
Mar 23 at 13:45
add a comment |
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1
$begingroup$
What is $P^e$? ...
$endgroup$
– Maria Mazur
Mar 23 at 11:33
$begingroup$
Is it $Ψ$ correctly defined?
$endgroup$
– Maria Mazur
Mar 23 at 11:34
1
$begingroup$
Surely you can answer some of these questions right?
$endgroup$
– Kavi Rama Murthy
Mar 23 at 11:35
$begingroup$
$p(x)in R(x)$, don't you mean $p(x)in R[x]$ ?
$endgroup$
– Jakobian
Mar 23 at 11:37
$begingroup$
i need its answer guys ...its urgent
$endgroup$
– nitz vision
Mar 23 at 11:57