Prove that $p ◦ p = p$. (Representing a Linear Transformation as a Matrix)












0












$begingroup$


Let $B = (1, X, X^2)$ be an ordered basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}(mathbb{R}_2[X])$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.



By definition of the task, I found that $A = M_B(p)$



Compute $A^2$. Prove that $p ◦ p = p$.



begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}



I computed the multiplication and it appears that $A^2$ = $A$. How should I prove it?






linear-algebra matrices linear-transformations change-of-basis projection-matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:45










  • $begingroup$
    Sure, that's correct.
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:47
















0












$begingroup$


Let $B = (1, X, X^2)$ be an ordered basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}(mathbb{R}_2[X])$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.



By definition of the task, I found that $A = M_B(p)$



Compute $A^2$. Prove that $p ◦ p = p$.



begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}



I computed the multiplication and it appears that $A^2$ = $A$. How should I prove it?






linear-algebra matrices linear-transformations change-of-basis projection-matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:45










  • $begingroup$
    Sure, that's correct.
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:47














0












0








0





$begingroup$


Let $B = (1, X, X^2)$ be an ordered basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}(mathbb{R}_2[X])$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.



By definition of the task, I found that $A = M_B(p)$



Compute $A^2$. Prove that $p ◦ p = p$.



begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}



I computed the multiplication and it appears that $A^2$ = $A$. How should I prove it?






linear-algebra matrices linear-transformations change-of-basis projection-matrices







share|cite|improve this question











$endgroup$




Let $B = (1, X, X^2)$ be an ordered basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}(mathbb{R}_2[X])$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.



By definition of the task, I found that $A = M_B(p)$



Compute $A^2$. Prove that $p ◦ p = p$.



begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}



I computed the multiplication and it appears that $A^2$ = $A$. How should I prove it?






linear-algebra matrices linear-transformations change-of-basis projection-matrices




linear-algebra matrices linear-transformations change-of-basis projection-matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 12:18









Batominovski

1




1










asked Dec 13 '18 at 16:40









nasdaqnasdaq

62




62












  • $begingroup$
    Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:45










  • $begingroup$
    Sure, that's correct.
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:47


















  • $begingroup$
    Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:45










  • $begingroup$
    Sure, that's correct.
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:47
















$begingroup$
Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
$endgroup$
– Shubham Johri
Dec 13 '18 at 16:45




$begingroup$
Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
$endgroup$
– Shubham Johri
Dec 13 '18 at 16:45












$begingroup$
Sure, that's correct.
$endgroup$
– nasdaq
Dec 13 '18 at 16:47




$begingroup$
Sure, that's correct.
$endgroup$
– nasdaq
Dec 13 '18 at 16:47










1 Answer
1






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oldest

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2












$begingroup$

If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=Acdot v forall vinBbb R_2[X]$.



$implies(pcirc p)(v)=p(p(v))=Acdot(p(v))=Acdot(Av)=(Acdot A)v=A^2cdot v$



This means the matrix corresponding to the linear operator $pcirc p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $pcirc p$ are identical, which means $p=pcirc p$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, that's very helpful!
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:52










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:55











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=Acdot v forall vinBbb R_2[X]$.



$implies(pcirc p)(v)=p(p(v))=Acdot(p(v))=Acdot(Av)=(Acdot A)v=A^2cdot v$



This means the matrix corresponding to the linear operator $pcirc p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $pcirc p$ are identical, which means $p=pcirc p$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, that's very helpful!
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:52










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:55
















2












$begingroup$

If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=Acdot v forall vinBbb R_2[X]$.



$implies(pcirc p)(v)=p(p(v))=Acdot(p(v))=Acdot(Av)=(Acdot A)v=A^2cdot v$



This means the matrix corresponding to the linear operator $pcirc p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $pcirc p$ are identical, which means $p=pcirc p$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, that's very helpful!
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:52










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:55














2












2








2





$begingroup$

If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=Acdot v forall vinBbb R_2[X]$.



$implies(pcirc p)(v)=p(p(v))=Acdot(p(v))=Acdot(Av)=(Acdot A)v=A^2cdot v$



This means the matrix corresponding to the linear operator $pcirc p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $pcirc p$ are identical, which means $p=pcirc p$.






share|cite|improve this answer











$endgroup$



If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=Acdot v forall vinBbb R_2[X]$.



$implies(pcirc p)(v)=p(p(v))=Acdot(p(v))=Acdot(Av)=(Acdot A)v=A^2cdot v$



This means the matrix corresponding to the linear operator $pcirc p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $pcirc p$ are identical, which means $p=pcirc p$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 16:54

























answered Dec 13 '18 at 16:50









Shubham JohriShubham Johri

4,666717




4,666717












  • $begingroup$
    Thanks, that's very helpful!
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:52










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:55


















  • $begingroup$
    Thanks, that's very helpful!
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:52










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:55
















$begingroup$
Thanks, that's very helpful!
$endgroup$
– nasdaq
Dec 13 '18 at 16:52




$begingroup$
Thanks, that's very helpful!
$endgroup$
– nasdaq
Dec 13 '18 at 16:52












$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 13 '18 at 16:55




$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 13 '18 at 16:55


















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