Mixing
Endomorphism on polynomial vector space $mathbb{R}_3[x]$
$begingroup$
Let $ mathbb{R}_3[x] $ be the vector space of the polynomials with the degree $ le 3$. Given the endomorphism on this vector space, $$ T:mathbb{R}_3[x] to mathbb{R}_3[x], T(f)(x) = f(x+1)-f(x), $$ calculate its eigenvalues, eigenvectors and also its matrix representation in the base $ {{1,x,x^2, x^3} } $ of this space.
So far I tried to calculate the eigenvalues using the basic formula $T(f)(x) = lambda x$. This resulted in an equation depending on $lambda : f(x+1)-f(x)(lambda-1)=0$. What should I do next?
Thank you!
linear-algebra vector-spaces eigenvalues-eigenvectors vector-space-isomorphism
asked Jan 21 at 10:17
Radu BouaruRadu Bouaru
747
$endgroup$
add a comment |
$begingroup$
Let $ mathbb{R}_3[x] $ be the vector space of the polynomials with the degree $ le 3$. Given the endomorphism on this vector space, $$ T:mathbb{R}_3[x] to mathbb{R}_3[x], T(f)(x) = f(x+1)-f(x), $$ calculate its eigenvalues, eigenvectors and also its matrix representation in the base $ {{1,x,x^2, x^3} } $ of this space.
So far I tried to calculate the eigenvalues using the basic formula $T(f)(x) = lambda x$. This resulted in an equation depending on $lambda : f(x+1)-f(x)(lambda-1)=0$. What should I do next?
Thank you!
linear-algebra vector-spaces eigenvalues-eigenvectors vector-space-isomorphism
asked Jan 21 at 10:17
Radu BouaruRadu Bouaru
747
$endgroup$
1
$begingroup$
Note that the formula you would get is $T(f) = lambda f$, or, equivalently, $T(f)(x) = (lambda f)(x) = lambda f(x)$. The obvious thing to do next is to write $f(x) = ax^3 + bx^2 + cx + d$, and then see what conditions you get on $a, b, c, d, lambda$ so that your equation holds. (Note that this is basically equivalent to writing $f$ using the basis coordinates ${1, x, x^2, x^3}$ and looking at $T$ as a mapping $mathbb R^4 to mathbb R^4$.)
$endgroup$
– Mees de Vries
Jan 21 at 10:29
add a comment |
$begingroup$
Let $ mathbb{R}_3[x] $ be the vector space of the polynomials with the degree $ le 3$. Given the endomorphism on this vector space, $$ T:mathbb{R}_3[x] to mathbb{R}_3[x], T(f)(x) = f(x+1)-f(x), $$ calculate its eigenvalues, eigenvectors and also its matrix representation in the base $ {{1,x,x^2, x^3} } $ of this space.
So far I tried to calculate the eigenvalues using the basic formula $T(f)(x) = lambda x$. This resulted in an equation depending on $lambda : f(x+1)-f(x)(lambda-1)=0$. What should I do next?
Thank you!
linear-algebra vector-spaces eigenvalues-eigenvectors vector-space-isomorphism
asked Jan 21 at 10:17
Radu BouaruRadu Bouaru
747
$endgroup$
Let $ mathbb{R}_3[x] $ be the vector space of the polynomials with the degree $ le 3$. Given the endomorphism on this vector space, $$ T:mathbb{R}_3[x] to mathbb{R}_3[x], T(f)(x) = f(x+1)-f(x), $$ calculate its eigenvalues, eigenvectors and also its matrix representation in the base $ {{1,x,x^2, x^3} } $ of this space.
So far I tried to calculate the eigenvalues using the basic formula $T(f)(x) = lambda x$. This resulted in an equation depending on $lambda : f(x+1)-f(x)(lambda-1)=0$. What should I do next?
Thank you!
linear-algebra vector-spaces eigenvalues-eigenvectors vector-space-isomorphism
linear-algebra vector-spaces eigenvalues-eigenvectors vector-space-isomorphism
asked Jan 21 at 10:17
Radu BouaruRadu Bouaru
747
asked Jan 21 at 10:17
Radu BouaruRadu Bouaru
747
asked Jan 21 at 10:17
Radu BouaruRadu Bouaru
747
asked Jan 21 at 10:17
Radu BouaruRadu Bouaru
747
asked Jan 21 at 10:17
Radu BouaruRadu Bouaru
747
747
1
$begingroup$
Note that the formula you would get is $T(f) = lambda f$, or, equivalently, $T(f)(x) = (lambda f)(x) = lambda f(x)$. The obvious thing to do next is to write $f(x) = ax^3 + bx^2 + cx + d$, and then see what conditions you get on $a, b, c, d, lambda$ so that your equation holds. (Note that this is basically equivalent to writing $f$ using the basis coordinates ${1, x, x^2, x^3}$ and looking at $T$ as a mapping $mathbb R^4 to mathbb R^4$.)
$endgroup$
– Mees de Vries
Jan 21 at 10:29
add a comment |
1
$begingroup$
Note that the formula you would get is $T(f) = lambda f$, or, equivalently, $T(f)(x) = (lambda f)(x) = lambda f(x)$. The obvious thing to do next is to write $f(x) = ax^3 + bx^2 + cx + d$, and then see what conditions you get on $a, b, c, d, lambda$ so that your equation holds. (Note that this is basically equivalent to writing $f$ using the basis coordinates ${1, x, x^2, x^3}$ and looking at $T$ as a mapping $mathbb R^4 to mathbb R^4$.)
$endgroup$
– Mees de Vries
Jan 21 at 10:29
1
1
$begingroup$
Note that the formula you would get is $T(f) = lambda f$, or, equivalently, $T(f)(x) = (lambda f)(x) = lambda f(x)$. The obvious thing to do next is to write $f(x) = ax^3 + bx^2 + cx + d$, and then see what conditions you get on $a, b, c, d, lambda$ so that your equation holds. (Note that this is basically equivalent to writing $f$ using the basis coordinates ${1, x, x^2, x^3}$ and looking at $T$ as a mapping $mathbb R^4 to mathbb R^4$.)
$endgroup$
– Mees de Vries
Jan 21 at 10:29
$begingroup$
Note that the formula you would get is $T(f) = lambda f$, or, equivalently, $T(f)(x) = (lambda f)(x) = lambda f(x)$. The obvious thing to do next is to write $f(x) = ax^3 + bx^2 + cx + d$, and then see what conditions you get on $a, b, c, d, lambda$ so that your equation holds. (Note that this is basically equivalent to writing $f$ using the basis coordinates ${1, x, x^2, x^3}$ and looking at $T$ as a mapping $mathbb R^4 to mathbb R^4$.)
$endgroup$
– Mees de Vries
Jan 21 at 10:29
add a comment |
1 Answer
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oldest
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$begingroup$
There are two ways to see that only $ 0 $ is an eigenvalue, with eigenvectors the constant polynomials :
Consider your polynomial $ P = aX^k + R $ where $ k > 0 $ and $ operatorname{deg} R < k $. We have
$$ begin{align} T(P)&= a((X+1)^k - X^k) + T(R) \
&= a(X+1 -X)sum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R)\
&= asum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R) end{align}$$
which is of degree $ < k $, and thus cannot be equal to $ P $Consider your polynomial $ P $ in $ mathbb{C}[X] $ instead, then suppose $ T(P) = lambda P $ and $ P $ non-constant. It has one root $ z $ by d'Alembert-Gauss, and the functional equation gives you that $ z + 1 $ is another root. By induction $ P $ has an infinite number of roots and $ P = 0 $.
Calculation of $ T $ on constant polynomials then gives you the desired conclusion.
As for the matrix representation, just use the identity $ (a + b)^n = sum_{k=0}^n binom{n}{k}a^kb^{n-k}.$
answered Jan 21 at 10:43
FreeSaladFreeSalad
479211
$endgroup$
add a comment |
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$begingroup$
There are two ways to see that only $ 0 $ is an eigenvalue, with eigenvectors the constant polynomials :
Consider your polynomial $ P = aX^k + R $ where $ k > 0 $ and $ operatorname{deg} R < k $. We have
$$ begin{align} T(P)&= a((X+1)^k - X^k) + T(R) \
&= a(X+1 -X)sum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R)\
&= asum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R) end{align}$$
which is of degree $ < k $, and thus cannot be equal to $ P $Consider your polynomial $ P $ in $ mathbb{C}[X] $ instead, then suppose $ T(P) = lambda P $ and $ P $ non-constant. It has one root $ z $ by d'Alembert-Gauss, and the functional equation gives you that $ z + 1 $ is another root. By induction $ P $ has an infinite number of roots and $ P = 0 $.
Calculation of $ T $ on constant polynomials then gives you the desired conclusion.
As for the matrix representation, just use the identity $ (a + b)^n = sum_{k=0}^n binom{n}{k}a^kb^{n-k}.$
answered Jan 21 at 10:43
FreeSaladFreeSalad
479211
$endgroup$
add a comment |
$begingroup$
There are two ways to see that only $ 0 $ is an eigenvalue, with eigenvectors the constant polynomials :
Consider your polynomial $ P = aX^k + R $ where $ k > 0 $ and $ operatorname{deg} R < k $. We have
$$ begin{align} T(P)&= a((X+1)^k - X^k) + T(R) \
&= a(X+1 -X)sum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R)\
&= asum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R) end{align}$$
which is of degree $ < k $, and thus cannot be equal to $ P $Consider your polynomial $ P $ in $ mathbb{C}[X] $ instead, then suppose $ T(P) = lambda P $ and $ P $ non-constant. It has one root $ z $ by d'Alembert-Gauss, and the functional equation gives you that $ z + 1 $ is another root. By induction $ P $ has an infinite number of roots and $ P = 0 $.
Calculation of $ T $ on constant polynomials then gives you the desired conclusion.
As for the matrix representation, just use the identity $ (a + b)^n = sum_{k=0}^n binom{n}{k}a^kb^{n-k}.$
answered Jan 21 at 10:43
FreeSaladFreeSalad
479211
$endgroup$
add a comment |
$begingroup$
There are two ways to see that only $ 0 $ is an eigenvalue, with eigenvectors the constant polynomials :
Consider your polynomial $ P = aX^k + R $ where $ k > 0 $ and $ operatorname{deg} R < k $. We have
$$ begin{align} T(P)&= a((X+1)^k - X^k) + T(R) \
&= a(X+1 -X)sum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R)\
&= asum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R) end{align}$$
which is of degree $ < k $, and thus cannot be equal to $ P $Consider your polynomial $ P $ in $ mathbb{C}[X] $ instead, then suppose $ T(P) = lambda P $ and $ P $ non-constant. It has one root $ z $ by d'Alembert-Gauss, and the functional equation gives you that $ z + 1 $ is another root. By induction $ P $ has an infinite number of roots and $ P = 0 $.
Calculation of $ T $ on constant polynomials then gives you the desired conclusion.
As for the matrix representation, just use the identity $ (a + b)^n = sum_{k=0}^n binom{n}{k}a^kb^{n-k}.$
answered Jan 21 at 10:43
FreeSaladFreeSalad
479211
$endgroup$
There are two ways to see that only $ 0 $ is an eigenvalue, with eigenvectors the constant polynomials :
Consider your polynomial $ P = aX^k + R $ where $ k > 0 $ and $ operatorname{deg} R < k $. We have
$$ begin{align} T(P)&= a((X+1)^k - X^k) + T(R) \
&= a(X+1 -X)sum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R)\
&= asum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R) end{align}$$
which is of degree $ < k $, and thus cannot be equal to $ P $Consider your polynomial $ P $ in $ mathbb{C}[X] $ instead, then suppose $ T(P) = lambda P $ and $ P $ non-constant. It has one root $ z $ by d'Alembert-Gauss, and the functional equation gives you that $ z + 1 $ is another root. By induction $ P $ has an infinite number of roots and $ P = 0 $.
Calculation of $ T $ on constant polynomials then gives you the desired conclusion.
As for the matrix representation, just use the identity $ (a + b)^n = sum_{k=0}^n binom{n}{k}a^kb^{n-k}.$
answered Jan 21 at 10:43
FreeSaladFreeSalad
479211
answered Jan 21 at 10:43
FreeSaladFreeSalad
479211
answered Jan 21 at 10:43
FreeSaladFreeSalad
479211
answered Jan 21 at 10:43
FreeSaladFreeSalad
479211
479211
add a comment |
add a comment |
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$begingroup$
Note that the formula you would get is $T(f) = lambda f$, or, equivalently, $T(f)(x) = (lambda f)(x) = lambda f(x)$. The obvious thing to do next is to write $f(x) = ax^3 + bx^2 + cx + d$, and then see what conditions you get on $a, b, c, d, lambda$ so that your equation holds. (Note that this is basically equivalent to writing $f$ using the basis coordinates ${1, x, x^2, x^3}$ and looking at $T$ as a mapping $mathbb R^4 to mathbb R^4$.)
$endgroup$
– Mees de Vries
Jan 21 at 10:29