Eigenvalue of a Polynomial [duplicate]
This question already has an answer here:
How to prove “eigenvalues of polynomial of matrix $A$ = polynomial of eigenvalues of matrix $A$ ”
2 answers
Let $A: mathbb C^4 to mathbb C^4$ be a linear operator and let $f(x)$ be a polynomial with complex coefficents. If $c$ is an eigenvalue for $f(A)$, does there exists a eigenvalue $a$ of $A$ such that $f(a) = c$?
Please, explain why this is true or false.
linear-algebra eigenvalues-eigenvectors
asked Nov 26 '18 at 19:42
LinearGuy
1437
marked as duplicate by Arnaud D., Mostafa Ayaz, amWhy linear-algebra
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Dec 14 '18 at 20:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
How to prove “eigenvalues of polynomial of matrix $A$ = polynomial of eigenvalues of matrix $A$ ”
2 answers
Let $A: mathbb C^4 to mathbb C^4$ be a linear operator and let $f(x)$ be a polynomial with complex coefficents. If $c$ is an eigenvalue for $f(A)$, does there exists a eigenvalue $a$ of $A$ such that $f(a) = c$?
Please, explain why this is true or false.
linear-algebra eigenvalues-eigenvectors
asked Nov 26 '18 at 19:42
LinearGuy
1437
marked as duplicate by Arnaud D., Mostafa Ayaz, amWhy linear-algebra
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Dec 14 '18 at 20:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 26 '18 at 19:45
What have you done so far? What is the condition for $c$ to be an eigenvalue of $f(A)$?
– SZN
Nov 26 '18 at 19:47
Hint: Express $A$ in Jordan normal form and then compute the diagonal elements of $f(A)$.
– Connor Harris
Nov 26 '18 at 20:03
add a comment |
This question already has an answer here:
How to prove “eigenvalues of polynomial of matrix $A$ = polynomial of eigenvalues of matrix $A$ ”
2 answers
Let $A: mathbb C^4 to mathbb C^4$ be a linear operator and let $f(x)$ be a polynomial with complex coefficents. If $c$ is an eigenvalue for $f(A)$, does there exists a eigenvalue $a$ of $A$ such that $f(a) = c$?
Please, explain why this is true or false.
linear-algebra eigenvalues-eigenvectors
asked Nov 26 '18 at 19:42
LinearGuy
1437
This question already has an answer here:
How to prove “eigenvalues of polynomial of matrix $A$ = polynomial of eigenvalues of matrix $A$ ”
2 answers
Let $A: mathbb C^4 to mathbb C^4$ be a linear operator and let $f(x)$ be a polynomial with complex coefficents. If $c$ is an eigenvalue for $f(A)$, does there exists a eigenvalue $a$ of $A$ such that $f(a) = c$?
Please, explain why this is true or false.
This question already has an answer here:
How to prove “eigenvalues of polynomial of matrix $A$ = polynomial of eigenvalues of matrix $A$ ”
2 answers
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Nov 26 '18 at 19:42
LinearGuy
1437
asked Nov 26 '18 at 19:42
LinearGuy
1437
asked Nov 26 '18 at 19:42
LinearGuy
1437
asked Nov 26 '18 at 19:42
LinearGuy
1437
asked Nov 26 '18 at 19:42
LinearGuy
1437
1437
marked as duplicate by Arnaud D., Mostafa Ayaz, amWhy linear-algebra
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Dec 14 '18 at 20:21
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Dec 14 '18 at 20:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 26 '18 at 19:45
What have you done so far? What is the condition for $c$ to be an eigenvalue of $f(A)$?
– SZN
Nov 26 '18 at 19:47
Hint: Express $A$ in Jordan normal form and then compute the diagonal elements of $f(A)$.
– Connor Harris
Nov 26 '18 at 20:03
add a comment |
1
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 26 '18 at 19:45
What have you done so far? What is the condition for $c$ to be an eigenvalue of $f(A)$?
– SZN
Nov 26 '18 at 19:47
Hint: Express $A$ in Jordan normal form and then compute the diagonal elements of $f(A)$.
– Connor Harris
Nov 26 '18 at 20:03
1
1
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 26 '18 at 19:45
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 26 '18 at 19:45
What have you done so far? What is the condition for $c$ to be an eigenvalue of $f(A)$?
– SZN
Nov 26 '18 at 19:47
What have you done so far? What is the condition for $c$ to be an eigenvalue of $f(A)$?
– SZN
Nov 26 '18 at 19:47
Hint: Express $A$ in Jordan normal form and then compute the diagonal elements of $f(A)$.
– Connor Harris
Nov 26 '18 at 20:03
Hint: Express $A$ in Jordan normal form and then compute the diagonal elements of $f(A)$.
– Connor Harris
Nov 26 '18 at 20:03
add a comment |
1 Answer
1
active
oldest
votes
The statement is true. One proof is as follows: let $f(z) - c = (z-z_1)cdots(z-z_d)$ be a factorization into linear factors. Each $z_i$ satisfies $f(z_i) = c$. If $f(A)$ has $c$ as an eigenvector, then $f(A) - cI$ is not invertible. Applying our above factorization, this means that the matrix product
$$
(A - z_1 I) cdots (A - z_d I)
$$
fails to be invertible. Thus, $(A - z_i I)$ is non-invertible for some $i$. That is, $z_i$ must be an eigenvector of $A$.
answered Nov 26 '18 at 19:58
Omnomnomnom
126k788176
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The statement is true. One proof is as follows: let $f(z) - c = (z-z_1)cdots(z-z_d)$ be a factorization into linear factors. Each $z_i$ satisfies $f(z_i) = c$. If $f(A)$ has $c$ as an eigenvector, then $f(A) - cI$ is not invertible. Applying our above factorization, this means that the matrix product
$$
(A - z_1 I) cdots (A - z_d I)
$$
fails to be invertible. Thus, $(A - z_i I)$ is non-invertible for some $i$. That is, $z_i$ must be an eigenvector of $A$.
answered Nov 26 '18 at 19:58
Omnomnomnom
126k788176
add a comment |
The statement is true. One proof is as follows: let $f(z) - c = (z-z_1)cdots(z-z_d)$ be a factorization into linear factors. Each $z_i$ satisfies $f(z_i) = c$. If $f(A)$ has $c$ as an eigenvector, then $f(A) - cI$ is not invertible. Applying our above factorization, this means that the matrix product
$$
(A - z_1 I) cdots (A - z_d I)
$$
fails to be invertible. Thus, $(A - z_i I)$ is non-invertible for some $i$. That is, $z_i$ must be an eigenvector of $A$.
answered Nov 26 '18 at 19:58
Omnomnomnom
126k788176
add a comment |
The statement is true. One proof is as follows: let $f(z) - c = (z-z_1)cdots(z-z_d)$ be a factorization into linear factors. Each $z_i$ satisfies $f(z_i) = c$. If $f(A)$ has $c$ as an eigenvector, then $f(A) - cI$ is not invertible. Applying our above factorization, this means that the matrix product
$$
(A - z_1 I) cdots (A - z_d I)
$$
fails to be invertible. Thus, $(A - z_i I)$ is non-invertible for some $i$. That is, $z_i$ must be an eigenvector of $A$.
answered Nov 26 '18 at 19:58
Omnomnomnom
126k788176
The statement is true. One proof is as follows: let $f(z) - c = (z-z_1)cdots(z-z_d)$ be a factorization into linear factors. Each $z_i$ satisfies $f(z_i) = c$. If $f(A)$ has $c$ as an eigenvector, then $f(A) - cI$ is not invertible. Applying our above factorization, this means that the matrix product
$$
(A - z_1 I) cdots (A - z_d I)
$$
fails to be invertible. Thus, $(A - z_i I)$ is non-invertible for some $i$. That is, $z_i$ must be an eigenvector of $A$.
answered Nov 26 '18 at 19:58
Omnomnomnom
126k788176
answered Nov 26 '18 at 19:58
Omnomnomnom
126k788176
answered Nov 26 '18 at 19:58
Omnomnomnom
126k788176
answered Nov 26 '18 at 19:58
Omnomnomnom
126k788176
126k788176
add a comment |
add a comment |
1
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 26 '18 at 19:45
What have you done so far? What is the condition for $c$ to be an eigenvalue of $f(A)$?
– SZN
Nov 26 '18 at 19:47
Hint: Express $A$ in Jordan normal form and then compute the diagonal elements of $f(A)$.
– Connor Harris
Nov 26 '18 at 20:03