Eigenvalue of polynomialsSuppose $T^2$ is diagonalizable and $kerT=0$, and every eigenvalue of $T^2$ is nonnegative. Show that $T$ is diagonalizable.prove that two linear maps over a finite dimensional vector space are conjugateConsider a linear operator $L$ and some polynomial of it, $L'=p(L)$. Show that the minimal polynomial of $L'$ has smaller degree than that of $L$.Eigenvalue for a conjugate operator.Finding the minimal polynomial of a linear operatorProve that operator L on $M_n(mathbb F)$ is diagonalizableRelation between left and right eigenvectors corresponding to the same eigenvalueIf $f$ is diagonalisable then its minimal polynomial is the product of distinct linear factorsExamples of real $2times2$ and complex $3times3$ matrices with minimal polynomial $t^2+1$Square of spectral radius and Frobenius norm
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Eigenvalue of polynomials
Suppose $T^2$ is diagonalizable and $kerT=0$, and every eigenvalue of $T^2$ is nonnegative. Show that $T$ is diagonalizable.prove that two linear maps over a finite dimensional vector space are conjugateConsider a linear operator $L$ and some polynomial of it, $L'=p(L)$. Show that the minimal polynomial of $L'$ has smaller degree than that of $L$.Eigenvalue for a conjugate operator.Finding the minimal polynomial of a linear operatorProve that operator L on $M_n(mathbb F)$ is diagonalizableRelation between left and right eigenvectors corresponding to the same eigenvalueIf $f$ is diagonalisable then its minimal polynomial is the product of distinct linear factorsExamples of real $2times2$ and complex $3times3$ matrices with minimal polynomial $t^2+1$Square of spectral radius and Frobenius norm
$begingroup$
Let $ P in mathbbF[X] $ a polynomial, $T:V to V$ a linear operator..
Prove or disprove:
$ lambda $ is an eigenvalue of $T$ iff $ P(lambda)$ is an eigenvalue of $P(T)$.
Given that $ lambda $ is an eigenvalue of $T$, its quite easy to prove the second half, however, given the second half gives me nothing to work with to prove the first half, so I'll assume its not correct.
What would be a way to construct a sufficient counterexample? Every polynomial I can find shows that this claim is true, but I can't manage to prove it. Could I somehow use the minimal polynomial as an example perhaps?
linear-algebra
asked May 1 at 19:17
TegernakoTegernako
1158
$endgroup$
add a comment |
$begingroup$
Let $ P in mathbbF[X] $ a polynomial, $T:V to V$ a linear operator..
Prove or disprove:
$ lambda $ is an eigenvalue of $T$ iff $ P(lambda)$ is an eigenvalue of $P(T)$.
Given that $ lambda $ is an eigenvalue of $T$, its quite easy to prove the second half, however, given the second half gives me nothing to work with to prove the first half, so I'll assume its not correct.
What would be a way to construct a sufficient counterexample? Every polynomial I can find shows that this claim is true, but I can't manage to prove it. Could I somehow use the minimal polynomial as an example perhaps?
linear-algebra
asked May 1 at 19:17
TegernakoTegernako
1158
$endgroup$
$begingroup$
Let $P$ be the zero polynomial ...
$endgroup$
– Hagen von Eitzen
May 1 at 19:43
add a comment |
$begingroup$
Let $ P in mathbbF[X] $ a polynomial, $T:V to V$ a linear operator..
Prove or disprove:
$ lambda $ is an eigenvalue of $T$ iff $ P(lambda)$ is an eigenvalue of $P(T)$.
Given that $ lambda $ is an eigenvalue of $T$, its quite easy to prove the second half, however, given the second half gives me nothing to work with to prove the first half, so I'll assume its not correct.
What would be a way to construct a sufficient counterexample? Every polynomial I can find shows that this claim is true, but I can't manage to prove it. Could I somehow use the minimal polynomial as an example perhaps?
linear-algebra
asked May 1 at 19:17
TegernakoTegernako
1158
$endgroup$
Let $ P in mathbbF[X] $ a polynomial, $T:V to V$ a linear operator..
Prove or disprove:
$ lambda $ is an eigenvalue of $T$ iff $ P(lambda)$ is an eigenvalue of $P(T)$.
Given that $ lambda $ is an eigenvalue of $T$, its quite easy to prove the second half, however, given the second half gives me nothing to work with to prove the first half, so I'll assume its not correct.
What would be a way to construct a sufficient counterexample? Every polynomial I can find shows that this claim is true, but I can't manage to prove it. Could I somehow use the minimal polynomial as an example perhaps?
linear-algebra
linear-algebra
asked May 1 at 19:17
TegernakoTegernako
1158
asked May 1 at 19:17
TegernakoTegernako
1158
asked May 1 at 19:17
TegernakoTegernako
1158
asked May 1 at 19:17
TegernakoTegernako
1158
asked May 1 at 19:17
TegernakoTegernako
1158
1158
$begingroup$
Let $P$ be the zero polynomial ...
$endgroup$
– Hagen von Eitzen
May 1 at 19:43
add a comment |
$begingroup$
Let $P$ be the zero polynomial ...
$endgroup$
– Hagen von Eitzen
May 1 at 19:43
$begingroup$
Let $P$ be the zero polynomial ...
$endgroup$
– Hagen von Eitzen
May 1 at 19:43
$begingroup$
Let $P$ be the zero polynomial ...
$endgroup$
– Hagen von Eitzen
May 1 at 19:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $T$ be the identity, and let $P(x)=x^2$.
Then $P(-1)=1$ is an eigenvalue of $T^2$, but $-1$ is not an eigenvalue of $T$.
answered May 1 at 19:26
TonyKTonyK
44.5k358137
$endgroup$
add a comment |
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$begingroup$
Let $T$ be the identity, and let $P(x)=x^2$.
Then $P(-1)=1$ is an eigenvalue of $T^2$, but $-1$ is not an eigenvalue of $T$.
answered May 1 at 19:26
TonyKTonyK
44.5k358137
$endgroup$
add a comment |
$begingroup$
Let $T$ be the identity, and let $P(x)=x^2$.
Then $P(-1)=1$ is an eigenvalue of $T^2$, but $-1$ is not an eigenvalue of $T$.
answered May 1 at 19:26
TonyKTonyK
44.5k358137
$endgroup$
add a comment |
$begingroup$
Let $T$ be the identity, and let $P(x)=x^2$.
Then $P(-1)=1$ is an eigenvalue of $T^2$, but $-1$ is not an eigenvalue of $T$.
answered May 1 at 19:26
TonyKTonyK
44.5k358137
$endgroup$
Let $T$ be the identity, and let $P(x)=x^2$.
Then $P(-1)=1$ is an eigenvalue of $T^2$, but $-1$ is not an eigenvalue of $T$.
answered May 1 at 19:26
TonyKTonyK
44.5k358137
answered May 1 at 19:26
TonyKTonyK
44.5k358137
answered May 1 at 19:26
TonyKTonyK
44.5k358137
answered May 1 at 19:26
TonyKTonyK
44.5k358137
44.5k358137
add a comment |
add a comment |
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$begingroup$
Let $P$ be the zero polynomial ...
$endgroup$
– Hagen von Eitzen
May 1 at 19:43