Eigenvalues and power of a operator
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If $lambda$ is an eigenvalue of $A^k$, prove that at least one of the $k^{th}$ roots of $lambda$ is an eigenvalue of $A$.
I know it is easy to prove when $A$ can be presented by a matrix, but $A$ is an operator on a complex inner product space, which may be infinite dimensional.
functional-analysis operator-theory
asked Dec 2 '18 at 14:54
ZengZeng
524
$endgroup$
add a comment |
$begingroup$
If $lambda$ is an eigenvalue of $A^k$, prove that at least one of the $k^{th}$ roots of $lambda$ is an eigenvalue of $A$.
I know it is easy to prove when $A$ can be presented by a matrix, but $A$ is an operator on a complex inner product space, which may be infinite dimensional.
functional-analysis operator-theory
asked Dec 2 '18 at 14:54
ZengZeng
524
$endgroup$
$begingroup$
Hint: Being over the field $mathbb{C}$ is critical.
$endgroup$
– qbert
Dec 2 '18 at 15:02
add a comment |
$begingroup$
If $lambda$ is an eigenvalue of $A^k$, prove that at least one of the $k^{th}$ roots of $lambda$ is an eigenvalue of $A$.
I know it is easy to prove when $A$ can be presented by a matrix, but $A$ is an operator on a complex inner product space, which may be infinite dimensional.
functional-analysis operator-theory
asked Dec 2 '18 at 14:54
ZengZeng
524
$endgroup$
If $lambda$ is an eigenvalue of $A^k$, prove that at least one of the $k^{th}$ roots of $lambda$ is an eigenvalue of $A$.
I know it is easy to prove when $A$ can be presented by a matrix, but $A$ is an operator on a complex inner product space, which may be infinite dimensional.
functional-analysis operator-theory
functional-analysis operator-theory
asked Dec 2 '18 at 14:54
ZengZeng
524
asked Dec 2 '18 at 14:54
ZengZeng
524
asked Dec 2 '18 at 14:54
ZengZeng
524
asked Dec 2 '18 at 14:54
ZengZeng
524
asked Dec 2 '18 at 14:54
ZengZeng
524
524
$begingroup$
Hint: Being over the field $mathbb{C}$ is critical.
$endgroup$
– qbert
Dec 2 '18 at 15:02
add a comment |
$begingroup$
Hint: Being over the field $mathbb{C}$ is critical.
$endgroup$
– qbert
Dec 2 '18 at 15:02
$begingroup$
Hint: Being over the field $mathbb{C}$ is critical.
$endgroup$
– qbert
Dec 2 '18 at 15:02
$begingroup$
Hint: Being over the field $mathbb{C}$ is critical.
$endgroup$
– qbert
Dec 2 '18 at 15:02
add a comment |
1 Answer
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$begingroup$
Let $mu_1,...,mu_k$ be the roots of order $k$ of $lambda$.
Clearly $mu_1,...,mu_k$ are solutions for the polynomial $(x^k-lambda)$. In particular we have $x^k-lambda = prod_{l=1}^k (x-mu_l)$.
We conclude that $(A^k-lambdacdot I) = prod_{l=1}^{k} (A-mu^lcdot I)$.
Can you finish now? (If $ker(S)=ker(T)={0}$ what can you say about $ker(ST)$?)
answered Dec 2 '18 at 14:59
YankoYanko
6,6841728
$endgroup$
$begingroup$
Thanks! Ker($ST$) is also {0} so there must be a $mu_l$ s.t. Ker($A-mu_l cdot I$) $neq$ {0}.
$endgroup$
– Zeng
Dec 2 '18 at 15:16
$begingroup$
@Zeng yes, exactly.
$endgroup$
– Yanko
Dec 2 '18 at 15:18
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $mu_1,...,mu_k$ be the roots of order $k$ of $lambda$.
Clearly $mu_1,...,mu_k$ are solutions for the polynomial $(x^k-lambda)$. In particular we have $x^k-lambda = prod_{l=1}^k (x-mu_l)$.
We conclude that $(A^k-lambdacdot I) = prod_{l=1}^{k} (A-mu^lcdot I)$.
Can you finish now? (If $ker(S)=ker(T)={0}$ what can you say about $ker(ST)$?)
answered Dec 2 '18 at 14:59
YankoYanko
6,6841728
$endgroup$
$begingroup$
Thanks! Ker($ST$) is also {0} so there must be a $mu_l$ s.t. Ker($A-mu_l cdot I$) $neq$ {0}.
$endgroup$
– Zeng
Dec 2 '18 at 15:16
$begingroup$
@Zeng yes, exactly.
$endgroup$
– Yanko
Dec 2 '18 at 15:18
add a comment |
$begingroup$
Let $mu_1,...,mu_k$ be the roots of order $k$ of $lambda$.
Clearly $mu_1,...,mu_k$ are solutions for the polynomial $(x^k-lambda)$. In particular we have $x^k-lambda = prod_{l=1}^k (x-mu_l)$.
We conclude that $(A^k-lambdacdot I) = prod_{l=1}^{k} (A-mu^lcdot I)$.
Can you finish now? (If $ker(S)=ker(T)={0}$ what can you say about $ker(ST)$?)
answered Dec 2 '18 at 14:59
YankoYanko
6,6841728
$endgroup$
$begingroup$
Thanks! Ker($ST$) is also {0} so there must be a $mu_l$ s.t. Ker($A-mu_l cdot I$) $neq$ {0}.
$endgroup$
– Zeng
Dec 2 '18 at 15:16
$begingroup$
@Zeng yes, exactly.
$endgroup$
– Yanko
Dec 2 '18 at 15:18
add a comment |
$begingroup$
Let $mu_1,...,mu_k$ be the roots of order $k$ of $lambda$.
Clearly $mu_1,...,mu_k$ are solutions for the polynomial $(x^k-lambda)$. In particular we have $x^k-lambda = prod_{l=1}^k (x-mu_l)$.
We conclude that $(A^k-lambdacdot I) = prod_{l=1}^{k} (A-mu^lcdot I)$.
Can you finish now? (If $ker(S)=ker(T)={0}$ what can you say about $ker(ST)$?)
answered Dec 2 '18 at 14:59
YankoYanko
6,6841728
$endgroup$
Let $mu_1,...,mu_k$ be the roots of order $k$ of $lambda$.
Clearly $mu_1,...,mu_k$ are solutions for the polynomial $(x^k-lambda)$. In particular we have $x^k-lambda = prod_{l=1}^k (x-mu_l)$.
We conclude that $(A^k-lambdacdot I) = prod_{l=1}^{k} (A-mu^lcdot I)$.
Can you finish now? (If $ker(S)=ker(T)={0}$ what can you say about $ker(ST)$?)
answered Dec 2 '18 at 14:59
YankoYanko
6,6841728
edited Dec 2 '18 at 15:08
answered Dec 2 '18 at 14:59
YankoYanko
6,6841728
answered Dec 2 '18 at 14:59
YankoYanko
6,6841728
answered Dec 2 '18 at 14:59
YankoYanko
6,6841728
6,6841728
$begingroup$
Thanks! Ker($ST$) is also {0} so there must be a $mu_l$ s.t. Ker($A-mu_l cdot I$) $neq$ {0}.
$endgroup$
– Zeng
Dec 2 '18 at 15:16
$begingroup$
@Zeng yes, exactly.
$endgroup$
– Yanko
Dec 2 '18 at 15:18
add a comment |
$begingroup$
Thanks! Ker($ST$) is also {0} so there must be a $mu_l$ s.t. Ker($A-mu_l cdot I$) $neq$ {0}.
$endgroup$
– Zeng
Dec 2 '18 at 15:16
$begingroup$
@Zeng yes, exactly.
$endgroup$
– Yanko
Dec 2 '18 at 15:18
$begingroup$
Thanks! Ker($ST$) is also {0} so there must be a $mu_l$ s.t. Ker($A-mu_l cdot I$) $neq$ {0}.
$endgroup$
– Zeng
Dec 2 '18 at 15:16
$begingroup$
Thanks! Ker($ST$) is also {0} so there must be a $mu_l$ s.t. Ker($A-mu_l cdot I$) $neq$ {0}.
$endgroup$
– Zeng
Dec 2 '18 at 15:16
$begingroup$
@Zeng yes, exactly.
$endgroup$
– Yanko
Dec 2 '18 at 15:18
$begingroup$
@Zeng yes, exactly.
$endgroup$
– Yanko
Dec 2 '18 at 15:18
add a comment |
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$begingroup$
Hint: Being over the field $mathbb{C}$ is critical.
$endgroup$
– qbert
Dec 2 '18 at 15:02