$sigma_p(f(A))=f(sigma_p(A))$?
Let $A$ be a bounded linear operator on a Hilbert space, $f$ be a complex polynomial and $sigma_p(A)$ be the point spectrum of $A$ (the set of $lambda$ for which $ker(A-lambda I)neq 0$). I am trying to find a counter example for
$sigma_p(f(A))=f(sigma_p(A))$.
I tried for $A$ the left and right shift operators on $l^2(N)$ with $f=x+1$, and it seems that $sigma_p(f(A))$ and $sigma_p(A)$ are both empty...
Is this correct? If so, how do I more cleverly approach this problem of finding a counter example?
functional-analysis operator-theory spectral-theory
asked Dec 10 '18 at 20:13
Meneer-Beer
1479
add a comment |
Let $A$ be a bounded linear operator on a Hilbert space, $f$ be a complex polynomial and $sigma_p(A)$ be the point spectrum of $A$ (the set of $lambda$ for which $ker(A-lambda I)neq 0$). I am trying to find a counter example for
$sigma_p(f(A))=f(sigma_p(A))$.
I tried for $A$ the left and right shift operators on $l^2(N)$ with $f=x+1$, and it seems that $sigma_p(f(A))$ and $sigma_p(A)$ are both empty...
Is this correct? If so, how do I more cleverly approach this problem of finding a counter example?
functional-analysis operator-theory spectral-theory
asked Dec 10 '18 at 20:13
Meneer-Beer
1479
add a comment |
Let $A$ be a bounded linear operator on a Hilbert space, $f$ be a complex polynomial and $sigma_p(A)$ be the point spectrum of $A$ (the set of $lambda$ for which $ker(A-lambda I)neq 0$). I am trying to find a counter example for
$sigma_p(f(A))=f(sigma_p(A))$.
I tried for $A$ the left and right shift operators on $l^2(N)$ with $f=x+1$, and it seems that $sigma_p(f(A))$ and $sigma_p(A)$ are both empty...
Is this correct? If so, how do I more cleverly approach this problem of finding a counter example?
functional-analysis operator-theory spectral-theory
asked Dec 10 '18 at 20:13
Meneer-Beer
1479
Let $A$ be a bounded linear operator on a Hilbert space, $f$ be a complex polynomial and $sigma_p(A)$ be the point spectrum of $A$ (the set of $lambda$ for which $ker(A-lambda I)neq 0$). I am trying to find a counter example for
$sigma_p(f(A))=f(sigma_p(A))$.
I tried for $A$ the left and right shift operators on $l^2(N)$ with $f=x+1$, and it seems that $sigma_p(f(A))$ and $sigma_p(A)$ are both empty...
Is this correct? If so, how do I more cleverly approach this problem of finding a counter example?
functional-analysis operator-theory spectral-theory
functional-analysis operator-theory spectral-theory
asked Dec 10 '18 at 20:13
Meneer-Beer
1479
asked Dec 10 '18 at 20:13
Meneer-Beer
1479
asked Dec 10 '18 at 20:13
Meneer-Beer
1479
asked Dec 10 '18 at 20:13
Meneer-Beer
1479
asked Dec 10 '18 at 20:13
Meneer-Beer
1479
1479
add a comment |
add a comment |
1 Answer
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Let $f$ be a constant polynomial. Then $f(A) = c I$ for some scalar $c$, so $sigma_p(f(A)) = {c}$. But
$sigma_p(A)$ might be empty.
But these are the only counterexamples.
If $lambda in sigma_p(A)$, there is nonzero $v in H$ such that
$A v = lambda v$, and then $f(A) v = f(lambda) v$, so $p(lambda) in sigma_p(f(A))$. Thus $f(sigma_p(A)) subseteq sigma_p(f(A))$.
On the other hand, suppose $lambda in sigma_p(f(A))$.
Thus there is nonzero $v in H$ such that $f(A) v = lambda v$.
If $f$ has degree $d > 0$, the polynomial $f(z) - lambda$ can be factored over the complex numbers as
$$f(z) - lambda = prod_{j=1}^d (z - alpha_j)$$
so that
$$ 0 = (f(A) - lambda) v = prod_{j=1}^d (A - alpha_j I) v $$
For some $k$, $1 le k le d$, we must have $prod_{j=k}^d (A - alpha_j I)v = 0$ but $prod_{j=k+1}^d (A - alpha_j I) v ne 0$, i.e. with $w = prod_{j=k+1}^d (A - alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $wne 0$ and
$(A - alpha_k I) w = 0$, so $alpha_k in sigma_p(A)$, and
$f(alpha_k) = lambda$. So $sigma_p(f(A)) subseteq f(sigma_p(A))$.
answered Dec 10 '18 at 20:33
Robert Israel
318k23208457
add a comment |
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Let $f$ be a constant polynomial. Then $f(A) = c I$ for some scalar $c$, so $sigma_p(f(A)) = {c}$. But
$sigma_p(A)$ might be empty.
But these are the only counterexamples.
If $lambda in sigma_p(A)$, there is nonzero $v in H$ such that
$A v = lambda v$, and then $f(A) v = f(lambda) v$, so $p(lambda) in sigma_p(f(A))$. Thus $f(sigma_p(A)) subseteq sigma_p(f(A))$.
On the other hand, suppose $lambda in sigma_p(f(A))$.
Thus there is nonzero $v in H$ such that $f(A) v = lambda v$.
If $f$ has degree $d > 0$, the polynomial $f(z) - lambda$ can be factored over the complex numbers as
$$f(z) - lambda = prod_{j=1}^d (z - alpha_j)$$
so that
$$ 0 = (f(A) - lambda) v = prod_{j=1}^d (A - alpha_j I) v $$
For some $k$, $1 le k le d$, we must have $prod_{j=k}^d (A - alpha_j I)v = 0$ but $prod_{j=k+1}^d (A - alpha_j I) v ne 0$, i.e. with $w = prod_{j=k+1}^d (A - alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $wne 0$ and
$(A - alpha_k I) w = 0$, so $alpha_k in sigma_p(A)$, and
$f(alpha_k) = lambda$. So $sigma_p(f(A)) subseteq f(sigma_p(A))$.
answered Dec 10 '18 at 20:33
Robert Israel
318k23208457
add a comment |
Let $f$ be a constant polynomial. Then $f(A) = c I$ for some scalar $c$, so $sigma_p(f(A)) = {c}$. But
$sigma_p(A)$ might be empty.
But these are the only counterexamples.
If $lambda in sigma_p(A)$, there is nonzero $v in H$ such that
$A v = lambda v$, and then $f(A) v = f(lambda) v$, so $p(lambda) in sigma_p(f(A))$. Thus $f(sigma_p(A)) subseteq sigma_p(f(A))$.
On the other hand, suppose $lambda in sigma_p(f(A))$.
Thus there is nonzero $v in H$ such that $f(A) v = lambda v$.
If $f$ has degree $d > 0$, the polynomial $f(z) - lambda$ can be factored over the complex numbers as
$$f(z) - lambda = prod_{j=1}^d (z - alpha_j)$$
so that
$$ 0 = (f(A) - lambda) v = prod_{j=1}^d (A - alpha_j I) v $$
For some $k$, $1 le k le d$, we must have $prod_{j=k}^d (A - alpha_j I)v = 0$ but $prod_{j=k+1}^d (A - alpha_j I) v ne 0$, i.e. with $w = prod_{j=k+1}^d (A - alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $wne 0$ and
$(A - alpha_k I) w = 0$, so $alpha_k in sigma_p(A)$, and
$f(alpha_k) = lambda$. So $sigma_p(f(A)) subseteq f(sigma_p(A))$.
answered Dec 10 '18 at 20:33
Robert Israel
318k23208457
add a comment |
Let $f$ be a constant polynomial. Then $f(A) = c I$ for some scalar $c$, so $sigma_p(f(A)) = {c}$. But
$sigma_p(A)$ might be empty.
But these are the only counterexamples.
If $lambda in sigma_p(A)$, there is nonzero $v in H$ such that
$A v = lambda v$, and then $f(A) v = f(lambda) v$, so $p(lambda) in sigma_p(f(A))$. Thus $f(sigma_p(A)) subseteq sigma_p(f(A))$.
On the other hand, suppose $lambda in sigma_p(f(A))$.
Thus there is nonzero $v in H$ such that $f(A) v = lambda v$.
If $f$ has degree $d > 0$, the polynomial $f(z) - lambda$ can be factored over the complex numbers as
$$f(z) - lambda = prod_{j=1}^d (z - alpha_j)$$
so that
$$ 0 = (f(A) - lambda) v = prod_{j=1}^d (A - alpha_j I) v $$
For some $k$, $1 le k le d$, we must have $prod_{j=k}^d (A - alpha_j I)v = 0$ but $prod_{j=k+1}^d (A - alpha_j I) v ne 0$, i.e. with $w = prod_{j=k+1}^d (A - alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $wne 0$ and
$(A - alpha_k I) w = 0$, so $alpha_k in sigma_p(A)$, and
$f(alpha_k) = lambda$. So $sigma_p(f(A)) subseteq f(sigma_p(A))$.
answered Dec 10 '18 at 20:33
Robert Israel
318k23208457
Let $f$ be a constant polynomial. Then $f(A) = c I$ for some scalar $c$, so $sigma_p(f(A)) = {c}$. But
$sigma_p(A)$ might be empty.
But these are the only counterexamples.
If $lambda in sigma_p(A)$, there is nonzero $v in H$ such that
$A v = lambda v$, and then $f(A) v = f(lambda) v$, so $p(lambda) in sigma_p(f(A))$. Thus $f(sigma_p(A)) subseteq sigma_p(f(A))$.
On the other hand, suppose $lambda in sigma_p(f(A))$.
Thus there is nonzero $v in H$ such that $f(A) v = lambda v$.
If $f$ has degree $d > 0$, the polynomial $f(z) - lambda$ can be factored over the complex numbers as
$$f(z) - lambda = prod_{j=1}^d (z - alpha_j)$$
so that
$$ 0 = (f(A) - lambda) v = prod_{j=1}^d (A - alpha_j I) v $$
For some $k$, $1 le k le d$, we must have $prod_{j=k}^d (A - alpha_j I)v = 0$ but $prod_{j=k+1}^d (A - alpha_j I) v ne 0$, i.e. with $w = prod_{j=k+1}^d (A - alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $wne 0$ and
$(A - alpha_k I) w = 0$, so $alpha_k in sigma_p(A)$, and
$f(alpha_k) = lambda$. So $sigma_p(f(A)) subseteq f(sigma_p(A))$.
answered Dec 10 '18 at 20:33
Robert Israel
318k23208457
edited Dec 10 '18 at 20:40
answered Dec 10 '18 at 20:33
Robert Israel
318k23208457
answered Dec 10 '18 at 20:33
Robert Israel
318k23208457
answered Dec 10 '18 at 20:33
Robert Israel
318k23208457
318k23208457
add a comment |
add a comment |
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