Does $D$ have real eigenvalues?
Let $V$ be the subspace of the real vector space of real valued functions
on $mathbb{R}$, spanned by $cos t$ and $sin t$. Let $D : V to V$ be the linear map sending $f(t) in V $ to $dfrac{ df(t)}{dt}$.
Prove that $D$ has a real eigenvalue.
What happens if we consider instead the complex vector space of complex valued functions? Does it have an eigenvalue?
My attempt: (I claim 1 is false)
Suppose $D$ has an eigenvalue.
Let $v$ be an eigenvector $v=a sin t + b cos t$
then by using definition I get
$$a(1-lambda) sin t = (1+ lambda) b cos t$$
This holds for all $t$, $tinmathbb{R}$. Put $t=0$, $b=0$ or $lambda=-1$ and both don't work. So it has no eigenvalue.
But I think my answer to 1 is concrete, I want to see a intuitive and rigorous proof.
And for 2, I know every operator over complex field has eigenvalue; is it real in this case?
linear-algebra eigenvalues-eigenvectors linear-transformations
edited Nov 24 at 17:33
José Carlos Santos
148k22117218
asked Nov 24 at 15:58
Cloud JR
806417
|
show 1 more comment
Let $V$ be the subspace of the real vector space of real valued functions
on $mathbb{R}$, spanned by $cos t$ and $sin t$. Let $D : V to V$ be the linear map sending $f(t) in V $ to $dfrac{ df(t)}{dt}$.
Prove that $D$ has a real eigenvalue.
What happens if we consider instead the complex vector space of complex valued functions? Does it have an eigenvalue?
My attempt: (I claim 1 is false)
Suppose $D$ has an eigenvalue.
Let $v$ be an eigenvector $v=a sin t + b cos t$
then by using definition I get
$$a(1-lambda) sin t = (1+ lambda) b cos t$$
This holds for all $t$, $tinmathbb{R}$. Put $t=0$, $b=0$ or $lambda=-1$ and both don't work. So it has no eigenvalue.
But I think my answer to 1 is concrete, I want to see a intuitive and rigorous proof.
And for 2, I know every operator over complex field has eigenvalue; is it real in this case?
linear-algebra eigenvalues-eigenvectors linear-transformations
edited Nov 24 at 17:33
José Carlos Santos
148k22117218
asked Nov 24 at 15:58
Cloud JR
806417
Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
– Yadati Kiran
Nov 24 at 16:18
Thanks it completes 2
– Cloud JR
Nov 24 at 16:24
By the way it is possible with $a sin t + b cos t$ also. Try it.
– Yadati Kiran
Nov 24 at 16:25
If scalars are real then we dont get eigen value
– Cloud JR
Nov 24 at 16:28
Yeah but you have to show it.
– Yadati Kiran
Nov 24 at 16:28
|
show 1 more comment
Let $V$ be the subspace of the real vector space of real valued functions
on $mathbb{R}$, spanned by $cos t$ and $sin t$. Let $D : V to V$ be the linear map sending $f(t) in V $ to $dfrac{ df(t)}{dt}$.
Prove that $D$ has a real eigenvalue.
What happens if we consider instead the complex vector space of complex valued functions? Does it have an eigenvalue?
My attempt: (I claim 1 is false)
Suppose $D$ has an eigenvalue.
Let $v$ be an eigenvector $v=a sin t + b cos t$
then by using definition I get
$$a(1-lambda) sin t = (1+ lambda) b cos t$$
This holds for all $t$, $tinmathbb{R}$. Put $t=0$, $b=0$ or $lambda=-1$ and both don't work. So it has no eigenvalue.
But I think my answer to 1 is concrete, I want to see a intuitive and rigorous proof.
And for 2, I know every operator over complex field has eigenvalue; is it real in this case?
linear-algebra eigenvalues-eigenvectors linear-transformations
edited Nov 24 at 17:33
José Carlos Santos
148k22117218
asked Nov 24 at 15:58
Cloud JR
806417
Let $V$ be the subspace of the real vector space of real valued functions
on $mathbb{R}$, spanned by $cos t$ and $sin t$. Let $D : V to V$ be the linear map sending $f(t) in V $ to $dfrac{ df(t)}{dt}$.
Prove that $D$ has a real eigenvalue.
What happens if we consider instead the complex vector space of complex valued functions? Does it have an eigenvalue?
My attempt: (I claim 1 is false)
Suppose $D$ has an eigenvalue.
Let $v$ be an eigenvector $v=a sin t + b cos t$
then by using definition I get
$$a(1-lambda) sin t = (1+ lambda) b cos t$$
This holds for all $t$, $tinmathbb{R}$. Put $t=0$, $b=0$ or $lambda=-1$ and both don't work. So it has no eigenvalue.
But I think my answer to 1 is concrete, I want to see a intuitive and rigorous proof.
And for 2, I know every operator over complex field has eigenvalue; is it real in this case?
linear-algebra eigenvalues-eigenvectors linear-transformations
linear-algebra eigenvalues-eigenvectors linear-transformations
edited Nov 24 at 17:33
José Carlos Santos
148k22117218
asked Nov 24 at 15:58
Cloud JR
806417
edited Nov 24 at 17:33
José Carlos Santos
148k22117218
asked Nov 24 at 15:58
Cloud JR
806417
edited Nov 24 at 17:33
José Carlos Santos
148k22117218
edited Nov 24 at 17:33
José Carlos Santos
148k22117218
edited Nov 24 at 17:33
José Carlos Santos
148k22117218
148k22117218
asked Nov 24 at 15:58
Cloud JR
806417
asked Nov 24 at 15:58
Cloud JR
806417
asked Nov 24 at 15:58
Cloud JR
806417
806417
Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
– Yadati Kiran
Nov 24 at 16:18
Thanks it completes 2
– Cloud JR
Nov 24 at 16:24
By the way it is possible with $a sin t + b cos t$ also. Try it.
– Yadati Kiran
Nov 24 at 16:25
If scalars are real then we dont get eigen value
– Cloud JR
Nov 24 at 16:28
Yeah but you have to show it.
– Yadati Kiran
Nov 24 at 16:28
|
show 1 more comment
Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
– Yadati Kiran
Nov 24 at 16:18
Thanks it completes 2
– Cloud JR
Nov 24 at 16:24
By the way it is possible with $a sin t + b cos t$ also. Try it.
– Yadati Kiran
Nov 24 at 16:25
If scalars are real then we dont get eigen value
– Cloud JR
Nov 24 at 16:28
Yeah but you have to show it.
– Yadati Kiran
Nov 24 at 16:28
Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
– Yadati Kiran
Nov 24 at 16:18
Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
– Yadati Kiran
Nov 24 at 16:18
Thanks it completes 2
– Cloud JR
Nov 24 at 16:24
Thanks it completes 2
– Cloud JR
Nov 24 at 16:24
By the way it is possible with $a sin t + b cos t$ also. Try it.
– Yadati Kiran
Nov 24 at 16:25
By the way it is possible with $a sin t + b cos t$ also. Try it.
– Yadati Kiran
Nov 24 at 16:25
If scalars are real then we dont get eigen value
– Cloud JR
Nov 24 at 16:28
If scalars are real then we dont get eigen value
– Cloud JR
Nov 24 at 16:28
Yeah but you have to show it.
– Yadati Kiran
Nov 24 at 16:28
Yeah but you have to show it.
– Yadati Kiran
Nov 24 at 16:28
|
show 1 more comment
2 Answers
2
active
oldest
votes
Let $mathscr{B}={cos t,sin t}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is
begin{bmatrix}
0 & -1 \
1 & 0
end{bmatrix}
Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.
answered Nov 24 at 16:50
egreg
177k1484198
This helps a lot ... Thanks
– Cloud JR
Nov 24 at 17:30
add a comment |
Your answer to 1) is fine. You can also say that if $fcolonmathbb{R}longrightarrowmathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $kinmathbb R$. SUch a function doesn't belong to your space.
For 2), use the fact that, if $f(t)=cos(t)+sin(t)i$, then$$f'(t)=-sin(t)+cos(t)i=if(t).$$
answered Nov 24 at 16:47
José Carlos Santos
148k22117218
add a comment |
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2 Answers
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2 Answers
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active
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Let $mathscr{B}={cos t,sin t}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is
begin{bmatrix}
0 & -1 \
1 & 0
end{bmatrix}
Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.
answered Nov 24 at 16:50
egreg
177k1484198
This helps a lot ... Thanks
– Cloud JR
Nov 24 at 17:30
add a comment |
Let $mathscr{B}={cos t,sin t}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is
begin{bmatrix}
0 & -1 \
1 & 0
end{bmatrix}
Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.
answered Nov 24 at 16:50
egreg
177k1484198
This helps a lot ... Thanks
– Cloud JR
Nov 24 at 17:30
add a comment |
Let $mathscr{B}={cos t,sin t}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is
begin{bmatrix}
0 & -1 \
1 & 0
end{bmatrix}
Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.
answered Nov 24 at 16:50
egreg
177k1484198
Let $mathscr{B}={cos t,sin t}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is
begin{bmatrix}
0 & -1 \
1 & 0
end{bmatrix}
Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.
answered Nov 24 at 16:50
egreg
177k1484198
answered Nov 24 at 16:50
egreg
177k1484198
answered Nov 24 at 16:50
egreg
177k1484198
answered Nov 24 at 16:50
egreg
177k1484198
177k1484198
This helps a lot ... Thanks
– Cloud JR
Nov 24 at 17:30
add a comment |
This helps a lot ... Thanks
– Cloud JR
Nov 24 at 17:30
This helps a lot ... Thanks
– Cloud JR
Nov 24 at 17:30
This helps a lot ... Thanks
– Cloud JR
Nov 24 at 17:30
add a comment |
Your answer to 1) is fine. You can also say that if $fcolonmathbb{R}longrightarrowmathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $kinmathbb R$. SUch a function doesn't belong to your space.
For 2), use the fact that, if $f(t)=cos(t)+sin(t)i$, then$$f'(t)=-sin(t)+cos(t)i=if(t).$$
answered Nov 24 at 16:47
José Carlos Santos
148k22117218
add a comment |
Your answer to 1) is fine. You can also say that if $fcolonmathbb{R}longrightarrowmathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $kinmathbb R$. SUch a function doesn't belong to your space.
For 2), use the fact that, if $f(t)=cos(t)+sin(t)i$, then$$f'(t)=-sin(t)+cos(t)i=if(t).$$
answered Nov 24 at 16:47
José Carlos Santos
148k22117218
add a comment |
Your answer to 1) is fine. You can also say that if $fcolonmathbb{R}longrightarrowmathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $kinmathbb R$. SUch a function doesn't belong to your space.
For 2), use the fact that, if $f(t)=cos(t)+sin(t)i$, then$$f'(t)=-sin(t)+cos(t)i=if(t).$$
answered Nov 24 at 16:47
José Carlos Santos
148k22117218
Your answer to 1) is fine. You can also say that if $fcolonmathbb{R}longrightarrowmathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $kinmathbb R$. SUch a function doesn't belong to your space.
For 2), use the fact that, if $f(t)=cos(t)+sin(t)i$, then$$f'(t)=-sin(t)+cos(t)i=if(t).$$
answered Nov 24 at 16:47
José Carlos Santos
148k22117218
answered Nov 24 at 16:47
José Carlos Santos
148k22117218
answered Nov 24 at 16:47
José Carlos Santos
148k22117218
answered Nov 24 at 16:47
José Carlos Santos
148k22117218
148k22117218
add a comment |
add a comment |
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Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
– Yadati Kiran
Nov 24 at 16:18
Thanks it completes 2
– Cloud JR
Nov 24 at 16:24
By the way it is possible with $a sin t + b cos t$ also. Try it.
– Yadati Kiran
Nov 24 at 16:25
If scalars are real then we dont get eigen value
– Cloud JR
Nov 24 at 16:28
Yeah but you have to show it.
– Yadati Kiran
Nov 24 at 16:28