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can a Linear operator on infinite dimensional vector space have infinite eigenvalues?

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1

1

$begingroup$

I know that for a finite dimensional vector space, the number of eigenvalues is at most the dimension of the vector space. Is there an example of an infinite dimensional vector space and a linear operator such that the linear operator allows infinite eigenvalues? I cannot think of such an example, but I was wondering if it is possible to construct one since I was also unable to prove why the number of eigenvalues would necessarily be finite in such a case.

linear-algebra vector-spaces eigenvalues-eigenvectors

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asked Mar 22 at 14:48

mmmmommmmo

1347

$endgroup$

add a comment | 

1

1

$begingroup$

I know that for a finite dimensional vector space, the number of eigenvalues is at most the dimension of the vector space. Is there an example of an infinite dimensional vector space and a linear operator such that the linear operator allows infinite eigenvalues? I cannot think of such an example, but I was wondering if it is possible to construct one since I was also unable to prove why the number of eigenvalues would necessarily be finite in such a case.

linear-algebra vector-spaces eigenvalues-eigenvectors

share|cite|improve this question

asked Mar 22 at 14:48

mmmmommmmo

1347

$endgroup$

add a comment | 

$begingroup$

I know that for a finite dimensional vector space, the number of eigenvalues is at most the dimension of the vector space. Is there an example of an infinite dimensional vector space and a linear operator such that the linear operator allows infinite eigenvalues? I cannot think of such an example, but I was wondering if it is possible to construct one since I was also unable to prove why the number of eigenvalues would necessarily be finite in such a case.

linear-algebra vector-spaces eigenvalues-eigenvectors

share|cite|improve this question

asked Mar 22 at 14:48

mmmmommmmo

1347

$endgroup$

share|cite|improve this question

asked Mar 22 at 14:48

mmmmommmmo

1347

share|cite|improve this question

asked Mar 22 at 14:48

mmmmommmmo

1347

asked Mar 22 at 14:48

mmmmommmmo

1347

asked Mar 22 at 14:48

mmmmommmmo

1347

3 Answers
3

active

oldest

votes

3

$begingroup$

Of course there is, as a linear map is given by the images of a basis.

So, let $(B_i mid i in mathbbN)$ be a basis of an infinite dimensional $mathbbR$-vector space $V$. Define a map $f : V to V$ by putting $f(B_i) = icdot B_i$ and you have infinitely many eigenvalues.

share|cite|improve this answer

answered Mar 22 at 14:53

DirkDirk

4,533219

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I quite like this example, elegant and quite obvious
  $endgroup$
  – mmmmo
  Mar 22 at 15:08
  

  
  
  
  

add a comment | 

2

$begingroup$

In $ell^2$, let$$A(x_1,x_2,x_3,x_4,ldots)=left(x_1,fracx_22,fracx_33,ldotsright).$$Then, for each $ninmathbb N$, $frac1n$ is an eigenvalue of $A$.

share|cite|improve this answer

answered Mar 22 at 14:52

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

2

$begingroup$

Another example from a different area just for the sake of building intuition. If you let $C^1(mathbbR)$ be the vector space of continuously differentiable functions $f:mathbbR to mathbbR$, and let $L$ be the derivative operator, mapping $L(f) = f'$, then the eigenvalues of $L$ satisfy
$$
kf = L(f) = f',
$$
in other words any real $k$ is an eigenvalue with the eigenvector
$$
f(x) = e^kx.
$$

So here is an example with not just infinite, but uncountable eigenvectors/eigenvalues.

share|cite|improve this answer

answered Mar 22 at 15:06

gt6989bgt6989b

35.6k22557

$endgroup$

add a comment | 

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3

$begingroup$

Of course there is, as a linear map is given by the images of a basis.

So, let $(B_i mid i in mathbbN)$ be a basis of an infinite dimensional $mathbbR$-vector space $V$. Define a map $f : V to V$ by putting $f(B_i) = icdot B_i$ and you have infinitely many eigenvalues.

share|cite|improve this answer

answered Mar 22 at 14:53

DirkDirk

4,533219

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I quite like this example, elegant and quite obvious
  $endgroup$
  – mmmmo
  Mar 22 at 15:08
  

  
  
  
  

add a comment | 

3

$begingroup$

Of course there is, as a linear map is given by the images of a basis.

So, let $(B_i mid i in mathbbN)$ be a basis of an infinite dimensional $mathbbR$-vector space $V$. Define a map $f : V to V$ by putting $f(B_i) = icdot B_i$ and you have infinitely many eigenvalues.

share|cite|improve this answer

answered Mar 22 at 14:53

DirkDirk

4,533219

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I quite like this example, elegant and quite obvious
  $endgroup$
  – mmmmo
  Mar 22 at 15:08
  

  
  
  
  

add a comment | 

$begingroup$

Of course there is, as a linear map is given by the images of a basis.

So, let $(B_i mid i in mathbbN)$ be a basis of an infinite dimensional $mathbbR$-vector space $V$. Define a map $f : V to V$ by putting $f(B_i) = icdot B_i$ and you have infinitely many eigenvalues.

share|cite|improve this answer

answered Mar 22 at 14:53

DirkDirk

4,533219

$endgroup$

share|cite|improve this answer

answered Mar 22 at 14:53

DirkDirk

4,533219

answered Mar 22 at 14:53

DirkDirk

4,533219

answered Mar 22 at 14:53

DirkDirk

4,533219

2

$begingroup$

In $ell^2$, let$$A(x_1,x_2,x_3,x_4,ldots)=left(x_1,fracx_22,fracx_33,ldotsright).$$Then, for each $ninmathbb N$, $frac1n$ is an eigenvalue of $A$.

share|cite|improve this answer

answered Mar 22 at 14:52

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

2

$begingroup$

In $ell^2$, let$$A(x_1,x_2,x_3,x_4,ldots)=left(x_1,fracx_22,fracx_33,ldotsright).$$Then, for each $ninmathbb N$, $frac1n$ is an eigenvalue of $A$.

share|cite|improve this answer

answered Mar 22 at 14:52

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

$begingroup$

In $ell^2$, let$$A(x_1,x_2,x_3,x_4,ldots)=left(x_1,fracx_22,fracx_33,ldotsright).$$Then, for each $ninmathbb N$, $frac1n$ is an eigenvalue of $A$.

share|cite|improve this answer

answered Mar 22 at 14:52

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

share|cite|improve this answer

answered Mar 22 at 14:52

José Carlos SantosJosé Carlos Santos

173k23133241

answered Mar 22 at 14:52

José Carlos SantosJosé Carlos Santos

173k23133241

answered Mar 22 at 14:52

José Carlos SantosJosé Carlos Santos

173k23133241

2

$begingroup$

Another example from a different area just for the sake of building intuition. If you let $C^1(mathbbR)$ be the vector space of continuously differentiable functions $f:mathbbR to mathbbR$, and let $L$ be the derivative operator, mapping $L(f) = f'$, then the eigenvalues of $L$ satisfy
$$
kf = L(f) = f',
$$
in other words any real $k$ is an eigenvalue with the eigenvector
$$
f(x) = e^kx.
$$

So here is an example with not just infinite, but uncountable eigenvectors/eigenvalues.

share|cite|improve this answer

answered Mar 22 at 15:06

gt6989bgt6989b

35.6k22557

$endgroup$

add a comment | 

2

$begingroup$

Another example from a different area just for the sake of building intuition. If you let $C^1(mathbbR)$ be the vector space of continuously differentiable functions $f:mathbbR to mathbbR$, and let $L$ be the derivative operator, mapping $L(f) = f'$, then the eigenvalues of $L$ satisfy
$$
kf = L(f) = f',
$$
in other words any real $k$ is an eigenvalue with the eigenvector
$$
f(x) = e^kx.
$$

So here is an example with not just infinite, but uncountable eigenvectors/eigenvalues.

share|cite|improve this answer

answered Mar 22 at 15:06

gt6989bgt6989b

35.6k22557

$endgroup$

add a comment | 

$begingroup$

Another example from a different area just for the sake of building intuition. If you let $C^1(mathbbR)$ be the vector space of continuously differentiable functions $f:mathbbR to mathbbR$, and let $L$ be the derivative operator, mapping $L(f) = f'$, then the eigenvalues of $L$ satisfy
$$
kf = L(f) = f',
$$
in other words any real $k$ is an eigenvalue with the eigenvector
$$
f(x) = e^kx.
$$

So here is an example with not just infinite, but uncountable eigenvectors/eigenvalues.

share|cite|improve this answer

answered Mar 22 at 15:06

gt6989bgt6989b

35.6k22557

$endgroup$

share|cite|improve this answer

answered Mar 22 at 15:06

gt6989bgt6989b

35.6k22557

answered Mar 22 at 15:06

gt6989bgt6989b

35.6k22557

answered Mar 22 at 15:06

gt6989bgt6989b

35.6k22557