|
|
| View previous topic :: View next topic |
| Author |
Message |
MCT
Guest
|
 Posted: Thu May 15, 2008 11:03 am Post subject: Linear algebra problem |
 |
|
Hi,
A friend of mine gave me this problem and I can't seem to find a way to
solve it:
Let V be a real finite-dimensional inner-product space. Let S be an
isometry on V. If S^2+aS+bI is not invertible, then b=1.
Regards,
Mats |
|
| Back to top |
|
 |
| |
Ads |
Advertising
Sponsor
|
|
MCT
Guest
|
| Posted: Thu May 15, 2008 11:03 am Post subject: Re: Linear algebra problem |
|
|
Jannick Asmus wrote:
| Quote: |
On 15.05.2008 12:23, MCT wrote:
Hi,
A friend of mine gave me this problem and I can't seem to find a way
to solve it:
Let V be a real finite-dimensional inner-product space. Let S be an
isometry on V. If S^2+aS+bI is not invertible, then b=1.
Counterexample: dim V = 1, S = ID_V, a=2, b=-3.
Best wishes,
J.
|
Sorry, forgot to add that S was not allowed to have an eigenvalue...
Regards,
Mats |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Jannick Asmus
Guest
|
| Posted: Thu May 15, 2008 11:03 am Post subject: Re: Linear algebra problem |
|
|
On 15.05.2008 12:23, MCT wrote:
| Quote: |
Hi,
A friend of mine gave me this problem and I can't seem to find a way to
solve it:
Let V be a real finite-dimensional inner-product space. Let S be an
isometry on V. If S^2+aS+bI is not invertible, then b=1.
|
Counterexample: dim V = 1, S = ID_V, a=2, b=-3.
Best wishes,
J. |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Jannick Asmus
Guest
|
| Posted: Thu May 15, 2008 11:03 am Post subject: Re: Linear algebra problem |
|
|
On 15.05.2008 12:32, MCT wrote:
| Quote: |
Jannick Asmus wrote:
On 15.05.2008 12:23, MCT wrote:
Hi,
A friend of mine gave me this problem and I can't seem to find a way
to solve it:
Let V be a real finite-dimensional inner-product space. Let S be an
isometry on V. If S^2+aS+bI is not invertible, then b=1.
Counterexample: dim V = 1, S = ID_V, a=2, b=-3.
Best wishes,
J.
Sorry, forgot to add that S was not allowed to have an eigenvalue...
|
.... over the reals I assume. ;)
If dim V = 2, then divide by the characteristic polynomial of S to
reduce to the case were the polynomial is not quadratic, but linear.
This could give you some more information of the image of S^2+aS+b.
The general case could drop out of the decomposition of V in
2-dimensional S-invariant subspaces.
HTH.
--
Best wishes,
J. |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
