Suppose A is an n by n matrix such that A^2 = A. By considering the eigenvalues of A, prove that either (i) detA = 1 and TrA = n, or (ii) detA = 0 and TrA is an integer less than n.
(I have done this part of the question. I am stuck on this next bit)
If Ax=y, what is Ay? What is the dimension of the space of nonvanishing vectors y for the two cases mentioned above?
Spoiler:ShowSo I figured the question was asking for the dimension of the image of A, or equivalently, for this particular case, the dimension of the eigenspace corresponding to eigenvalue 1. I fiddled around a little and couldn't make any progress. I anticipate that the answer will be different depending on the TrA  that is, depending on the number of 0 eigenvalues and the number of 1 eigenvalues  but I don't know how to incorporate this into a solution. Maybe I need to use the number of linearly independent columns?

TBendy
 Follow
 0 followers
 0 badges
 Send a private message to TBendy
 Thread Starter
Offline0ReputationRep: Follow
 1
 09042013 14:09
Last edited by TBendy; 09042013 at 14:12. 
 Follow
 2
 09042013 14:48
(Original post by TBendy)
Suppose A is an n by n matrix such that A^2 = A. By considering the eigenvalues of A, prove that either (i) detA = 1 and TrA = n, or (ii) detA = 0 and TrA is an integer less than n.
(I have done this part of the question. I am stuck on this next bit)
If Ax=y, what is Ay? What is the dimension of the space of nonvanishing vectors y for the two cases mentioned above?
Spoiler:ShowSo I figured the question was asking for the dimension of the image of A, or equivalently, for this particular case, the dimension of the eigenspace corresponding to eigenvalue 1. I fiddled around a little and couldn't make any progress. I anticipate that the answer will be different depending on the TrA  that is, depending on the number of 0 eigenvalues and the number of 1 eigenvalues  but I don't know how to incorporate this into a solution. Maybe I need to use the number of linearly independent columns? 
TBendy
 Follow
 0 followers
 0 badges
 Send a private message to TBendy
 Thread Starter
Offline0ReputationRep: Follow
 3
 09042013 18:09
(Original post by Mark13)
You can work out the algebraic multiplicity of the eigenvalues 1 and 0 from the trace of A, and then you can use the form of the minimal polynomial of A to work out the corresponding geometric multiplicities. Let's say the algebraic multiplicity of the eigenvalue 1 is k  do you know what the minimal polynomial of A would look like if the geometric multiplicity of the eigenvalue 1 wasn't also k?
Thanks for your help. 
 Follow
 4
 09042013 18:19
(Original post by TBendy)
No I don't  I have not met the concept of 'minimal polynomial'! Could you please explain to me what it is and how it applies to this question?
Thanks for your help.
If you haven't met the idea of a minimal polynomial, you can do the question in other ways  it's just the first way I thought to do it.
First off, are you familiar with the idea of a generalised eigenspace? 
TBendy
 Follow
 0 followers
 0 badges
 Send a private message to TBendy
 Thread Starter
Offline0ReputationRep: Follow
 5
 09042013 19:04
(Original post by Mark13)
No worries.
If you haven't met the idea of a minimal polynomial, you can do the question in other ways  it's just the first way I thought to do it.
First off, are you familiar with the idea of a generalised eigenspace?
Thanks 
 Follow
 6
 09042013 19:29
(Original post by TBendy)
I am aware of the definition, but not familiar.. This question may assume knowledge I do not have, it was not made by the same people who decided on my course's syllabus  so it may be that the concepts of minimal polynomials / generalised eigenspaces are necessary in order to tackle this question. However, my limited knowledge on the subject of linear algebra leads me to ask whether you could find the most elementary way of solving this problem. If this most elementary way does require such concepts as the above then I am more than willing to learn about them.
Thanks
You've already identified that the image of A consists of eigenvectors of eigenvalue 1.
You can show that the image of A must consist of all eigenvectors of eigenvalue 1.
You can find the dimension of the generalised eigenspace corresponding to 1 in terms of the trace of A.
You can show that the eigenspace corresponding to 1 is actually the generalised eigenspace corresponding to 1 i.e. if for some positive integer r then , and from all the above steps, you'll get the dimension of the image of A.
Let me know if any of that's unclear! 
TBendy
 Follow
 0 followers
 0 badges
 Send a private message to TBendy
 Thread Starter
Offline0ReputationRep: Follow
 7
 09042013 19:48
(Original post by Mark13)
The way I'd approach the question is:
You've already identified that the image of A consists of eigenvectors of eigenvalue 1.
You can show that the image of A must consist of all eigenvectors of eigenvalue 1.
You can find the dimension of the generalised eigenspace corresponding to 1 in terms of the trace of A.
You can show that the eigenspace corresponding to 1 is actually the generalised eigenspace corresponding to 1 i.e. if for some positive integer r then , and from all the above steps, you'll get the dimension of the image of A.
Let me know if any of that's unclear!
Cheers 
 Follow
 8
 09042013 20:00
(Original post by TBendy)
OK, so how do you find the dimension of the generalised 1eigenspace using the trace?
Cheers
So the problem is reduced to finding the multiplicity of 1 as a root of the characteristic equation.
Since A^2=A, A can only have 2 distinct eigenvalues  can you see what they must be? Based on this, you can work out from the trace of A what the multiplicity of the e'val 1 is, since the trace is the sum of the eigenvalues. 
TBendy
 Follow
 0 followers
 0 badges
 Send a private message to TBendy
 Thread Starter
Offline0ReputationRep: Follow
 9
 09042013 20:22
(Original post by Mark13)
The dimension of the generalised 1eigenspace is the same as the multiplicity of 1 as a root of the characteristic equation (I tend to think about these kind of things with Jordan Normal form in mind, but you might not have heard of that, so I'm not sure what the best way to prove this is from your standpoint if you haven't seen this before).
So the problem is reduced to finding the multiplicity of 1 as a root of the characteristic equation.
Since A^2=A, A can only have 2 distinct eigenvalues  can you see what they must be? Based on this, you can work out from the trace of A what the multiplicity of the e'val 1 is, since the trace is the sum of the eigenvalues.
Then we know that if a vector is in the generalised 1eigenspace it is a solution of A^n x = x (for some positive integer n), but since A^n = n for all positive integer n then such a vector is in the 1eigenspace, and hence the image. So the dimension of the image is TrA... And no appeal to the determinant is required.
Is this right? If so, can I ask how we conclude that the dimension of the generalised 1eigenspace is equal to the algebraic multiplicity of the eigenvalue 1?
Thank you. 
 Follow
 10
 09042013 20:37
(Original post by TBendy)
Then we know that if a vector is in the generalised 1eigenspace it is a solution of A^n x = x
Is this right? If so, can I ask how we conclude that the dimension of the generalised 1eigenspace is equal to the algebraic multiplicity of the eigenvalue 1?Last edited by Mark13; 10042013 at 00:15. 
TBendy
 Follow
 0 followers
 0 badges
 Send a private message to TBendy
 Thread Starter
Offline0ReputationRep: Follow
 11
 09042013 22:43
(Original post by Mark13)
That's not quite right  a vector x is in the generalised eigenspace of an eigenvalue lambda if for some positive integer [\lambda]. You're on the right lines with the rest of it though.
It comes from a more general result about the decomposition of a vector space into generalised eigenspaces of a linear map  it's been a couple of years since I've seen the proof so I can't remember it offhand, but I would think that most first courses in linear algebra would address this kind of thing, so it might be worth looking ahead in the notes you're working from. 
 Follow
 12
 10042013 00:25
(Original post by TBendy)
Ah OK, thank you. My final question then is how we conclude that the 1eigenspace is the same as the 1generalised eigenspace?
(AI)^r (AI)x = 0
for some integer r, but since A (AI)x=0, this simplifies to say
(I)^r (AI)x = 0
so (AI)x=0, so x is in the 1eigenspace. Therefore the generalised 1eigenspace is the same as the 1eigenspace. 
TBendy
 Follow
 0 followers
 0 badges
 Send a private message to TBendy
 Thread Starter
Offline0ReputationRep: Follow
 13
 10042013 01:37
(Original post by Mark13)
Since A^2=A, you know that 0=A^2A=A(AI). So if x is in the generalised 1eigenspace, then (AI)x is in the 0eigenspace, and also the generalised 1eigenspace. Since its in the generalised 1eigenspace, we have
(AI)^r (AI)x = 0
for some integer r, but since A (AI)x=0, this simplifies to say
(I)^r (AI)x = 0
so (AI)x=0, so x is in the 1eigenspace. Therefore the generalised 1eigenspace is the same as the 1eigenspace.
Cheers again
Reply
Submit reply
Related discussions:
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
 SherlockHolmes
 Notnek
 charco
 Mr M
 TSR Moderator
 Nirgilis
 usycool1
 Changing Skies
 James A
 rayquaza17
 RDKGames
 randdom
 davros
 Gingerbread101
 Kvothe the Arcane
 The Financier
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 Reality Check
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 LeCroissant
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Moltenmo
Updated: April 10, 2013
Share this discussion:
Tweet