# Eigenvalues

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#1
I was wondering if a real matrix (a matrix will all real entries) can ever have both: real eigenvalues and complex eigenvalues, or will it either have:
- all real eigenvalues
OR - all complex eigenvalues (no real eigenvalues).
0
1 year ago
#2
Any of those cases are possible.
1
1 year ago
#3
(Original post by Takeover Season)
I was wondering if a real matrix (a matrix will all real entries) can ever have both: real eigenvalues and complex eigenvalues, or will it either have:
- all real eigenvalues
OR - all complex eigenvalues (no real eigenvalues).
As mentioned above, any of those can happen. The only restriction is that the complex ones will occur in conjugate pairs.
0
1 year ago
#4
The eigenvalues are given by the roots of the characteristic polynomial - and for any given polynomial you could construct a matrix whose characteristic polynomial is that polynomial - so all those cases are possible
0
1 year ago
#5
(Original post by Takeover Season)
I was wondering if a real matrix (a matrix will all real entries) can ever have both: real eigenvalues and complex eigenvalues, or will it either have:
- all real eigenvalues
OR - all complex eigenvalues (no real eigenvalues).
For 2x2 matrices, you can have at most 2 eigenvalues, and if the entries are real, the characteristic polynomial has real-coefficient and so any roots (eigenvalues) if complex, will occur in complex conjugate pairs. So if you have complex eigenvalues, they'll occur in complex conjugate pairs. Which means they are either both real or both complex.

If you have larger matrices then it's easy to construct ones in whose characteristic polynomial is (x-a)(x-z)(x-z*) for z complex.
1
#6
(Original post by Ano123)
Any of those cases are possible.
(Original post by mqb2766)
As mentioned above, any of those can happen. The only restriction is that the complex ones will occur in conjugate pairs.
(Original post by nimon)
The eigenvalues are given by the roots of the characteristic polynomial - and for any given polynomial you could construct a matrix whose characteristic polynomial is that polynomial - so all those cases are possible
Ok, thank you everyone. So, just to clarify, for a REAL matrix i.e. a matrix with only REAL entries (all its entries are real numbers), you can get both REAL and COMPLEX eigenvalues, but the COMPLEX eigenvalues have to occur in COMPLEX CONJUGATE pairs. Is this correct? I am just making sure my question was interpreted correctly!
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#7
(Original post by Zacken)
For 2x2 matrices, you can have at most 2 eigenvalues, and if the entries are real, the characteristic polynomial has real-coefficient and so any roots (eigenvalues) if complex, will occur in complex conjugate pairs. So if you have complex eigenvalues, they'll occur in complex conjugate pairs. Which means they are either both real or both complex.

If you have larger matrices then it's easy to construct ones in whose characteristic polynomial is (x-a)(x-z)(x-z*) for z complex.
Oh okay, that explains what I asked just now, thank you so much!
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1 year ago
#8
(Original post by Takeover Season)
Ok, thank you everyone. So, just to clarify, for a REAL matrix i.e. a matrix with only REAL entries (all its entries are real numbers), you can get both REAL and COMPLEX eigenvalues, but the COMPLEX eigenvalues have to occur in COMPLEX CONJUGATE pairs. Is this correct? I am just making sure my question was interpreted correctly!
That's correct.
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#9
(Original post by mqb2766)
That's correct.
Thanks a lot!
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#10
(Original post by mqb2766)
That's correct.
Just one final question sorry. If I had a matrix A and I want to diagonalise it. But, say it has a combination of REAL AND COMPLEX eigenvalues, then can I be sure to say that we can only diagonalise A using a COMPLEX diagonal matrix D i.e. a matrix with complex numbers (where you can have real and complex entries since real numbers are also complex numbers), i.e. A is not diagonalise over the real numbers, it is only diagonalisable over the complex numbers?
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1 year ago
#11
(Original post by Takeover Season)
Just one final question sorry. If I had a matrix A and I want to diagonalise it. But, say it has a combination of REAL AND COMPLEX eigenvalues, then can I be sure to say that we can only diagonalise A using a COMPLEX diagonal matrix D i.e. a matrix with complex numbers (where you can have real and complex entries since real numbers are also complex numbers), i.e. A is not diagonalise over the real numbers, it is only diagonalisable over the complex numbers?
Yeah if at least one eigenvalue is complex then A can be diagonalised over but not over 0
#12
Thanks a lot!
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#13
Just one final question: RDKGames mqb2766, if we have a complex square matrix of order n, does it mean that it'll always have n eigenvalues, by that, I mean including repeated ones, so if you have an eigenvalue of algebraic multiplicity 'a', that eigenvalue appears 'a' times e.g. not 1 time.

So, by that I mean, if we have a n x n complex matrix, will it always have n eigenvalues in total, considering all e.g. distinct ones, same ones, all of them?
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1 year ago
#14
(Original post by Takeover Season)
Just one final question: RDKGames mqb2766, if we have a complex square matrix of order n, does it mean that it'll always have n eigenvalues, by that, I mean including repeated ones, so if you have an eigenvalue of algebraic multiplicity 'a', that eigenvalue appears 'a' times e.g. not 1 time.

So, by that I mean, if we have a n x n complex matrix, will it always have n eigenvalues in total, considering all e.g. distinct ones, same ones, all of them?
Sure. A real n*n matrix would have n eigenvals as well. Its a property of the characteristic polynomial.
0
1 year ago
#15
(Original post by Takeover Season)
Just one final question: RDKGames mqb2766, if we have a complex square matrix of order n, does it mean that it'll always have n eigenvalues, by that, I mean including repeated ones, so if you have an eigenvalue of algebraic multiplicity 'a', that eigenvalue appears 'a' times e.g. not 1 time.

So, by that I mean, if we have a n x n complex matrix, will it always have n eigenvalues in total, considering all e.g. distinct ones, same ones, all of them?
Yes. Fundamental Theorem of Algebra.
Last edited by RDKGames; 1 year ago
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1 year ago
#16
Just to be clear, so that this comment isn't misunderstood, not all real matrices are diagonalizable over or A separate comment: depending on your definitions, it may be that eigenvalues of a real matrix are necessarily real because you need to have a real eigenvector. The characteristic polynomial may well have complex roots, but no eigenvector for those roots would be real; there would though be complex eigenvectors.
Last edited by RichE; 1 year ago
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1 year ago
#17
Ah yes, confused confusion
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#18
(Original post by mqb2766)
Sure. A real n*n matrix would have n eigenvals as well. Its a property of the characteristic polynomial.
Hi, yes thank you. Perfect! I forgot to mention I consider real matrices as a subset of complex matrices. They are special cases where all the entries of the complex matrix are real numbers. So, by asking about complex matrices I automatically know it's true for real matrices too.
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#19
(Original post by RDKGames)
Yes. Fundamental Theorem of Algebra.
Thanks!
(Original post by RichE)
Just to be clear, so that this comment isn't misunderstood, not all real matrices are diagonalizable over or A separate comment: depending on your definitions, it may be that eigenvalues of a real matrix are necessarily real because you need to have a real eigenvector. The characteristic polynomial may well have complex roots, but no eigenvector for those roots would be real; there would though be complex eigenvectors.
Hi, just wondering so if we have real matrices, you are saying every complex eigenvalue has only complex eigenvectors (no real eigenvectors). If so, does the same apply to complex matrices with complex eigenvalues? Thanks
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1 year ago
#20
(Original post by Takeover Season)
Thanks!

Hi, just wondering so if we have real matrices, you are saying every complex eigenvalue has only complex eigenvectors (no real eigenvectors). If so, does the same apply to complex matrices with complex eigenvalues? Thanks
Well yes, to the first question. If Av = cv, and A is a real matrix, v is a real eigenvector, then c cannot be complex as the LHS is a real vector so the RHS needs to be too.

If c and A are complex then v might be purely real. But this isn't a particularly important observation. Maybe I'm not understanding your second question.
Last edited by RichE; 1 year ago
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