# 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

- all real eigenvalues

OR - all complex eigenvalues (no real eigenvalues).

**and**complex eigenvalues, or will it either have:- all real eigenvalues

OR - all complex eigenvalues (no real eigenvalues).

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Ano123

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#2

mqb2766

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#3

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#3

(Original post by

I was wondering if a real matrix (a matrix will all real entries) can ever have both: real eigenvalues

- all real eigenvalues

OR - all complex eigenvalues (no real eigenvalues).

**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).

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nimon

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#4

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#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

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Zacken

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#5

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#5

**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).

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.

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#6

(Original post by

Any of those cases are possible.

**Ano123**)Any of those cases are possible.

(Original post by

As mentioned above, any of those can happen. The only restriction is that the complex ones will occur in conjugate pairs.

**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

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

**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

**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

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.

**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.

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mqb2766

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#8

(Original post by

Ok, thank you everyone. So, just to clarify, for a

**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!
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#9

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#10

(Original post by

That's correct.

**mqb2766**)That's correct.

**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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RDKGames

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#11

(Original post by

Just one final question sorry. If I had a matrix A and I want to diagonalise it. But, say it has a combination of

**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?
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#12

(Original post by

Yeah if at least one eigenvalue is complex then A can be diagonalised over but not over

**RDKGames**)Yeah if at least one eigenvalue is complex then A can be diagonalised over but not over

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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?

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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mqb2766

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(Original post by

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?

**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?

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RDKGames

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#15

**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?

Last edited by RDKGames; 1 year ago

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RichE

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#16

**RDKGames**)

Yeah if at least one eigenvalue is complex then A can be diagonalised over but not over

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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JJJJJAAAAMES

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#18

(Original post by

Sure. A real n*n matrix would have n eigenvals as well. Its a property of the characteristic polynomial.

**mqb2766**)Sure. A real n*n matrix would have n eigenvals as well. Its a property of the characteristic polynomial.

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#19

(Original post by

Yes. Fundamental Theorem of Algebra.

**RDKGames**)Yes. Fundamental Theorem of Algebra.

(Original post by

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.

**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.

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RichE

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#20

(Original post by

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

**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

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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