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Real eigenvalues of a matrix over a field of Complex Numbers
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I'm trying to solve a question (exam prep) which asks for Real eigenvalues of a matrix. My problem is that the Matrix is given over a field of Complex numbers (as shown on the picture) Matrix
Is there a special method for finding eigenvalues for matrices over a field of complex numbers? or can i use the normal method of finding Det(Matrix - IX)v =0 roots. If i use the normal method i get eigenvalue of -1.
linear-algebra complex-numbers eigenvalues-eigenvectors
asked Jul 28 at 15:24
Asi Orasi
32
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down vote
favorite
I'm trying to solve a question (exam prep) which asks for Real eigenvalues of a matrix. My problem is that the Matrix is given over a field of Complex numbers (as shown on the picture) Matrix
Is there a special method for finding eigenvalues for matrices over a field of complex numbers? or can i use the normal method of finding Det(Matrix - IX)v =0 roots. If i use the normal method i get eigenvalue of -1.
linear-algebra complex-numbers eigenvalues-eigenvectors
asked Jul 28 at 15:24
Asi Orasi
32
1
That's actually a real matrix. Also please see math.meta.stackexchange.com/questions/5020 for tips on how to write maths, including matrices.
– Lord Shark the Unknown
Jul 28 at 15:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to solve a question (exam prep) which asks for Real eigenvalues of a matrix. My problem is that the Matrix is given over a field of Complex numbers (as shown on the picture) Matrix
Is there a special method for finding eigenvalues for matrices over a field of complex numbers? or can i use the normal method of finding Det(Matrix - IX)v =0 roots. If i use the normal method i get eigenvalue of -1.
linear-algebra complex-numbers eigenvalues-eigenvectors
asked Jul 28 at 15:24
Asi Orasi
32
I'm trying to solve a question (exam prep) which asks for Real eigenvalues of a matrix. My problem is that the Matrix is given over a field of Complex numbers (as shown on the picture) Matrix
Is there a special method for finding eigenvalues for matrices over a field of complex numbers? or can i use the normal method of finding Det(Matrix - IX)v =0 roots. If i use the normal method i get eigenvalue of -1.
linear-algebra complex-numbers eigenvalues-eigenvectors
asked Jul 28 at 15:24
Asi Orasi
32
asked Jul 28 at 15:24
Asi Orasi
32
asked Jul 28 at 15:24
Asi Orasi
32
asked Jul 28 at 15:24
Asi Orasi
32
32
1
That's actually a real matrix. Also please see math.meta.stackexchange.com/questions/5020 for tips on how to write maths, including matrices.
– Lord Shark the Unknown
Jul 28 at 15:28
add a comment |Â
1
That's actually a real matrix. Also please see math.meta.stackexchange.com/questions/5020 for tips on how to write maths, including matrices.
– Lord Shark the Unknown
Jul 28 at 15:28
1
1
That's actually a real matrix. Also please see math.meta.stackexchange.com/questions/5020 for tips on how to write maths, including matrices.
– Lord Shark the Unknown
Jul 28 at 15:28
That's actually a real matrix. Also please see math.meta.stackexchange.com/questions/5020 for tips on how to write maths, including matrices.
– Lord Shark the Unknown
Jul 28 at 15:28
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The only why of finding eigenvalues is solving the characteristic polynomial, doesn't matter over what field your matrices are, they can be over anything. In your case is even easier because the matrix is effectively a real matrix, the fact that is over the complex field means that even if you have some complex eigenvalues you can still diagonalise the matrix. In anyway, you find the eigenvalues by the roots of the characteristic polynomial $$det(A-lambdamathbbI)=0$$ The problem asks you to take only the eigenvalues s.t. $lambdainmathbbR$, so in this case, being the characteristic polynomial as follows $$-(lambda+1)(lambda^2+1)=0implieslambda_1 =-1;;lambda_2=i;;lambda_3=-i$$
the answer to your problem is $lambda_1=-1$.
One way for a complex matrix to have all real eigenvalues is if the matrix is hermitian for the spectral theorem. See as a simple example the Puli matrices. But it's not the only way. As pointed out by user Lord Shark the Unknown even the matrix $$left(beginmatrix1&i\0&-1endmatrixright)$$ has two real eigenvalues.
answered Jul 28 at 15:32
Davide Morgante
1,724220
Complex matrices may have real eigenvalues without being Hermitian: How about $pmatrix1&i\0&-1$?
– Lord Shark the Unknown
Jul 28 at 15:51
@LordSharktheUnknown Yes, that's true! I shouldn't say "the only way" but instead "one way", thanks for letting me know
– Davide Morgante
Jul 28 at 15:52
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The only why of finding eigenvalues is solving the characteristic polynomial, doesn't matter over what field your matrices are, they can be over anything. In your case is even easier because the matrix is effectively a real matrix, the fact that is over the complex field means that even if you have some complex eigenvalues you can still diagonalise the matrix. In anyway, you find the eigenvalues by the roots of the characteristic polynomial $$det(A-lambdamathbbI)=0$$ The problem asks you to take only the eigenvalues s.t. $lambdainmathbbR$, so in this case, being the characteristic polynomial as follows $$-(lambda+1)(lambda^2+1)=0implieslambda_1 =-1;;lambda_2=i;;lambda_3=-i$$
the answer to your problem is $lambda_1=-1$.
One way for a complex matrix to have all real eigenvalues is if the matrix is hermitian for the spectral theorem. See as a simple example the Puli matrices. But it's not the only way. As pointed out by user Lord Shark the Unknown even the matrix $$left(beginmatrix1&i\0&-1endmatrixright)$$ has two real eigenvalues.
answered Jul 28 at 15:32
Davide Morgante
1,724220
Complex matrices may have real eigenvalues without being Hermitian: How about $pmatrix1&i\0&-1$?
– Lord Shark the Unknown
Jul 28 at 15:51
@LordSharktheUnknown Yes, that's true! I shouldn't say "the only way" but instead "one way", thanks for letting me know
– Davide Morgante
Jul 28 at 15:52
add a comment |Â
up vote
2
down vote
accepted
The only why of finding eigenvalues is solving the characteristic polynomial, doesn't matter over what field your matrices are, they can be over anything. In your case is even easier because the matrix is effectively a real matrix, the fact that is over the complex field means that even if you have some complex eigenvalues you can still diagonalise the matrix. In anyway, you find the eigenvalues by the roots of the characteristic polynomial $$det(A-lambdamathbbI)=0$$ The problem asks you to take only the eigenvalues s.t. $lambdainmathbbR$, so in this case, being the characteristic polynomial as follows $$-(lambda+1)(lambda^2+1)=0implieslambda_1 =-1;;lambda_2=i;;lambda_3=-i$$
the answer to your problem is $lambda_1=-1$.
One way for a complex matrix to have all real eigenvalues is if the matrix is hermitian for the spectral theorem. See as a simple example the Puli matrices. But it's not the only way. As pointed out by user Lord Shark the Unknown even the matrix $$left(beginmatrix1&i\0&-1endmatrixright)$$ has two real eigenvalues.
answered Jul 28 at 15:32
Davide Morgante
1,724220
Complex matrices may have real eigenvalues without being Hermitian: How about $pmatrix1&i\0&-1$?
– Lord Shark the Unknown
Jul 28 at 15:51
@LordSharktheUnknown Yes, that's true! I shouldn't say "the only way" but instead "one way", thanks for letting me know
– Davide Morgante
Jul 28 at 15:52
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The only why of finding eigenvalues is solving the characteristic polynomial, doesn't matter over what field your matrices are, they can be over anything. In your case is even easier because the matrix is effectively a real matrix, the fact that is over the complex field means that even if you have some complex eigenvalues you can still diagonalise the matrix. In anyway, you find the eigenvalues by the roots of the characteristic polynomial $$det(A-lambdamathbbI)=0$$ The problem asks you to take only the eigenvalues s.t. $lambdainmathbbR$, so in this case, being the characteristic polynomial as follows $$-(lambda+1)(lambda^2+1)=0implieslambda_1 =-1;;lambda_2=i;;lambda_3=-i$$
the answer to your problem is $lambda_1=-1$.
One way for a complex matrix to have all real eigenvalues is if the matrix is hermitian for the spectral theorem. See as a simple example the Puli matrices. But it's not the only way. As pointed out by user Lord Shark the Unknown even the matrix $$left(beginmatrix1&i\0&-1endmatrixright)$$ has two real eigenvalues.
answered Jul 28 at 15:32
Davide Morgante
1,724220
The only why of finding eigenvalues is solving the characteristic polynomial, doesn't matter over what field your matrices are, they can be over anything. In your case is even easier because the matrix is effectively a real matrix, the fact that is over the complex field means that even if you have some complex eigenvalues you can still diagonalise the matrix. In anyway, you find the eigenvalues by the roots of the characteristic polynomial $$det(A-lambdamathbbI)=0$$ The problem asks you to take only the eigenvalues s.t. $lambdainmathbbR$, so in this case, being the characteristic polynomial as follows $$-(lambda+1)(lambda^2+1)=0implieslambda_1 =-1;;lambda_2=i;;lambda_3=-i$$
the answer to your problem is $lambda_1=-1$.
One way for a complex matrix to have all real eigenvalues is if the matrix is hermitian for the spectral theorem. See as a simple example the Puli matrices. But it's not the only way. As pointed out by user Lord Shark the Unknown even the matrix $$left(beginmatrix1&i\0&-1endmatrixright)$$ has two real eigenvalues.
answered Jul 28 at 15:32
Davide Morgante
1,724220
edited Jul 28 at 15:56
answered Jul 28 at 15:32
Davide Morgante
1,724220
answered Jul 28 at 15:32
Davide Morgante
1,724220
answered Jul 28 at 15:32
Davide Morgante
1,724220
1,724220
Complex matrices may have real eigenvalues without being Hermitian: How about $pmatrix1&i\0&-1$?
– Lord Shark the Unknown
Jul 28 at 15:51
@LordSharktheUnknown Yes, that's true! I shouldn't say "the only way" but instead "one way", thanks for letting me know
– Davide Morgante
Jul 28 at 15:52
add a comment |Â
Complex matrices may have real eigenvalues without being Hermitian: How about $pmatrix1&i\0&-1$?
– Lord Shark the Unknown
Jul 28 at 15:51
@LordSharktheUnknown Yes, that's true! I shouldn't say "the only way" but instead "one way", thanks for letting me know
– Davide Morgante
Jul 28 at 15:52
Complex matrices may have real eigenvalues without being Hermitian: How about $pmatrix1&i\0&-1$?
– Lord Shark the Unknown
Jul 28 at 15:51
Complex matrices may have real eigenvalues without being Hermitian: How about $pmatrix1&i\0&-1$?
– Lord Shark the Unknown
Jul 28 at 15:51
@LordSharktheUnknown Yes, that's true! I shouldn't say "the only way" but instead "one way", thanks for letting me know
– Davide Morgante
Jul 28 at 15:52
@LordSharktheUnknown Yes, that's true! I shouldn't say "the only way" but instead "one way", thanks for letting me know
– Davide Morgante
Jul 28 at 15:52
add a comment |Â
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1
That's actually a real matrix. Also please see math.meta.stackexchange.com/questions/5020 for tips on how to write maths, including matrices.
– Lord Shark the Unknown
Jul 28 at 15:28