Let $Ainmathbb{R}^{ntimes n}$ such that $A^2=-I$. Find all eigenvalues of A.
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The conclusion I got was A has no real eigenvalues.
Can anyone tell me if it's right or what are these eigenvalues?
linear-algebra eigenvalues-eigenvectors
asked Dec 2 at 20:14
Juliana de Souza
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The conclusion I got was A has no real eigenvalues.
Can anyone tell me if it's right or what are these eigenvalues?
linear-algebra eigenvalues-eigenvectors
asked Dec 2 at 20:14
Juliana de Souza
656
New contributor
Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
You're right: the eigenvalues are not in $mathbb R$. However, $A$ does have complex eigenvalues; see greedoid's answer. It is actually often the case that a real $ntimes n$ matrix will have complex eigenvalues.
– Dave
Dec 2 at 20:30
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The conclusion I got was A has no real eigenvalues.
Can anyone tell me if it's right or what are these eigenvalues?
linear-algebra eigenvalues-eigenvectors
asked Dec 2 at 20:14
Juliana de Souza
656
New contributor
Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The conclusion I got was A has no real eigenvalues.
Can anyone tell me if it's right or what are these eigenvalues?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Dec 2 at 20:14
Juliana de Souza
656
New contributor
Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 2 at 20:14
Juliana de Souza
656
New contributor
Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 2 at 20:14
Juliana de Souza
656
New contributor
Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 2 at 20:14
Juliana de Souza
656
asked Dec 2 at 20:14
Juliana de Souza
656
656
New contributor
Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
You're right: the eigenvalues are not in $mathbb R$. However, $A$ does have complex eigenvalues; see greedoid's answer. It is actually often the case that a real $ntimes n$ matrix will have complex eigenvalues.
– Dave
Dec 2 at 20:30
add a comment |
2
You're right: the eigenvalues are not in $mathbb R$. However, $A$ does have complex eigenvalues; see greedoid's answer. It is actually often the case that a real $ntimes n$ matrix will have complex eigenvalues.
– Dave
Dec 2 at 20:30
2
2
You're right: the eigenvalues are not in $mathbb R$. However, $A$ does have complex eigenvalues; see greedoid's answer. It is actually often the case that a real $ntimes n$ matrix will have complex eigenvalues.
– Dave
Dec 2 at 20:30
You're right: the eigenvalues are not in $mathbb R$. However, $A$ does have complex eigenvalues; see greedoid's answer. It is actually often the case that a real $ntimes n$ matrix will have complex eigenvalues.
– Dave
Dec 2 at 20:30
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If $Av = lambda v$ then $$ A^2v = A( lambda v) = lambda (Av) = lambda^2v $$
But $$A^2v = -Iv=-v implies ( lambda^2+1)v = 0$$
So $ lambda =pm i$.
answered Dec 2 at 20:18
greedoid
36.2k114591
I do this. If $Ainmathbb{R}^{ntimes n}$ there is no problem of the eigenvalues being in $mathbb{C}$?
– Juliana de Souza
Dec 2 at 20:22
2
No, ther is no problem with $mathbb{C}$. It is the same as if you have a polynomial with real coeficients. It can have comples roots.
– greedoid
Dec 2 at 20:32
1
You can see something similary here math.stackexchange.com/questions/2314703/…
– greedoid
Dec 2 at 20:33
Yes, there is a problem. Look at the case $n=3$.
– loup blanc
Dec 2 at 21:47
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If $Av = lambda v$ then $$ A^2v = A( lambda v) = lambda (Av) = lambda^2v $$
But $$A^2v = -Iv=-v implies ( lambda^2+1)v = 0$$
So $ lambda =pm i$.
answered Dec 2 at 20:18
greedoid
36.2k114591
I do this. If $Ainmathbb{R}^{ntimes n}$ there is no problem of the eigenvalues being in $mathbb{C}$?
– Juliana de Souza
Dec 2 at 20:22
2
No, ther is no problem with $mathbb{C}$. It is the same as if you have a polynomial with real coeficients. It can have comples roots.
– greedoid
Dec 2 at 20:32
1
You can see something similary here math.stackexchange.com/questions/2314703/…
– greedoid
Dec 2 at 20:33
Yes, there is a problem. Look at the case $n=3$.
– loup blanc
Dec 2 at 21:47
add a comment |
up vote
2
down vote
accepted
If $Av = lambda v$ then $$ A^2v = A( lambda v) = lambda (Av) = lambda^2v $$
But $$A^2v = -Iv=-v implies ( lambda^2+1)v = 0$$
So $ lambda =pm i$.
answered Dec 2 at 20:18
greedoid
36.2k114591
I do this. If $Ainmathbb{R}^{ntimes n}$ there is no problem of the eigenvalues being in $mathbb{C}$?
– Juliana de Souza
Dec 2 at 20:22
2
No, ther is no problem with $mathbb{C}$. It is the same as if you have a polynomial with real coeficients. It can have comples roots.
– greedoid
Dec 2 at 20:32
1
You can see something similary here math.stackexchange.com/questions/2314703/…
– greedoid
Dec 2 at 20:33
Yes, there is a problem. Look at the case $n=3$.
– loup blanc
Dec 2 at 21:47
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If $Av = lambda v$ then $$ A^2v = A( lambda v) = lambda (Av) = lambda^2v $$
But $$A^2v = -Iv=-v implies ( lambda^2+1)v = 0$$
So $ lambda =pm i$.
answered Dec 2 at 20:18
greedoid
36.2k114591
If $Av = lambda v$ then $$ A^2v = A( lambda v) = lambda (Av) = lambda^2v $$
But $$A^2v = -Iv=-v implies ( lambda^2+1)v = 0$$
So $ lambda =pm i$.
answered Dec 2 at 20:18
greedoid
36.2k114591
answered Dec 2 at 20:18
greedoid
36.2k114591
answered Dec 2 at 20:18
greedoid
36.2k114591
answered Dec 2 at 20:18
greedoid
36.2k114591
36.2k114591
I do this. If $Ainmathbb{R}^{ntimes n}$ there is no problem of the eigenvalues being in $mathbb{C}$?
– Juliana de Souza
Dec 2 at 20:22
2
No, ther is no problem with $mathbb{C}$. It is the same as if you have a polynomial with real coeficients. It can have comples roots.
– greedoid
Dec 2 at 20:32
1
You can see something similary here math.stackexchange.com/questions/2314703/…
– greedoid
Dec 2 at 20:33
Yes, there is a problem. Look at the case $n=3$.
– loup blanc
Dec 2 at 21:47
add a comment |
I do this. If $Ainmathbb{R}^{ntimes n}$ there is no problem of the eigenvalues being in $mathbb{C}$?
– Juliana de Souza
Dec 2 at 20:22
2
No, ther is no problem with $mathbb{C}$. It is the same as if you have a polynomial with real coeficients. It can have comples roots.
– greedoid
Dec 2 at 20:32
1
You can see something similary here math.stackexchange.com/questions/2314703/…
– greedoid
Dec 2 at 20:33
Yes, there is a problem. Look at the case $n=3$.
– loup blanc
Dec 2 at 21:47
I do this. If $Ainmathbb{R}^{ntimes n}$ there is no problem of the eigenvalues being in $mathbb{C}$?
– Juliana de Souza
Dec 2 at 20:22
I do this. If $Ainmathbb{R}^{ntimes n}$ there is no problem of the eigenvalues being in $mathbb{C}$?
– Juliana de Souza
Dec 2 at 20:22
2
2
No, ther is no problem with $mathbb{C}$. It is the same as if you have a polynomial with real coeficients. It can have comples roots.
– greedoid
Dec 2 at 20:32
No, ther is no problem with $mathbb{C}$. It is the same as if you have a polynomial with real coeficients. It can have comples roots.
– greedoid
Dec 2 at 20:32
1
1
You can see something similary here math.stackexchange.com/questions/2314703/…
– greedoid
Dec 2 at 20:33
You can see something similary here math.stackexchange.com/questions/2314703/…
– greedoid
Dec 2 at 20:33
Yes, there is a problem. Look at the case $n=3$.
– loup blanc
Dec 2 at 21:47
Yes, there is a problem. Look at the case $n=3$.
– loup blanc
Dec 2 at 21:47
add a comment |
Juliana de Souza is a new contributor. Be nice, and check out our Code of Conduct.
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You're right: the eigenvalues are not in $mathbb R$. However, $A$ does have complex eigenvalues; see greedoid's answer. It is actually often the case that a real $ntimes n$ matrix will have complex eigenvalues.
– Dave
Dec 2 at 20:30