Why aren't these eigenvectors orthogonal?
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As far as I know, the matrix: $$M = begin{pmatrix}1 & 0 & isqrt{3} \ 0 & 2 & 0 \ -isqrt{3} & 0 & 3 end{pmatrix} $$
Is hermitian with eigenvalues: $lambda_1 = 0, lambda_2 = 2, lambda_3 = 4$.
And corresponding eigenvectors:
$V_{lambda_1 = 0} = begin{pmatrix}-isqrt{3} \ 0 \ 1end{pmatrix} $,
$V_{lambda_2 = 2} = begin{pmatrix}0\ 1 \ 0end{pmatrix} $,
$V_{lambda_3 = 4} = begin{pmatrix}dfrac{i}{sqrt{3}}\ 0 \ 1end{pmatrix} $
Since these are eigenvectors of distinct eigenvalues they should be orthogonal but the first and third are not orthogonal.
Why is this?
eigenvalues-eigenvectors
asked Nov 19 at 16:52
Rzmwood
61
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0
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As far as I know, the matrix: $$M = begin{pmatrix}1 & 0 & isqrt{3} \ 0 & 2 & 0 \ -isqrt{3} & 0 & 3 end{pmatrix} $$
Is hermitian with eigenvalues: $lambda_1 = 0, lambda_2 = 2, lambda_3 = 4$.
And corresponding eigenvectors:
$V_{lambda_1 = 0} = begin{pmatrix}-isqrt{3} \ 0 \ 1end{pmatrix} $,
$V_{lambda_2 = 2} = begin{pmatrix}0\ 1 \ 0end{pmatrix} $,
$V_{lambda_3 = 4} = begin{pmatrix}dfrac{i}{sqrt{3}}\ 0 \ 1end{pmatrix} $
Since these are eigenvectors of distinct eigenvalues they should be orthogonal but the first and third are not orthogonal.
Why is this?
eigenvalues-eigenvectors
asked Nov 19 at 16:52
Rzmwood
61
1
Maybe they are hermitian orthogonal.
– Charlie Frohman
Nov 19 at 16:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As far as I know, the matrix: $$M = begin{pmatrix}1 & 0 & isqrt{3} \ 0 & 2 & 0 \ -isqrt{3} & 0 & 3 end{pmatrix} $$
Is hermitian with eigenvalues: $lambda_1 = 0, lambda_2 = 2, lambda_3 = 4$.
And corresponding eigenvectors:
$V_{lambda_1 = 0} = begin{pmatrix}-isqrt{3} \ 0 \ 1end{pmatrix} $,
$V_{lambda_2 = 2} = begin{pmatrix}0\ 1 \ 0end{pmatrix} $,
$V_{lambda_3 = 4} = begin{pmatrix}dfrac{i}{sqrt{3}}\ 0 \ 1end{pmatrix} $
Since these are eigenvectors of distinct eigenvalues they should be orthogonal but the first and third are not orthogonal.
Why is this?
eigenvalues-eigenvectors
asked Nov 19 at 16:52
Rzmwood
61
As far as I know, the matrix: $$M = begin{pmatrix}1 & 0 & isqrt{3} \ 0 & 2 & 0 \ -isqrt{3} & 0 & 3 end{pmatrix} $$
Is hermitian with eigenvalues: $lambda_1 = 0, lambda_2 = 2, lambda_3 = 4$.
And corresponding eigenvectors:
$V_{lambda_1 = 0} = begin{pmatrix}-isqrt{3} \ 0 \ 1end{pmatrix} $,
$V_{lambda_2 = 2} = begin{pmatrix}0\ 1 \ 0end{pmatrix} $,
$V_{lambda_3 = 4} = begin{pmatrix}dfrac{i}{sqrt{3}}\ 0 \ 1end{pmatrix} $
Since these are eigenvectors of distinct eigenvalues they should be orthogonal but the first and third are not orthogonal.
Why is this?
eigenvalues-eigenvectors
eigenvalues-eigenvectors
asked Nov 19 at 16:52
Rzmwood
61
asked Nov 19 at 16:52
Rzmwood
61
asked Nov 19 at 16:52
Rzmwood
61
asked Nov 19 at 16:52
Rzmwood
61
asked Nov 19 at 16:52
Rzmwood
61
61
1
Maybe they are hermitian orthogonal.
– Charlie Frohman
Nov 19 at 16:55
add a comment |
1
Maybe they are hermitian orthogonal.
– Charlie Frohman
Nov 19 at 16:55
1
1
Maybe they are hermitian orthogonal.
– Charlie Frohman
Nov 19 at 16:55
Maybe they are hermitian orthogonal.
– Charlie Frohman
Nov 19 at 16:55
add a comment |
1 Answer
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They are orthogonal. begin{align}V_0^HV_2&=0\ V_2^HV_4&=0\ V_0^HV_4&=begin{pmatrix}isqrt3&0&1end{pmatrix}cdotbegin{pmatrix}frac i{sqrt3}\ 0\ 1end{pmatrix}=0end{align}
answered Nov 19 at 16:59
Saucy O'Path
5,6161626
I did not check that the original eigenvectors are OK, but there is a sign difference in the top term of your vertical vector.
– Andrei
Nov 19 at 17:00
@Andrei Like the one you do when you take the Hermitian of something?
– Saucy O'Path
Nov 19 at 17:01
Ah! Thank you so much
– Rzmwood
Nov 19 at 17:02
@SaucyO'Path thanks. I somehow missed that.
– Andrei
Nov 19 at 17:07
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
They are orthogonal. begin{align}V_0^HV_2&=0\ V_2^HV_4&=0\ V_0^HV_4&=begin{pmatrix}isqrt3&0&1end{pmatrix}cdotbegin{pmatrix}frac i{sqrt3}\ 0\ 1end{pmatrix}=0end{align}
answered Nov 19 at 16:59
Saucy O'Path
5,6161626
I did not check that the original eigenvectors are OK, but there is a sign difference in the top term of your vertical vector.
– Andrei
Nov 19 at 17:00
@Andrei Like the one you do when you take the Hermitian of something?
– Saucy O'Path
Nov 19 at 17:01
Ah! Thank you so much
– Rzmwood
Nov 19 at 17:02
@SaucyO'Path thanks. I somehow missed that.
– Andrei
Nov 19 at 17:07
add a comment |
up vote
0
down vote
They are orthogonal. begin{align}V_0^HV_2&=0\ V_2^HV_4&=0\ V_0^HV_4&=begin{pmatrix}isqrt3&0&1end{pmatrix}cdotbegin{pmatrix}frac i{sqrt3}\ 0\ 1end{pmatrix}=0end{align}
answered Nov 19 at 16:59
Saucy O'Path
5,6161626
I did not check that the original eigenvectors are OK, but there is a sign difference in the top term of your vertical vector.
– Andrei
Nov 19 at 17:00
@Andrei Like the one you do when you take the Hermitian of something?
– Saucy O'Path
Nov 19 at 17:01
Ah! Thank you so much
– Rzmwood
Nov 19 at 17:02
@SaucyO'Path thanks. I somehow missed that.
– Andrei
Nov 19 at 17:07
add a comment |
up vote
0
down vote
up vote
0
down vote
They are orthogonal. begin{align}V_0^HV_2&=0\ V_2^HV_4&=0\ V_0^HV_4&=begin{pmatrix}isqrt3&0&1end{pmatrix}cdotbegin{pmatrix}frac i{sqrt3}\ 0\ 1end{pmatrix}=0end{align}
answered Nov 19 at 16:59
Saucy O'Path
5,6161626
They are orthogonal. begin{align}V_0^HV_2&=0\ V_2^HV_4&=0\ V_0^HV_4&=begin{pmatrix}isqrt3&0&1end{pmatrix}cdotbegin{pmatrix}frac i{sqrt3}\ 0\ 1end{pmatrix}=0end{align}
answered Nov 19 at 16:59
Saucy O'Path
5,6161626
answered Nov 19 at 16:59
Saucy O'Path
5,6161626
answered Nov 19 at 16:59
Saucy O'Path
5,6161626
answered Nov 19 at 16:59
Saucy O'Path
5,6161626
5,6161626
I did not check that the original eigenvectors are OK, but there is a sign difference in the top term of your vertical vector.
– Andrei
Nov 19 at 17:00
@Andrei Like the one you do when you take the Hermitian of something?
– Saucy O'Path
Nov 19 at 17:01
Ah! Thank you so much
– Rzmwood
Nov 19 at 17:02
@SaucyO'Path thanks. I somehow missed that.
– Andrei
Nov 19 at 17:07
add a comment |
I did not check that the original eigenvectors are OK, but there is a sign difference in the top term of your vertical vector.
– Andrei
Nov 19 at 17:00
@Andrei Like the one you do when you take the Hermitian of something?
– Saucy O'Path
Nov 19 at 17:01
Ah! Thank you so much
– Rzmwood
Nov 19 at 17:02
@SaucyO'Path thanks. I somehow missed that.
– Andrei
Nov 19 at 17:07
I did not check that the original eigenvectors are OK, but there is a sign difference in the top term of your vertical vector.
– Andrei
Nov 19 at 17:00
I did not check that the original eigenvectors are OK, but there is a sign difference in the top term of your vertical vector.
– Andrei
Nov 19 at 17:00
@Andrei Like the one you do when you take the Hermitian of something?
– Saucy O'Path
Nov 19 at 17:01
@Andrei Like the one you do when you take the Hermitian of something?
– Saucy O'Path
Nov 19 at 17:01
Ah! Thank you so much
– Rzmwood
Nov 19 at 17:02
Ah! Thank you so much
– Rzmwood
Nov 19 at 17:02
@SaucyO'Path thanks. I somehow missed that.
– Andrei
Nov 19 at 17:07
@SaucyO'Path thanks. I somehow missed that.
– Andrei
Nov 19 at 17:07
add a comment |
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1
Maybe they are hermitian orthogonal.
– Charlie Frohman
Nov 19 at 16:55