Decomposition of the same matrix with different unitary matrices
Let matrix $M$ be Hermitian
$$M = U_1Lambda U^*_1 qquad qquad (1)$$
$$M = U_2Lambda U^*_2 qquad qquad (2)$$
$U_1$ and $U_2$ are unitary matrices.
How can we prove that we can have $U_1 ne U_2$ , with exactly the same diagonal $Lambda$?
Example
MATLAB code:
M = [4 3-i; 3+i 10]; %Hermitian
[U1, diag]= eig(M); % i.e. M = U1*diag*U1'
U2 = U1;
U2(:,1) = -U2(:,1);
U1*diag*U1'; % gives M
U2*diag*U2'; % also gives M
% U1 is not equal to U2
% U1 =
% 0.8716 - 0.2905i 0.3746 - 0.1249i
% -0.3948 + 0.0000i 0.9188 + 0.0000i
% U2 =
% -0.8716 + 0.2905i 0.3746 - 0.1249i
% 0.3948 + 0.0000i 0.9188 + 0.0000i
% U1*diag*U1' and U2*diag*U2' % they both return M
% ans =
% 4.0000 - 0.0000i 3.0000 - 1.0000i
% 3.0000 + 1.0000i 10.0000 + 0.0000i
eigenvalues-eigenvectors matrix-decomposition
asked Nov 29 at 21:37
Kay
113
add a comment |
Let matrix $M$ be Hermitian
$$M = U_1Lambda U^*_1 qquad qquad (1)$$
$$M = U_2Lambda U^*_2 qquad qquad (2)$$
$U_1$ and $U_2$ are unitary matrices.
How can we prove that we can have $U_1 ne U_2$ , with exactly the same diagonal $Lambda$?
Example
MATLAB code:
M = [4 3-i; 3+i 10]; %Hermitian
[U1, diag]= eig(M); % i.e. M = U1*diag*U1'
U2 = U1;
U2(:,1) = -U2(:,1);
U1*diag*U1'; % gives M
U2*diag*U2'; % also gives M
% U1 is not equal to U2
% U1 =
% 0.8716 - 0.2905i 0.3746 - 0.1249i
% -0.3948 + 0.0000i 0.9188 + 0.0000i
% U2 =
% -0.8716 + 0.2905i 0.3746 - 0.1249i
% 0.3948 + 0.0000i 0.9188 + 0.0000i
% U1*diag*U1' and U2*diag*U2' % they both return M
% ans =
% 4.0000 - 0.0000i 3.0000 - 1.0000i
% 3.0000 + 1.0000i 10.0000 + 0.0000i
eigenvalues-eigenvectors matrix-decomposition
asked Nov 29 at 21:37
Kay
113
1
Hint: an eigenvector is still an eigenvector if you multiply it by $-1$
– N74
Nov 29 at 21:43
That's right. I thought of that clause, but I wondered if the explanation is sufficient. Well, it should. Thanks.
– Kay
Nov 29 at 23:13
add a comment |
Let matrix $M$ be Hermitian
$$M = U_1Lambda U^*_1 qquad qquad (1)$$
$$M = U_2Lambda U^*_2 qquad qquad (2)$$
$U_1$ and $U_2$ are unitary matrices.
How can we prove that we can have $U_1 ne U_2$ , with exactly the same diagonal $Lambda$?
Example
MATLAB code:
M = [4 3-i; 3+i 10]; %Hermitian
[U1, diag]= eig(M); % i.e. M = U1*diag*U1'
U2 = U1;
U2(:,1) = -U2(:,1);
U1*diag*U1'; % gives M
U2*diag*U2'; % also gives M
% U1 is not equal to U2
% U1 =
% 0.8716 - 0.2905i 0.3746 - 0.1249i
% -0.3948 + 0.0000i 0.9188 + 0.0000i
% U2 =
% -0.8716 + 0.2905i 0.3746 - 0.1249i
% 0.3948 + 0.0000i 0.9188 + 0.0000i
% U1*diag*U1' and U2*diag*U2' % they both return M
% ans =
% 4.0000 - 0.0000i 3.0000 - 1.0000i
% 3.0000 + 1.0000i 10.0000 + 0.0000i
eigenvalues-eigenvectors matrix-decomposition
asked Nov 29 at 21:37
Kay
113
Let matrix $M$ be Hermitian
$$M = U_1Lambda U^*_1 qquad qquad (1)$$
$$M = U_2Lambda U^*_2 qquad qquad (2)$$
$U_1$ and $U_2$ are unitary matrices.
How can we prove that we can have $U_1 ne U_2$ , with exactly the same diagonal $Lambda$?
Example
MATLAB code:
M = [4 3-i; 3+i 10]; %Hermitian
[U1, diag]= eig(M); % i.e. M = U1*diag*U1'
U2 = U1;
U2(:,1) = -U2(:,1);
U1*diag*U1'; % gives M
U2*diag*U2'; % also gives M
% U1 is not equal to U2
% U1 =
% 0.8716 - 0.2905i 0.3746 - 0.1249i
% -0.3948 + 0.0000i 0.9188 + 0.0000i
% U2 =
% -0.8716 + 0.2905i 0.3746 - 0.1249i
% 0.3948 + 0.0000i 0.9188 + 0.0000i
% U1*diag*U1' and U2*diag*U2' % they both return M
% ans =
% 4.0000 - 0.0000i 3.0000 - 1.0000i
% 3.0000 + 1.0000i 10.0000 + 0.0000i
eigenvalues-eigenvectors matrix-decomposition
eigenvalues-eigenvectors matrix-decomposition
asked Nov 29 at 21:37
Kay
113
asked Nov 29 at 21:37
Kay
113
asked Nov 29 at 21:37
Kay
113
asked Nov 29 at 21:37
Kay
113
asked Nov 29 at 21:37
Kay
113
113
1
Hint: an eigenvector is still an eigenvector if you multiply it by $-1$
– N74
Nov 29 at 21:43
That's right. I thought of that clause, but I wondered if the explanation is sufficient. Well, it should. Thanks.
– Kay
Nov 29 at 23:13
add a comment |
1
Hint: an eigenvector is still an eigenvector if you multiply it by $-1$
– N74
Nov 29 at 21:43
That's right. I thought of that clause, but I wondered if the explanation is sufficient. Well, it should. Thanks.
– Kay
Nov 29 at 23:13
1
1
Hint: an eigenvector is still an eigenvector if you multiply it by $-1$
– N74
Nov 29 at 21:43
Hint: an eigenvector is still an eigenvector if you multiply it by $-1$
– N74
Nov 29 at 21:43
That's right. I thought of that clause, but I wondered if the explanation is sufficient. Well, it should. Thanks.
– Kay
Nov 29 at 23:13
That's right. I thought of that clause, but I wondered if the explanation is sufficient. Well, it should. Thanks.
– Kay
Nov 29 at 23:13
add a comment |
1 Answer
1
active
oldest
votes
The eigenvectors of a matrix a determined up to a phase $e^{iphi}$, indeed if ${bf u}$ is an eigenvector of $A$ with eigenvalue $lambda$ then
begin{eqnarray}
A {bf u} &=& lambda {bf u} \
Rightarrow~~~A(e^{iphi}{bf u}) &=& lambda (e^{iphi}{bf u})
end{eqnarray}
that means that $e^{iphi}{bf u}$ is also an eigenvector of $A$ with eigenvalue $lambda$. Also note both ${bf u}$ and $e^{iphi}{bf u}$ have the same norm
$$
require{cancel}
|e^{iphi}{bf u}| = cancelto{1}{|e^{iphi} |}| {bf u}| = | {bf u}|
$$
So if the eigenvectors are normalized (which is actually your case), then after multiplying them by an arbitrary phase they remain normalized!
answered Nov 29 at 23:25
caverac
13.2k21029
This does the explanation. I appreciate.
– Kay
Nov 29 at 23:53
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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oldest
votes
The eigenvectors of a matrix a determined up to a phase $e^{iphi}$, indeed if ${bf u}$ is an eigenvector of $A$ with eigenvalue $lambda$ then
begin{eqnarray}
A {bf u} &=& lambda {bf u} \
Rightarrow~~~A(e^{iphi}{bf u}) &=& lambda (e^{iphi}{bf u})
end{eqnarray}
that means that $e^{iphi}{bf u}$ is also an eigenvector of $A$ with eigenvalue $lambda$. Also note both ${bf u}$ and $e^{iphi}{bf u}$ have the same norm
$$
require{cancel}
|e^{iphi}{bf u}| = cancelto{1}{|e^{iphi} |}| {bf u}| = | {bf u}|
$$
So if the eigenvectors are normalized (which is actually your case), then after multiplying them by an arbitrary phase they remain normalized!
answered Nov 29 at 23:25
caverac
13.2k21029
This does the explanation. I appreciate.
– Kay
Nov 29 at 23:53
add a comment |
The eigenvectors of a matrix a determined up to a phase $e^{iphi}$, indeed if ${bf u}$ is an eigenvector of $A$ with eigenvalue $lambda$ then
begin{eqnarray}
A {bf u} &=& lambda {bf u} \
Rightarrow~~~A(e^{iphi}{bf u}) &=& lambda (e^{iphi}{bf u})
end{eqnarray}
that means that $e^{iphi}{bf u}$ is also an eigenvector of $A$ with eigenvalue $lambda$. Also note both ${bf u}$ and $e^{iphi}{bf u}$ have the same norm
$$
require{cancel}
|e^{iphi}{bf u}| = cancelto{1}{|e^{iphi} |}| {bf u}| = | {bf u}|
$$
So if the eigenvectors are normalized (which is actually your case), then after multiplying them by an arbitrary phase they remain normalized!
answered Nov 29 at 23:25
caverac
13.2k21029
This does the explanation. I appreciate.
– Kay
Nov 29 at 23:53
add a comment |
The eigenvectors of a matrix a determined up to a phase $e^{iphi}$, indeed if ${bf u}$ is an eigenvector of $A$ with eigenvalue $lambda$ then
begin{eqnarray}
A {bf u} &=& lambda {bf u} \
Rightarrow~~~A(e^{iphi}{bf u}) &=& lambda (e^{iphi}{bf u})
end{eqnarray}
that means that $e^{iphi}{bf u}$ is also an eigenvector of $A$ with eigenvalue $lambda$. Also note both ${bf u}$ and $e^{iphi}{bf u}$ have the same norm
$$
require{cancel}
|e^{iphi}{bf u}| = cancelto{1}{|e^{iphi} |}| {bf u}| = | {bf u}|
$$
So if the eigenvectors are normalized (which is actually your case), then after multiplying them by an arbitrary phase they remain normalized!
answered Nov 29 at 23:25
caverac
13.2k21029
The eigenvectors of a matrix a determined up to a phase $e^{iphi}$, indeed if ${bf u}$ is an eigenvector of $A$ with eigenvalue $lambda$ then
begin{eqnarray}
A {bf u} &=& lambda {bf u} \
Rightarrow~~~A(e^{iphi}{bf u}) &=& lambda (e^{iphi}{bf u})
end{eqnarray}
that means that $e^{iphi}{bf u}$ is also an eigenvector of $A$ with eigenvalue $lambda$. Also note both ${bf u}$ and $e^{iphi}{bf u}$ have the same norm
$$
require{cancel}
|e^{iphi}{bf u}| = cancelto{1}{|e^{iphi} |}| {bf u}| = | {bf u}|
$$
So if the eigenvectors are normalized (which is actually your case), then after multiplying them by an arbitrary phase they remain normalized!
answered Nov 29 at 23:25
caverac
13.2k21029
answered Nov 29 at 23:25
caverac
13.2k21029
answered Nov 29 at 23:25
caverac
13.2k21029
answered Nov 29 at 23:25
caverac
13.2k21029
13.2k21029
This does the explanation. I appreciate.
– Kay
Nov 29 at 23:53
add a comment |
This does the explanation. I appreciate.
– Kay
Nov 29 at 23:53
This does the explanation. I appreciate.
– Kay
Nov 29 at 23:53
This does the explanation. I appreciate.
– Kay
Nov 29 at 23:53
add a comment |
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Hint: an eigenvector is still an eigenvector if you multiply it by $-1$
– N74
Nov 29 at 21:43
That's right. I thought of that clause, but I wondered if the explanation is sufficient. Well, it should. Thanks.
– Kay
Nov 29 at 23:13