Help with Autonne-Takagi factorization of a complex symmetric matrix.
Let $A=A_1i+A_2$ with $A$ non singular. Now let
$$B =begin{bmatrix}
A_1 & A_2\
A_2 & -A_1
end{bmatrix}$$
With $A_1$, $A_2$ and $B$ symmetric. Is it true that:
1) $B$ is non singular
2) $B begin{bmatrix}
x \
-y
end{bmatrix}=lambda begin{bmatrix}
x \
-y
end{bmatrix}$ if and only if $B begin{bmatrix}
x \
-y
end{bmatrix}=-lambda begin{bmatrix}
x \
-y
end{bmatrix}$ so the
eigenvalues of $B$ appear in $+-$ pairs.
3) Let $begin{bmatrix}
x_1 \
-y_1
end{bmatrix},dots, begin{bmatrix}
x_n \
-y_n
end{bmatrix}$ be the orthonormal eigenvectors of $B$ associated
with its positive eigenvalues $lambda_1,dots,lambda_n$. Let $X=begin{bmatrix}
x_1 & dots & x_n
end{bmatrix}$, $Y=begin{bmatrix}
y_1 & dots & y_n
end{bmatrix}$, $Sigma=diag(lambda_1,dots,lambda_n)$, $V =begin{bmatrix}
X & Y\
-Y & X
end{bmatrix}$ and $Lambda = Sigma oplus (-Sigma)$. Then $V$ is real orthogonal and $B = VLambda V^T$. Let $U = X - iY$.
Explain why $U$ is unitary and show that $USigma U ^T=A$.
matrices factoring symmetry
edited Nov 25 '18 at 17:19
A.Γ.
22k32455
asked Apr 19 '14 at 8:39
Paula
312
add a comment |
Let $A=A_1i+A_2$ with $A$ non singular. Now let
$$B =begin{bmatrix}
A_1 & A_2\
A_2 & -A_1
end{bmatrix}$$
With $A_1$, $A_2$ and $B$ symmetric. Is it true that:
1) $B$ is non singular
2) $B begin{bmatrix}
x \
-y
end{bmatrix}=lambda begin{bmatrix}
x \
-y
end{bmatrix}$ if and only if $B begin{bmatrix}
x \
-y
end{bmatrix}=-lambda begin{bmatrix}
x \
-y
end{bmatrix}$ so the
eigenvalues of $B$ appear in $+-$ pairs.
3) Let $begin{bmatrix}
x_1 \
-y_1
end{bmatrix},dots, begin{bmatrix}
x_n \
-y_n
end{bmatrix}$ be the orthonormal eigenvectors of $B$ associated
with its positive eigenvalues $lambda_1,dots,lambda_n$. Let $X=begin{bmatrix}
x_1 & dots & x_n
end{bmatrix}$, $Y=begin{bmatrix}
y_1 & dots & y_n
end{bmatrix}$, $Sigma=diag(lambda_1,dots,lambda_n)$, $V =begin{bmatrix}
X & Y\
-Y & X
end{bmatrix}$ and $Lambda = Sigma oplus (-Sigma)$. Then $V$ is real orthogonal and $B = VLambda V^T$. Let $U = X - iY$.
Explain why $U$ is unitary and show that $USigma U ^T=A$.
matrices factoring symmetry
edited Nov 25 '18 at 17:19
A.Γ.
22k32455
asked Apr 19 '14 at 8:39
Paula
312
Can you show your ideas about the problem?
– egreg
Apr 19 '14 at 9:05
1)I think that DET(B)=DET($-A_1^2$-$A_2^2$) since A is non singular $-A_1^2-A_2^2 neq 0$ then B is non sigular. 2) can be seen by doing the plain calculation but I don't know if there is another way 3) Since A can be written as $U Sigma U$ and $Sigma$ has the eigenvalues of A, according to the takagi factorization U is unitary. But it sounds still vague...
– Paula
Apr 19 '14 at 9:32
add a comment |
Let $A=A_1i+A_2$ with $A$ non singular. Now let
$$B =begin{bmatrix}
A_1 & A_2\
A_2 & -A_1
end{bmatrix}$$
With $A_1$, $A_2$ and $B$ symmetric. Is it true that:
1) $B$ is non singular
2) $B begin{bmatrix}
x \
-y
end{bmatrix}=lambda begin{bmatrix}
x \
-y
end{bmatrix}$ if and only if $B begin{bmatrix}
x \
-y
end{bmatrix}=-lambda begin{bmatrix}
x \
-y
end{bmatrix}$ so the
eigenvalues of $B$ appear in $+-$ pairs.
3) Let $begin{bmatrix}
x_1 \
-y_1
end{bmatrix},dots, begin{bmatrix}
x_n \
-y_n
end{bmatrix}$ be the orthonormal eigenvectors of $B$ associated
with its positive eigenvalues $lambda_1,dots,lambda_n$. Let $X=begin{bmatrix}
x_1 & dots & x_n
end{bmatrix}$, $Y=begin{bmatrix}
y_1 & dots & y_n
end{bmatrix}$, $Sigma=diag(lambda_1,dots,lambda_n)$, $V =begin{bmatrix}
X & Y\
-Y & X
end{bmatrix}$ and $Lambda = Sigma oplus (-Sigma)$. Then $V$ is real orthogonal and $B = VLambda V^T$. Let $U = X - iY$.
Explain why $U$ is unitary and show that $USigma U ^T=A$.
matrices factoring symmetry
edited Nov 25 '18 at 17:19
A.Γ.
22k32455
asked Apr 19 '14 at 8:39
Paula
312
Let $A=A_1i+A_2$ with $A$ non singular. Now let
$$B =begin{bmatrix}
A_1 & A_2\
A_2 & -A_1
end{bmatrix}$$
With $A_1$, $A_2$ and $B$ symmetric. Is it true that:
1) $B$ is non singular
2) $B begin{bmatrix}
x \
-y
end{bmatrix}=lambda begin{bmatrix}
x \
-y
end{bmatrix}$ if and only if $B begin{bmatrix}
x \
-y
end{bmatrix}=-lambda begin{bmatrix}
x \
-y
end{bmatrix}$ so the
eigenvalues of $B$ appear in $+-$ pairs.
3) Let $begin{bmatrix}
x_1 \
-y_1
end{bmatrix},dots, begin{bmatrix}
x_n \
-y_n
end{bmatrix}$ be the orthonormal eigenvectors of $B$ associated
with its positive eigenvalues $lambda_1,dots,lambda_n$. Let $X=begin{bmatrix}
x_1 & dots & x_n
end{bmatrix}$, $Y=begin{bmatrix}
y_1 & dots & y_n
end{bmatrix}$, $Sigma=diag(lambda_1,dots,lambda_n)$, $V =begin{bmatrix}
X & Y\
-Y & X
end{bmatrix}$ and $Lambda = Sigma oplus (-Sigma)$. Then $V$ is real orthogonal and $B = VLambda V^T$. Let $U = X - iY$.
Explain why $U$ is unitary and show that $USigma U ^T=A$.
matrices factoring symmetry
matrices factoring symmetry
edited Nov 25 '18 at 17:19
A.Γ.
22k32455
asked Apr 19 '14 at 8:39
Paula
312
edited Nov 25 '18 at 17:19
A.Γ.
22k32455
asked Apr 19 '14 at 8:39
Paula
312
edited Nov 25 '18 at 17:19
A.Γ.
22k32455
edited Nov 25 '18 at 17:19
A.Γ.
22k32455
edited Nov 25 '18 at 17:19
A.Γ.
22k32455
22k32455
asked Apr 19 '14 at 8:39
Paula
312
asked Apr 19 '14 at 8:39
Paula
312
asked Apr 19 '14 at 8:39
Paula
312
312
Can you show your ideas about the problem?
– egreg
Apr 19 '14 at 9:05
1)I think that DET(B)=DET($-A_1^2$-$A_2^2$) since A is non singular $-A_1^2-A_2^2 neq 0$ then B is non sigular. 2) can be seen by doing the plain calculation but I don't know if there is another way 3) Since A can be written as $U Sigma U$ and $Sigma$ has the eigenvalues of A, according to the takagi factorization U is unitary. But it sounds still vague...
– Paula
Apr 19 '14 at 9:32
add a comment |
Can you show your ideas about the problem?
– egreg
Apr 19 '14 at 9:05
1)I think that DET(B)=DET($-A_1^2$-$A_2^2$) since A is non singular $-A_1^2-A_2^2 neq 0$ then B is non sigular. 2) can be seen by doing the plain calculation but I don't know if there is another way 3) Since A can be written as $U Sigma U$ and $Sigma$ has the eigenvalues of A, according to the takagi factorization U is unitary. But it sounds still vague...
– Paula
Apr 19 '14 at 9:32
Can you show your ideas about the problem?
– egreg
Apr 19 '14 at 9:05
Can you show your ideas about the problem?
– egreg
Apr 19 '14 at 9:05
1)I think that DET(B)=DET($-A_1^2$-$A_2^2$) since A is non singular $-A_1^2-A_2^2 neq 0$ then B is non sigular. 2) can be seen by doing the plain calculation but I don't know if there is another way 3) Since A can be written as $U Sigma U$ and $Sigma$ has the eigenvalues of A, according to the takagi factorization U is unitary. But it sounds still vague...
– Paula
Apr 19 '14 at 9:32
1)I think that DET(B)=DET($-A_1^2$-$A_2^2$) since A is non singular $-A_1^2-A_2^2 neq 0$ then B is non sigular. 2) can be seen by doing the plain calculation but I don't know if there is another way 3) Since A can be written as $U Sigma U$ and $Sigma$ has the eigenvalues of A, according to the takagi factorization U is unitary. But it sounds still vague...
– Paula
Apr 19 '14 at 9:32
add a comment |
1 Answer
1
active
oldest
votes
I am not going to do the homework for you. I have two remarks, though.
- Part (1) is actually true for any real matrices $A_1$ and $A_2$. It has nothing to do with matrix symmetry. Note that
$$
pmatrix{I&-iI\ 0&I} pmatrix{A_1&A_2\ A_2&-A_1} pmatrix{I&0\ iI&I}
= pmatrix{0&A_2+iA_1\ A_2-iA_1&-A_1}
$$
and
$$
det(A_2+iA_1),det(A_2-iA_1)=det(A_2+iA_1),overline{det(A_2+iA_1)}=
|det(A_2+iA_1)|^2.
$$ - Part (2) as it stands is false. Since $B$ is nonsingular, $lambdane0$. Therefore $lambda$ and $-lambda$ are different eigenvalues and they cannot share the same eigenvector $pmatrix{x\ -y}$. The correct statement should be this: if $lambda$ and $pmatrix{x\ -y}$ form an eigenpair of $B$, then $-lambda$ and $pmatrix{y\ x}$ is also an eigenpair of $B$. Therefore, eigenvalues of $B$ occur in pairs of the form $pmlambda$.
answered Apr 19 '14 at 12:14
user1551
71.5k566125
add a comment |
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active
oldest
votes
I am not going to do the homework for you. I have two remarks, though.
- Part (1) is actually true for any real matrices $A_1$ and $A_2$. It has nothing to do with matrix symmetry. Note that
$$
pmatrix{I&-iI\ 0&I} pmatrix{A_1&A_2\ A_2&-A_1} pmatrix{I&0\ iI&I}
= pmatrix{0&A_2+iA_1\ A_2-iA_1&-A_1}
$$
and
$$
det(A_2+iA_1),det(A_2-iA_1)=det(A_2+iA_1),overline{det(A_2+iA_1)}=
|det(A_2+iA_1)|^2.
$$ - Part (2) as it stands is false. Since $B$ is nonsingular, $lambdane0$. Therefore $lambda$ and $-lambda$ are different eigenvalues and they cannot share the same eigenvector $pmatrix{x\ -y}$. The correct statement should be this: if $lambda$ and $pmatrix{x\ -y}$ form an eigenpair of $B$, then $-lambda$ and $pmatrix{y\ x}$ is also an eigenpair of $B$. Therefore, eigenvalues of $B$ occur in pairs of the form $pmlambda$.
answered Apr 19 '14 at 12:14
user1551
71.5k566125
add a comment |
I am not going to do the homework for you. I have two remarks, though.
- Part (1) is actually true for any real matrices $A_1$ and $A_2$. It has nothing to do with matrix symmetry. Note that
$$
pmatrix{I&-iI\ 0&I} pmatrix{A_1&A_2\ A_2&-A_1} pmatrix{I&0\ iI&I}
= pmatrix{0&A_2+iA_1\ A_2-iA_1&-A_1}
$$
and
$$
det(A_2+iA_1),det(A_2-iA_1)=det(A_2+iA_1),overline{det(A_2+iA_1)}=
|det(A_2+iA_1)|^2.
$$ - Part (2) as it stands is false. Since $B$ is nonsingular, $lambdane0$. Therefore $lambda$ and $-lambda$ are different eigenvalues and they cannot share the same eigenvector $pmatrix{x\ -y}$. The correct statement should be this: if $lambda$ and $pmatrix{x\ -y}$ form an eigenpair of $B$, then $-lambda$ and $pmatrix{y\ x}$ is also an eigenpair of $B$. Therefore, eigenvalues of $B$ occur in pairs of the form $pmlambda$.
answered Apr 19 '14 at 12:14
user1551
71.5k566125
add a comment |
I am not going to do the homework for you. I have two remarks, though.
- Part (1) is actually true for any real matrices $A_1$ and $A_2$. It has nothing to do with matrix symmetry. Note that
$$
pmatrix{I&-iI\ 0&I} pmatrix{A_1&A_2\ A_2&-A_1} pmatrix{I&0\ iI&I}
= pmatrix{0&A_2+iA_1\ A_2-iA_1&-A_1}
$$
and
$$
det(A_2+iA_1),det(A_2-iA_1)=det(A_2+iA_1),overline{det(A_2+iA_1)}=
|det(A_2+iA_1)|^2.
$$ - Part (2) as it stands is false. Since $B$ is nonsingular, $lambdane0$. Therefore $lambda$ and $-lambda$ are different eigenvalues and they cannot share the same eigenvector $pmatrix{x\ -y}$. The correct statement should be this: if $lambda$ and $pmatrix{x\ -y}$ form an eigenpair of $B$, then $-lambda$ and $pmatrix{y\ x}$ is also an eigenpair of $B$. Therefore, eigenvalues of $B$ occur in pairs of the form $pmlambda$.
answered Apr 19 '14 at 12:14
user1551
71.5k566125
I am not going to do the homework for you. I have two remarks, though.
- Part (1) is actually true for any real matrices $A_1$ and $A_2$. It has nothing to do with matrix symmetry. Note that
$$
pmatrix{I&-iI\ 0&I} pmatrix{A_1&A_2\ A_2&-A_1} pmatrix{I&0\ iI&I}
= pmatrix{0&A_2+iA_1\ A_2-iA_1&-A_1}
$$
and
$$
det(A_2+iA_1),det(A_2-iA_1)=det(A_2+iA_1),overline{det(A_2+iA_1)}=
|det(A_2+iA_1)|^2.
$$ - Part (2) as it stands is false. Since $B$ is nonsingular, $lambdane0$. Therefore $lambda$ and $-lambda$ are different eigenvalues and they cannot share the same eigenvector $pmatrix{x\ -y}$. The correct statement should be this: if $lambda$ and $pmatrix{x\ -y}$ form an eigenpair of $B$, then $-lambda$ and $pmatrix{y\ x}$ is also an eigenpair of $B$. Therefore, eigenvalues of $B$ occur in pairs of the form $pmlambda$.
answered Apr 19 '14 at 12:14
user1551
71.5k566125
answered Apr 19 '14 at 12:14
user1551
71.5k566125
answered Apr 19 '14 at 12:14
user1551
71.5k566125
answered Apr 19 '14 at 12:14
user1551
71.5k566125
71.5k566125
add a comment |
add a comment |
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Can you show your ideas about the problem?
– egreg
Apr 19 '14 at 9:05
1)I think that DET(B)=DET($-A_1^2$-$A_2^2$) since A is non singular $-A_1^2-A_2^2 neq 0$ then B is non sigular. 2) can be seen by doing the plain calculation but I don't know if there is another way 3) Since A can be written as $U Sigma U$ and $Sigma$ has the eigenvalues of A, according to the takagi factorization U is unitary. But it sounds still vague...
– Paula
Apr 19 '14 at 9:32