Prove that $sum_{i,j=1}^Nx_ialpha_{ij}x_j>0$ for a non-symmetric matrix
$begingroup$
Let $A=(alpha)_{i,j=1}^N$ a real square matrix. The matrix $A$ is not necessary symmetric. I want to prove that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j>0,;;;forall (x_1,ldots,x_N)inmathbb{R}^Nbackslashleftlbrace 0rightrbrace.$$
If matrix $A$ was symmetric, one way to do this, is to prove that all eigenvalues are positive. How can I deal with this case?
linear-algebra matrices
asked Dec 28 '18 at 19:31
math_lovermath_lover
9710
$endgroup$
add a comment |
$begingroup$
Let $A=(alpha)_{i,j=1}^N$ a real square matrix. The matrix $A$ is not necessary symmetric. I want to prove that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j>0,;;;forall (x_1,ldots,x_N)inmathbb{R}^Nbackslashleftlbrace 0rightrbrace.$$
If matrix $A$ was symmetric, one way to do this, is to prove that all eigenvalues are positive. How can I deal with this case?
linear-algebra matrices
asked Dec 28 '18 at 19:31
math_lovermath_lover
9710
$endgroup$
1
$begingroup$
Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
$endgroup$
– anomaly
Dec 28 '18 at 21:22
add a comment |
$begingroup$
Let $A=(alpha)_{i,j=1}^N$ a real square matrix. The matrix $A$ is not necessary symmetric. I want to prove that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j>0,;;;forall (x_1,ldots,x_N)inmathbb{R}^Nbackslashleftlbrace 0rightrbrace.$$
If matrix $A$ was symmetric, one way to do this, is to prove that all eigenvalues are positive. How can I deal with this case?
linear-algebra matrices
asked Dec 28 '18 at 19:31
math_lovermath_lover
9710
$endgroup$
Let $A=(alpha)_{i,j=1}^N$ a real square matrix. The matrix $A$ is not necessary symmetric. I want to prove that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j>0,;;;forall (x_1,ldots,x_N)inmathbb{R}^Nbackslashleftlbrace 0rightrbrace.$$
If matrix $A$ was symmetric, one way to do this, is to prove that all eigenvalues are positive. How can I deal with this case?
linear-algebra matrices
linear-algebra matrices
asked Dec 28 '18 at 19:31
math_lovermath_lover
9710
asked Dec 28 '18 at 19:31
math_lovermath_lover
9710
asked Dec 28 '18 at 19:31
math_lovermath_lover
9710
asked Dec 28 '18 at 19:31
math_lovermath_lover
9710
asked Dec 28 '18 at 19:31
math_lovermath_lover
9710
9710
1
$begingroup$
Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
$endgroup$
– anomaly
Dec 28 '18 at 21:22
add a comment |
1
$begingroup$
Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
$endgroup$
– anomaly
Dec 28 '18 at 21:22
1
1
$begingroup$
Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
$endgroup$
– anomaly
Dec 28 '18 at 21:22
$begingroup$
Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
$endgroup$
– anomaly
Dec 28 '18 at 21:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Any real square matrix $A$ can be written (uniquely) as the sum of a symmetric matrix and a skew-symmetric matrix
$$A = A_{text{sym}} + A_{text{sq}}$$
where
$$A_{text{sym}} = frac{1}{2} ( A + A^t) \
A_{text{sq}} = frac{1}{2} ( A - A^t) $$
Say $A_{text{sym}}= (tilde alpha_{ij})$. It is easy to see that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j=sum_{i,j=1}^Nx_itildealpha_{ij}x_j $$ ( that is, $A$ and $A_{text{sym}}$ produce the same quadratic form). Now you are reduced to checking whether the real symmetric matrix $A_{text{sym}}$ is positive definite, which is standard.
answered Dec 28 '18 at 21:14
Orest BucicovschiOrest Bucicovschi
28.6k31748
$endgroup$
add a comment |
$begingroup$
This is false. Consider $A=-I_N$ and $x_1=x_2=cdots=x_N=1$. Then
$$sum_{i,j=1}^Nx_ia_{i,j}x_j=-N.$$
answered Dec 28 '18 at 19:35
Ben WBen W
2,734918
$endgroup$
$begingroup$
It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
$endgroup$
– anomaly
Dec 28 '18 at 21:23
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any real square matrix $A$ can be written (uniquely) as the sum of a symmetric matrix and a skew-symmetric matrix
$$A = A_{text{sym}} + A_{text{sq}}$$
where
$$A_{text{sym}} = frac{1}{2} ( A + A^t) \
A_{text{sq}} = frac{1}{2} ( A - A^t) $$
Say $A_{text{sym}}= (tilde alpha_{ij})$. It is easy to see that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j=sum_{i,j=1}^Nx_itildealpha_{ij}x_j $$ ( that is, $A$ and $A_{text{sym}}$ produce the same quadratic form). Now you are reduced to checking whether the real symmetric matrix $A_{text{sym}}$ is positive definite, which is standard.
answered Dec 28 '18 at 21:14
Orest BucicovschiOrest Bucicovschi
28.6k31748
$endgroup$
add a comment |
$begingroup$
Any real square matrix $A$ can be written (uniquely) as the sum of a symmetric matrix and a skew-symmetric matrix
$$A = A_{text{sym}} + A_{text{sq}}$$
where
$$A_{text{sym}} = frac{1}{2} ( A + A^t) \
A_{text{sq}} = frac{1}{2} ( A - A^t) $$
Say $A_{text{sym}}= (tilde alpha_{ij})$. It is easy to see that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j=sum_{i,j=1}^Nx_itildealpha_{ij}x_j $$ ( that is, $A$ and $A_{text{sym}}$ produce the same quadratic form). Now you are reduced to checking whether the real symmetric matrix $A_{text{sym}}$ is positive definite, which is standard.
answered Dec 28 '18 at 21:14
Orest BucicovschiOrest Bucicovschi
28.6k31748
$endgroup$
add a comment |
$begingroup$
Any real square matrix $A$ can be written (uniquely) as the sum of a symmetric matrix and a skew-symmetric matrix
$$A = A_{text{sym}} + A_{text{sq}}$$
where
$$A_{text{sym}} = frac{1}{2} ( A + A^t) \
A_{text{sq}} = frac{1}{2} ( A - A^t) $$
Say $A_{text{sym}}= (tilde alpha_{ij})$. It is easy to see that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j=sum_{i,j=1}^Nx_itildealpha_{ij}x_j $$ ( that is, $A$ and $A_{text{sym}}$ produce the same quadratic form). Now you are reduced to checking whether the real symmetric matrix $A_{text{sym}}$ is positive definite, which is standard.
answered Dec 28 '18 at 21:14
Orest BucicovschiOrest Bucicovschi
28.6k31748
$endgroup$
Any real square matrix $A$ can be written (uniquely) as the sum of a symmetric matrix and a skew-symmetric matrix
$$A = A_{text{sym}} + A_{text{sq}}$$
where
$$A_{text{sym}} = frac{1}{2} ( A + A^t) \
A_{text{sq}} = frac{1}{2} ( A - A^t) $$
Say $A_{text{sym}}= (tilde alpha_{ij})$. It is easy to see that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j=sum_{i,j=1}^Nx_itildealpha_{ij}x_j $$ ( that is, $A$ and $A_{text{sym}}$ produce the same quadratic form). Now you are reduced to checking whether the real symmetric matrix $A_{text{sym}}$ is positive definite, which is standard.
answered Dec 28 '18 at 21:14
Orest BucicovschiOrest Bucicovschi
28.6k31748
answered Dec 28 '18 at 21:14
Orest BucicovschiOrest Bucicovschi
28.6k31748
answered Dec 28 '18 at 21:14
Orest BucicovschiOrest Bucicovschi
28.6k31748
answered Dec 28 '18 at 21:14
Orest BucicovschiOrest Bucicovschi
28.6k31748
28.6k31748
add a comment |
add a comment |
$begingroup$
This is false. Consider $A=-I_N$ and $x_1=x_2=cdots=x_N=1$. Then
$$sum_{i,j=1}^Nx_ia_{i,j}x_j=-N.$$
answered Dec 28 '18 at 19:35
Ben WBen W
2,734918
$endgroup$
$begingroup$
It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
$endgroup$
– anomaly
Dec 28 '18 at 21:23
add a comment |
$begingroup$
This is false. Consider $A=-I_N$ and $x_1=x_2=cdots=x_N=1$. Then
$$sum_{i,j=1}^Nx_ia_{i,j}x_j=-N.$$
answered Dec 28 '18 at 19:35
Ben WBen W
2,734918
$endgroup$
$begingroup$
It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
$endgroup$
– anomaly
Dec 28 '18 at 21:23
add a comment |
$begingroup$
This is false. Consider $A=-I_N$ and $x_1=x_2=cdots=x_N=1$. Then
$$sum_{i,j=1}^Nx_ia_{i,j}x_j=-N.$$
answered Dec 28 '18 at 19:35
Ben WBen W
2,734918
$endgroup$
This is false. Consider $A=-I_N$ and $x_1=x_2=cdots=x_N=1$. Then
$$sum_{i,j=1}^Nx_ia_{i,j}x_j=-N.$$
answered Dec 28 '18 at 19:35
Ben WBen W
2,734918
answered Dec 28 '18 at 19:35
Ben WBen W
2,734918
answered Dec 28 '18 at 19:35
Ben WBen W
2,734918
answered Dec 28 '18 at 19:35
Ben WBen W
2,734918
2,734918
$begingroup$
It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
$endgroup$
– anomaly
Dec 28 '18 at 21:23
add a comment |
$begingroup$
It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
$endgroup$
– anomaly
Dec 28 '18 at 21:23
$begingroup$
It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
$endgroup$
– anomaly
Dec 28 '18 at 21:23
$begingroup$
It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
$endgroup$
– anomaly
Dec 28 '18 at 21:23
add a comment |
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$begingroup$
Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
$endgroup$
– anomaly
Dec 28 '18 at 21:22