If the off-diagonal entries of a positive-definite symmetric matrix $A$ are $leq 0$, then $A^{-1}$ has...
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Let $A in operatorname{GL}_n(mathbb R)$ be a symmetric matrix which is positive definite, i.e. $A = Q^tQ$ for some invertible matrix $Q$. Suppose that the off diagonal entries of $A$ are $leq 0$. Must all the entries of $A^{-1}$ be positive?
I am asking because the Cartan matrix $A = (langle alpha_i, alpha_j^{vee} rangle)$ of any reduced root system with base $alpha_1, ... , alpha_l$ is, after factoring out positive scalars out of each column, a symmetric positive definite matrix with this property. The inverse of this matrix gives a formula for how to write the fundamental weights $omega_i$ (that is, the dual basis to the simple coroots) as linear combinations of simple roots. I want to say that each fundamental weight is a positive linear combination of simple roots, which is equivalent to the assertion that $A^{-1}$ has all positive entries.
linear-algebra representation-theory lie-algebras matrix-decomposition root-systems
asked Dec 8 '18 at 22:05
D_SD_S
13.5k51551
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$begingroup$
Let $A in operatorname{GL}_n(mathbb R)$ be a symmetric matrix which is positive definite, i.e. $A = Q^tQ$ for some invertible matrix $Q$. Suppose that the off diagonal entries of $A$ are $leq 0$. Must all the entries of $A^{-1}$ be positive?
I am asking because the Cartan matrix $A = (langle alpha_i, alpha_j^{vee} rangle)$ of any reduced root system with base $alpha_1, ... , alpha_l$ is, after factoring out positive scalars out of each column, a symmetric positive definite matrix with this property. The inverse of this matrix gives a formula for how to write the fundamental weights $omega_i$ (that is, the dual basis to the simple coroots) as linear combinations of simple roots. I want to say that each fundamental weight is a positive linear combination of simple roots, which is equivalent to the assertion that $A^{-1}$ has all positive entries.
linear-algebra representation-theory lie-algebras matrix-decomposition root-systems
asked Dec 8 '18 at 22:05
D_SD_S
13.5k51551
$endgroup$
add a comment |
$begingroup$
Let $A in operatorname{GL}_n(mathbb R)$ be a symmetric matrix which is positive definite, i.e. $A = Q^tQ$ for some invertible matrix $Q$. Suppose that the off diagonal entries of $A$ are $leq 0$. Must all the entries of $A^{-1}$ be positive?
I am asking because the Cartan matrix $A = (langle alpha_i, alpha_j^{vee} rangle)$ of any reduced root system with base $alpha_1, ... , alpha_l$ is, after factoring out positive scalars out of each column, a symmetric positive definite matrix with this property. The inverse of this matrix gives a formula for how to write the fundamental weights $omega_i$ (that is, the dual basis to the simple coroots) as linear combinations of simple roots. I want to say that each fundamental weight is a positive linear combination of simple roots, which is equivalent to the assertion that $A^{-1}$ has all positive entries.
linear-algebra representation-theory lie-algebras matrix-decomposition root-systems
asked Dec 8 '18 at 22:05
D_SD_S
13.5k51551
$endgroup$
Let $A in operatorname{GL}_n(mathbb R)$ be a symmetric matrix which is positive definite, i.e. $A = Q^tQ$ for some invertible matrix $Q$. Suppose that the off diagonal entries of $A$ are $leq 0$. Must all the entries of $A^{-1}$ be positive?
I am asking because the Cartan matrix $A = (langle alpha_i, alpha_j^{vee} rangle)$ of any reduced root system with base $alpha_1, ... , alpha_l$ is, after factoring out positive scalars out of each column, a symmetric positive definite matrix with this property. The inverse of this matrix gives a formula for how to write the fundamental weights $omega_i$ (that is, the dual basis to the simple coroots) as linear combinations of simple roots. I want to say that each fundamental weight is a positive linear combination of simple roots, which is equivalent to the assertion that $A^{-1}$ has all positive entries.
linear-algebra representation-theory lie-algebras matrix-decomposition root-systems
linear-algebra representation-theory lie-algebras matrix-decomposition root-systems
asked Dec 8 '18 at 22:05
D_SD_S
13.5k51551
asked Dec 8 '18 at 22:05
D_SD_S
13.5k51551
asked Dec 8 '18 at 22:05
D_SD_S
13.5k51551
asked Dec 8 '18 at 22:05
D_SD_S
13.5k51551
asked Dec 8 '18 at 22:05
D_SD_S
13.5k51551
13.5k51551
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Yes. If you partition $A$ as $pmatrix{a&b^T\ b&D}$, then $A^{-1}=pmatrix{s^{-1}&-s^{-1}b^TD^{-1}\ ast&ast}$,
where $s=a-b^TD^{-1}b$ is the Schur complement of $D$ in $A$. Since $A$ is positive definite, $s^{-1}$ must be positive. Therefore, by mathematical induction, $D^{-1}$ and in turn $-s^{-1}b^TD^{-1}$ are also entrywise positive. Hence all entries on the first row of $A^{-1}$ are positive. Similarly, the entries on other rows are positive too.
answered Dec 8 '18 at 23:47
user1551user1551
72.7k566127
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Yes. If you partition $A$ as $pmatrix{a&b^T\ b&D}$, then $A^{-1}=pmatrix{s^{-1}&-s^{-1}b^TD^{-1}\ ast&ast}$,
where $s=a-b^TD^{-1}b$ is the Schur complement of $D$ in $A$. Since $A$ is positive definite, $s^{-1}$ must be positive. Therefore, by mathematical induction, $D^{-1}$ and in turn $-s^{-1}b^TD^{-1}$ are also entrywise positive. Hence all entries on the first row of $A^{-1}$ are positive. Similarly, the entries on other rows are positive too.
answered Dec 8 '18 at 23:47
user1551user1551
72.7k566127
$endgroup$
add a comment |
$begingroup$
Yes. If you partition $A$ as $pmatrix{a&b^T\ b&D}$, then $A^{-1}=pmatrix{s^{-1}&-s^{-1}b^TD^{-1}\ ast&ast}$,
where $s=a-b^TD^{-1}b$ is the Schur complement of $D$ in $A$. Since $A$ is positive definite, $s^{-1}$ must be positive. Therefore, by mathematical induction, $D^{-1}$ and in turn $-s^{-1}b^TD^{-1}$ are also entrywise positive. Hence all entries on the first row of $A^{-1}$ are positive. Similarly, the entries on other rows are positive too.
answered Dec 8 '18 at 23:47
user1551user1551
72.7k566127
$endgroup$
add a comment |
$begingroup$
Yes. If you partition $A$ as $pmatrix{a&b^T\ b&D}$, then $A^{-1}=pmatrix{s^{-1}&-s^{-1}b^TD^{-1}\ ast&ast}$,
where $s=a-b^TD^{-1}b$ is the Schur complement of $D$ in $A$. Since $A$ is positive definite, $s^{-1}$ must be positive. Therefore, by mathematical induction, $D^{-1}$ and in turn $-s^{-1}b^TD^{-1}$ are also entrywise positive. Hence all entries on the first row of $A^{-1}$ are positive. Similarly, the entries on other rows are positive too.
answered Dec 8 '18 at 23:47
user1551user1551
72.7k566127
$endgroup$
Yes. If you partition $A$ as $pmatrix{a&b^T\ b&D}$, then $A^{-1}=pmatrix{s^{-1}&-s^{-1}b^TD^{-1}\ ast&ast}$,
where $s=a-b^TD^{-1}b$ is the Schur complement of $D$ in $A$. Since $A$ is positive definite, $s^{-1}$ must be positive. Therefore, by mathematical induction, $D^{-1}$ and in turn $-s^{-1}b^TD^{-1}$ are also entrywise positive. Hence all entries on the first row of $A^{-1}$ are positive. Similarly, the entries on other rows are positive too.
answered Dec 8 '18 at 23:47
user1551user1551
72.7k566127
answered Dec 8 '18 at 23:47
user1551user1551
72.7k566127
answered Dec 8 '18 at 23:47
user1551user1551
72.7k566127
answered Dec 8 '18 at 23:47
user1551user1551
72.7k566127
72.7k566127
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