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Schur's Theorem about immanants
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$DeclareMathOperatorImmImm$I am looking for a proof in English or French of Schur's theorem that, for every $H$ in the space $mathbb H_n^+$ of positive semi-definite Hermitian matrices, and every irreducible character $chi$ of $mathfrak S_n$, $chi(e)det HleImm_chi(H)$, where the immanant $Imm_chi$ is defined by
$$Imm_chi(H):=sum_sigmachi(sigma)prod_i=1^nh_isigma(i).$$
Notice that the original paper I. Schur, "Über endlicher Gruppen und Hermiteschen Formen" Math. Z., 1 (1918) pp. 184–207, is in German.
By the way, it seems that many authors relate Schur's theorem to symmetric polynomials. Is there any purely representation-theoretic proof of the inequality above? Let $(rho,V)$ be a unitary representation whose character is $chi$. We may associate to $Imm_chi(H)$ a Hermitian matrix over $V$ by
$$K_rho:=sum_sigmaleft(prod_i=1^nh_isigma(i)right)rho(sigma).$$
It would be sufficient to prove that $Kge(det H)I_V$, where $I_V$ denotes the matrix of the scalar product. Because of Frobenius's theorem about the orthogonal decomposition of the regular representation, this amounts to proving that the analogous sum, where $rho$ is replaced by the regular representation, satisfies the same estimate. In other words, Schur's theorem would be implied by the inequality
$$forall xiinmathbb C^frak S_n,,forall Hinmathbb H_n^+,qquad |xi|^2det Hlesum_sigma,thetabarxi_sigmaxi_thetaprod_ih_sigma(i)theta(i).$$
Is this inequality true?
reference-request rt.representation-theory
edited 2 hours ago
LSpice
2,71822126
asked 4 hours ago
Denis Serre
28.7k791193
add a comment |Â
up vote
8
down vote
favorite
$DeclareMathOperatorImmImm$I am looking for a proof in English or French of Schur's theorem that, for every $H$ in the space $mathbb H_n^+$ of positive semi-definite Hermitian matrices, and every irreducible character $chi$ of $mathfrak S_n$, $chi(e)det HleImm_chi(H)$, where the immanant $Imm_chi$ is defined by
$$Imm_chi(H):=sum_sigmachi(sigma)prod_i=1^nh_isigma(i).$$
Notice that the original paper I. Schur, "Über endlicher Gruppen und Hermiteschen Formen" Math. Z., 1 (1918) pp. 184–207, is in German.
By the way, it seems that many authors relate Schur's theorem to symmetric polynomials. Is there any purely representation-theoretic proof of the inequality above? Let $(rho,V)$ be a unitary representation whose character is $chi$. We may associate to $Imm_chi(H)$ a Hermitian matrix over $V$ by
$$K_rho:=sum_sigmaleft(prod_i=1^nh_isigma(i)right)rho(sigma).$$
It would be sufficient to prove that $Kge(det H)I_V$, where $I_V$ denotes the matrix of the scalar product. Because of Frobenius's theorem about the orthogonal decomposition of the regular representation, this amounts to proving that the analogous sum, where $rho$ is replaced by the regular representation, satisfies the same estimate. In other words, Schur's theorem would be implied by the inequality
$$forall xiinmathbb C^frak S_n,,forall Hinmathbb H_n^+,qquad |xi|^2det Hlesum_sigma,thetabarxi_sigmaxi_thetaprod_ih_sigma(i)theta(i).$$
Is this inequality true?
reference-request rt.representation-theory
edited 2 hours ago
LSpice
2,71822126
asked 4 hours ago
Denis Serre
28.7k791193
1
Would you say what you denote by $mathbbH_n^+$?
– YCor
4 hours ago
@YCor. The cone of positive semi-definite Hermitian matrices.
– Denis Serre
3 hours ago
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
$DeclareMathOperatorImmImm$I am looking for a proof in English or French of Schur's theorem that, for every $H$ in the space $mathbb H_n^+$ of positive semi-definite Hermitian matrices, and every irreducible character $chi$ of $mathfrak S_n$, $chi(e)det HleImm_chi(H)$, where the immanant $Imm_chi$ is defined by
$$Imm_chi(H):=sum_sigmachi(sigma)prod_i=1^nh_isigma(i).$$
Notice that the original paper I. Schur, "Über endlicher Gruppen und Hermiteschen Formen" Math. Z., 1 (1918) pp. 184–207, is in German.
By the way, it seems that many authors relate Schur's theorem to symmetric polynomials. Is there any purely representation-theoretic proof of the inequality above? Let $(rho,V)$ be a unitary representation whose character is $chi$. We may associate to $Imm_chi(H)$ a Hermitian matrix over $V$ by
$$K_rho:=sum_sigmaleft(prod_i=1^nh_isigma(i)right)rho(sigma).$$
It would be sufficient to prove that $Kge(det H)I_V$, where $I_V$ denotes the matrix of the scalar product. Because of Frobenius's theorem about the orthogonal decomposition of the regular representation, this amounts to proving that the analogous sum, where $rho$ is replaced by the regular representation, satisfies the same estimate. In other words, Schur's theorem would be implied by the inequality
$$forall xiinmathbb C^frak S_n,,forall Hinmathbb H_n^+,qquad |xi|^2det Hlesum_sigma,thetabarxi_sigmaxi_thetaprod_ih_sigma(i)theta(i).$$
Is this inequality true?
reference-request rt.representation-theory
edited 2 hours ago
LSpice
2,71822126
asked 4 hours ago
Denis Serre
28.7k791193
$DeclareMathOperatorImmImm$I am looking for a proof in English or French of Schur's theorem that, for every $H$ in the space $mathbb H_n^+$ of positive semi-definite Hermitian matrices, and every irreducible character $chi$ of $mathfrak S_n$, $chi(e)det HleImm_chi(H)$, where the immanant $Imm_chi$ is defined by
$$Imm_chi(H):=sum_sigmachi(sigma)prod_i=1^nh_isigma(i).$$
Notice that the original paper I. Schur, "Über endlicher Gruppen und Hermiteschen Formen" Math. Z., 1 (1918) pp. 184–207, is in German.
By the way, it seems that many authors relate Schur's theorem to symmetric polynomials. Is there any purely representation-theoretic proof of the inequality above? Let $(rho,V)$ be a unitary representation whose character is $chi$. We may associate to $Imm_chi(H)$ a Hermitian matrix over $V$ by
$$K_rho:=sum_sigmaleft(prod_i=1^nh_isigma(i)right)rho(sigma).$$
It would be sufficient to prove that $Kge(det H)I_V$, where $I_V$ denotes the matrix of the scalar product. Because of Frobenius's theorem about the orthogonal decomposition of the regular representation, this amounts to proving that the analogous sum, where $rho$ is replaced by the regular representation, satisfies the same estimate. In other words, Schur's theorem would be implied by the inequality
$$forall xiinmathbb C^frak S_n,,forall Hinmathbb H_n^+,qquad |xi|^2det Hlesum_sigma,thetabarxi_sigmaxi_thetaprod_ih_sigma(i)theta(i).$$
Is this inequality true?
reference-request rt.representation-theory
reference-request rt.representation-theory
edited 2 hours ago
LSpice
2,71822126
asked 4 hours ago
Denis Serre
28.7k791193
edited 2 hours ago
LSpice
2,71822126
asked 4 hours ago
Denis Serre
28.7k791193
edited 2 hours ago
LSpice
2,71822126
edited 2 hours ago
LSpice
2,71822126
edited 2 hours ago
LSpice
2,71822126
2,71822126
asked 4 hours ago
Denis Serre
28.7k791193
asked 4 hours ago
Denis Serre
28.7k791193
asked 4 hours ago
Denis Serre
28.7k791193
28.7k791193
1
Would you say what you denote by $mathbbH_n^+$?
– YCor
4 hours ago
@YCor. The cone of positive semi-definite Hermitian matrices.
– Denis Serre
3 hours ago
add a comment |Â
1
Would you say what you denote by $mathbbH_n^+$?
– YCor
4 hours ago
@YCor. The cone of positive semi-definite Hermitian matrices.
– Denis Serre
3 hours ago
1
1
Would you say what you denote by $mathbbH_n^+$?
– YCor
4 hours ago
Would you say what you denote by $mathbbH_n^+$?
– YCor
4 hours ago
@YCor. The cone of positive semi-definite Hermitian matrices.
– Denis Serre
3 hours ago
@YCor. The cone of positive semi-definite Hermitian matrices.
– Denis Serre
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
The crux of the proof is to indeed consider the Schur power matrix. Let $A$ be an $ntimes n$ matrix. Then consider the $n! times n!$ matrix $S(A)$ indexed by permutations $sigma, tau$ such that the $(sigma,tau)$-entry is given by $prod_i=1^n a_sigma(i),tau(i)$ for $sigma, tau in mathfrakS_n$.
Schur's theorem. Let $A$ be a positive semidefinite matrix. Then, for any vector $x$, $x^TS(A)x ge det(A)x^Tx$.
It turns out that actually the determinant of $A$ is also an eigenvalue of $S(A)$ (and from the above theorem, it is the smallest eigenvalue). From this the Immanantal inequality also follows as a corollary. Let me repeat a proof of this claim below (this is not my original proof, obviously).
Proof. Consider the tensor product matrix $otimes^n A$, and augment $x$ (which is a vector of dim $n!$) by padding with zeros appropriately to obtain a vector $z$ that is of dim $n^n$, so that $$x^TS(A)x= z^T(otimes^n A)z.~~~~~~~~~~~~~~~(*)$$ Since $A$ is psd, we can write $A=B^TB$ for a lower-triangular matrix $B$. Thus, we have (using simple properties of the tensor product) $$[z^T(otimes^n A)z = z^T(otimes^n B^TB)z = z^T(otimes^n B^T)(otimes^n B)z.$$
Now by Cauchy-Schwarz we obtain
beginalign*
x^TS(A)x & le left[z^T(otimes^n B^T)^2zright]^1/2left[z^T(otimes^n B)^2zright]^1/2\
&= left[z^T(otimes^n (B^T)^2)zright]^1/2left[z^T(otimes^n B^2)zright]^1/2.
endalign*
Using the relation $(*)$ twice on the rhs above we thus obtain
beginequation*
x^TS(A)x le [x^TS(B^2)^Tx]^1/2[x^TS(B^2)x]^1/2.
endequation*
Since $B^2$ is lower triangular, the diagonal entries of $S(B^2)$ are all easily seen to be actually equal to $det(B^2)=det(B)^2$. Moreover, $S(B^2)$ itself is lower-triangular, so that $x^TS(B^2)x = det(B)^2x^Tx$ which equals $det(A)x^Tx$. Substituting this in the final inequality above, the proof is complete.
answered 3 hours ago
Suvrit
23.6k659119
To get the immanent inequality from Schur's Theorem, take $x_sigma = chi(sigma)$. The coefficient of $a_1rho(1)ldots a_nrho(n)$ in the left-hand side is then $sum_sigma, tau : tausigma^-1 = rho chi(sigma) chi(tau) = sum_sigma chi(sigma)chi(rhosigma) = sum_sigma chi(sigma^-1)chi(rhosigma) = |G| chi(rho) / chi(1)$ by an orthogonality relation. So the left-hand side is $|G|/chi(1)$ times the immanent sum, and the right-hand side is $|G| mathrmdet(A)$ again by character orthogonality.
– Mark Wildon
25 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The crux of the proof is to indeed consider the Schur power matrix. Let $A$ be an $ntimes n$ matrix. Then consider the $n! times n!$ matrix $S(A)$ indexed by permutations $sigma, tau$ such that the $(sigma,tau)$-entry is given by $prod_i=1^n a_sigma(i),tau(i)$ for $sigma, tau in mathfrakS_n$.
Schur's theorem. Let $A$ be a positive semidefinite matrix. Then, for any vector $x$, $x^TS(A)x ge det(A)x^Tx$.
It turns out that actually the determinant of $A$ is also an eigenvalue of $S(A)$ (and from the above theorem, it is the smallest eigenvalue). From this the Immanantal inequality also follows as a corollary. Let me repeat a proof of this claim below (this is not my original proof, obviously).
Proof. Consider the tensor product matrix $otimes^n A$, and augment $x$ (which is a vector of dim $n!$) by padding with zeros appropriately to obtain a vector $z$ that is of dim $n^n$, so that $$x^TS(A)x= z^T(otimes^n A)z.~~~~~~~~~~~~~~~(*)$$ Since $A$ is psd, we can write $A=B^TB$ for a lower-triangular matrix $B$. Thus, we have (using simple properties of the tensor product) $$[z^T(otimes^n A)z = z^T(otimes^n B^TB)z = z^T(otimes^n B^T)(otimes^n B)z.$$
Now by Cauchy-Schwarz we obtain
beginalign*
x^TS(A)x & le left[z^T(otimes^n B^T)^2zright]^1/2left[z^T(otimes^n B)^2zright]^1/2\
&= left[z^T(otimes^n (B^T)^2)zright]^1/2left[z^T(otimes^n B^2)zright]^1/2.
endalign*
Using the relation $(*)$ twice on the rhs above we thus obtain
beginequation*
x^TS(A)x le [x^TS(B^2)^Tx]^1/2[x^TS(B^2)x]^1/2.
endequation*
Since $B^2$ is lower triangular, the diagonal entries of $S(B^2)$ are all easily seen to be actually equal to $det(B^2)=det(B)^2$. Moreover, $S(B^2)$ itself is lower-triangular, so that $x^TS(B^2)x = det(B)^2x^Tx$ which equals $det(A)x^Tx$. Substituting this in the final inequality above, the proof is complete.
answered 3 hours ago
Suvrit
23.6k659119
To get the immanent inequality from Schur's Theorem, take $x_sigma = chi(sigma)$. The coefficient of $a_1rho(1)ldots a_nrho(n)$ in the left-hand side is then $sum_sigma, tau : tausigma^-1 = rho chi(sigma) chi(tau) = sum_sigma chi(sigma)chi(rhosigma) = sum_sigma chi(sigma^-1)chi(rhosigma) = |G| chi(rho) / chi(1)$ by an orthogonality relation. So the left-hand side is $|G|/chi(1)$ times the immanent sum, and the right-hand side is $|G| mathrmdet(A)$ again by character orthogonality.
– Mark Wildon
25 mins ago
add a comment |Â
up vote
4
down vote
The crux of the proof is to indeed consider the Schur power matrix. Let $A$ be an $ntimes n$ matrix. Then consider the $n! times n!$ matrix $S(A)$ indexed by permutations $sigma, tau$ such that the $(sigma,tau)$-entry is given by $prod_i=1^n a_sigma(i),tau(i)$ for $sigma, tau in mathfrakS_n$.
Schur's theorem. Let $A$ be a positive semidefinite matrix. Then, for any vector $x$, $x^TS(A)x ge det(A)x^Tx$.
It turns out that actually the determinant of $A$ is also an eigenvalue of $S(A)$ (and from the above theorem, it is the smallest eigenvalue). From this the Immanantal inequality also follows as a corollary. Let me repeat a proof of this claim below (this is not my original proof, obviously).
Proof. Consider the tensor product matrix $otimes^n A$, and augment $x$ (which is a vector of dim $n!$) by padding with zeros appropriately to obtain a vector $z$ that is of dim $n^n$, so that $$x^TS(A)x= z^T(otimes^n A)z.~~~~~~~~~~~~~~~(*)$$ Since $A$ is psd, we can write $A=B^TB$ for a lower-triangular matrix $B$. Thus, we have (using simple properties of the tensor product) $$[z^T(otimes^n A)z = z^T(otimes^n B^TB)z = z^T(otimes^n B^T)(otimes^n B)z.$$
Now by Cauchy-Schwarz we obtain
beginalign*
x^TS(A)x & le left[z^T(otimes^n B^T)^2zright]^1/2left[z^T(otimes^n B)^2zright]^1/2\
&= left[z^T(otimes^n (B^T)^2)zright]^1/2left[z^T(otimes^n B^2)zright]^1/2.
endalign*
Using the relation $(*)$ twice on the rhs above we thus obtain
beginequation*
x^TS(A)x le [x^TS(B^2)^Tx]^1/2[x^TS(B^2)x]^1/2.
endequation*
Since $B^2$ is lower triangular, the diagonal entries of $S(B^2)$ are all easily seen to be actually equal to $det(B^2)=det(B)^2$. Moreover, $S(B^2)$ itself is lower-triangular, so that $x^TS(B^2)x = det(B)^2x^Tx$ which equals $det(A)x^Tx$. Substituting this in the final inequality above, the proof is complete.
answered 3 hours ago
Suvrit
23.6k659119
To get the immanent inequality from Schur's Theorem, take $x_sigma = chi(sigma)$. The coefficient of $a_1rho(1)ldots a_nrho(n)$ in the left-hand side is then $sum_sigma, tau : tausigma^-1 = rho chi(sigma) chi(tau) = sum_sigma chi(sigma)chi(rhosigma) = sum_sigma chi(sigma^-1)chi(rhosigma) = |G| chi(rho) / chi(1)$ by an orthogonality relation. So the left-hand side is $|G|/chi(1)$ times the immanent sum, and the right-hand side is $|G| mathrmdet(A)$ again by character orthogonality.
– Mark Wildon
25 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The crux of the proof is to indeed consider the Schur power matrix. Let $A$ be an $ntimes n$ matrix. Then consider the $n! times n!$ matrix $S(A)$ indexed by permutations $sigma, tau$ such that the $(sigma,tau)$-entry is given by $prod_i=1^n a_sigma(i),tau(i)$ for $sigma, tau in mathfrakS_n$.
Schur's theorem. Let $A$ be a positive semidefinite matrix. Then, for any vector $x$, $x^TS(A)x ge det(A)x^Tx$.
It turns out that actually the determinant of $A$ is also an eigenvalue of $S(A)$ (and from the above theorem, it is the smallest eigenvalue). From this the Immanantal inequality also follows as a corollary. Let me repeat a proof of this claim below (this is not my original proof, obviously).
Proof. Consider the tensor product matrix $otimes^n A$, and augment $x$ (which is a vector of dim $n!$) by padding with zeros appropriately to obtain a vector $z$ that is of dim $n^n$, so that $$x^TS(A)x= z^T(otimes^n A)z.~~~~~~~~~~~~~~~(*)$$ Since $A$ is psd, we can write $A=B^TB$ for a lower-triangular matrix $B$. Thus, we have (using simple properties of the tensor product) $$[z^T(otimes^n A)z = z^T(otimes^n B^TB)z = z^T(otimes^n B^T)(otimes^n B)z.$$
Now by Cauchy-Schwarz we obtain
beginalign*
x^TS(A)x & le left[z^T(otimes^n B^T)^2zright]^1/2left[z^T(otimes^n B)^2zright]^1/2\
&= left[z^T(otimes^n (B^T)^2)zright]^1/2left[z^T(otimes^n B^2)zright]^1/2.
endalign*
Using the relation $(*)$ twice on the rhs above we thus obtain
beginequation*
x^TS(A)x le [x^TS(B^2)^Tx]^1/2[x^TS(B^2)x]^1/2.
endequation*
Since $B^2$ is lower triangular, the diagonal entries of $S(B^2)$ are all easily seen to be actually equal to $det(B^2)=det(B)^2$. Moreover, $S(B^2)$ itself is lower-triangular, so that $x^TS(B^2)x = det(B)^2x^Tx$ which equals $det(A)x^Tx$. Substituting this in the final inequality above, the proof is complete.
answered 3 hours ago
Suvrit
23.6k659119
The crux of the proof is to indeed consider the Schur power matrix. Let $A$ be an $ntimes n$ matrix. Then consider the $n! times n!$ matrix $S(A)$ indexed by permutations $sigma, tau$ such that the $(sigma,tau)$-entry is given by $prod_i=1^n a_sigma(i),tau(i)$ for $sigma, tau in mathfrakS_n$.
Schur's theorem. Let $A$ be a positive semidefinite matrix. Then, for any vector $x$, $x^TS(A)x ge det(A)x^Tx$.
It turns out that actually the determinant of $A$ is also an eigenvalue of $S(A)$ (and from the above theorem, it is the smallest eigenvalue). From this the Immanantal inequality also follows as a corollary. Let me repeat a proof of this claim below (this is not my original proof, obviously).
Proof. Consider the tensor product matrix $otimes^n A$, and augment $x$ (which is a vector of dim $n!$) by padding with zeros appropriately to obtain a vector $z$ that is of dim $n^n$, so that $$x^TS(A)x= z^T(otimes^n A)z.~~~~~~~~~~~~~~~(*)$$ Since $A$ is psd, we can write $A=B^TB$ for a lower-triangular matrix $B$. Thus, we have (using simple properties of the tensor product) $$[z^T(otimes^n A)z = z^T(otimes^n B^TB)z = z^T(otimes^n B^T)(otimes^n B)z.$$
Now by Cauchy-Schwarz we obtain
beginalign*
x^TS(A)x & le left[z^T(otimes^n B^T)^2zright]^1/2left[z^T(otimes^n B)^2zright]^1/2\
&= left[z^T(otimes^n (B^T)^2)zright]^1/2left[z^T(otimes^n B^2)zright]^1/2.
endalign*
Using the relation $(*)$ twice on the rhs above we thus obtain
beginequation*
x^TS(A)x le [x^TS(B^2)^Tx]^1/2[x^TS(B^2)x]^1/2.
endequation*
Since $B^2$ is lower triangular, the diagonal entries of $S(B^2)$ are all easily seen to be actually equal to $det(B^2)=det(B)^2$. Moreover, $S(B^2)$ itself is lower-triangular, so that $x^TS(B^2)x = det(B)^2x^Tx$ which equals $det(A)x^Tx$. Substituting this in the final inequality above, the proof is complete.
answered 3 hours ago
Suvrit
23.6k659119
answered 3 hours ago
Suvrit
23.6k659119
answered 3 hours ago
Suvrit
23.6k659119
answered 3 hours ago
Suvrit
23.6k659119
23.6k659119
To get the immanent inequality from Schur's Theorem, take $x_sigma = chi(sigma)$. The coefficient of $a_1rho(1)ldots a_nrho(n)$ in the left-hand side is then $sum_sigma, tau : tausigma^-1 = rho chi(sigma) chi(tau) = sum_sigma chi(sigma)chi(rhosigma) = sum_sigma chi(sigma^-1)chi(rhosigma) = |G| chi(rho) / chi(1)$ by an orthogonality relation. So the left-hand side is $|G|/chi(1)$ times the immanent sum, and the right-hand side is $|G| mathrmdet(A)$ again by character orthogonality.
– Mark Wildon
25 mins ago
add a comment |Â
To get the immanent inequality from Schur's Theorem, take $x_sigma = chi(sigma)$. The coefficient of $a_1rho(1)ldots a_nrho(n)$ in the left-hand side is then $sum_sigma, tau : tausigma^-1 = rho chi(sigma) chi(tau) = sum_sigma chi(sigma)chi(rhosigma) = sum_sigma chi(sigma^-1)chi(rhosigma) = |G| chi(rho) / chi(1)$ by an orthogonality relation. So the left-hand side is $|G|/chi(1)$ times the immanent sum, and the right-hand side is $|G| mathrmdet(A)$ again by character orthogonality.
– Mark Wildon
25 mins ago
To get the immanent inequality from Schur's Theorem, take $x_sigma = chi(sigma)$. The coefficient of $a_1rho(1)ldots a_nrho(n)$ in the left-hand side is then $sum_sigma, tau : tausigma^-1 = rho chi(sigma) chi(tau) = sum_sigma chi(sigma)chi(rhosigma) = sum_sigma chi(sigma^-1)chi(rhosigma) = |G| chi(rho) / chi(1)$ by an orthogonality relation. So the left-hand side is $|G|/chi(1)$ times the immanent sum, and the right-hand side is $|G| mathrmdet(A)$ again by character orthogonality.
– Mark Wildon
25 mins ago
To get the immanent inequality from Schur's Theorem, take $x_sigma = chi(sigma)$. The coefficient of $a_1rho(1)ldots a_nrho(n)$ in the left-hand side is then $sum_sigma, tau : tausigma^-1 = rho chi(sigma) chi(tau) = sum_sigma chi(sigma)chi(rhosigma) = sum_sigma chi(sigma^-1)chi(rhosigma) = |G| chi(rho) / chi(1)$ by an orthogonality relation. So the left-hand side is $|G|/chi(1)$ times the immanent sum, and the right-hand side is $|G| mathrmdet(A)$ again by character orthogonality.
– Mark Wildon
25 mins ago
add a comment |Â
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1
Would you say what you denote by $mathbbH_n^+$?
– YCor
4 hours ago
@YCor. The cone of positive semi-definite Hermitian matrices.
– Denis Serre
3 hours ago