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I love mountainbike. I hate the Matrix Cookbook.
6h

comment 
Lie's Theorem in characteristic $p$
Ditrich, thanks for the comment. I note that Lie's and Jacobson's theorems are valid when $char(K)>n$. I find it curious that this is rarely explicitly stated in most of the references; do there exist other theorems concerning lie algebras with this property? 
6h

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Lie's Theorem in characteristic $p$
Torsten, thanks for the reference. In fact Seligman is not clear about whether Lie's theorem (or another one) is still valid if the characteristic is greater than $n$. For instance, he gives a counterexample when $char(K)=n$, using the fact that there are $A,B$ s.t. $ABBA=I$ (that is not a scoop); moreover he gives negative results but, really, no explicit positive results, that is discouraging. 
2d

asked  Lie's Theorem in characteristic $p$ 
Apr 1 
revised 
Solving $P=AB,Q=BA$, in the unknowns $A,B$
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Apr 1 
asked  Solving $P=AB,Q=BA$, in the unknowns $A,B$ 
Mar 10 
comment 
Separating the eigenvalues of a Hermitian matrix with a special block structure
@Sam, can you show that $\det(JxI)=\det((AxI)(AxI\overline{B}(AxI)^{1}B))$ ? 
Mar 10 
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Separating the eigenvalues of a Hermitian matrix with a special block structure
Sam, I think that you did not work really about your question. In particular, $J$ is "almost" a generic hermitian matrix and then, without supplementary hypothesis about $A,B$ (positive, definite ?), why this matrix $J$ would have a signature equal to $(n,n)$ under the hypothesis: $B$ is a real matrix ? It is obviously false: when $n=1$, the eigenvalues are $a\pmb$. Moreover, assume that $J$ has signature $(n,n)$ ; if $B$ is SLIGHTLY perturbed in the complex domain, then its signature is invariant. Observe seriously the case when $B$ is real. 
Mar 7 
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Matrix equation XAX=B where the solution must be diagonal
Yes Suvrit, I agree. If $X$ is only assumed to be symmetric and $A,B$ are generic, then the minimum is $0$. That shows the interest to assume that $X$ is a diagonal matrix. Yet, I do not know if this hypothesis on $X$ is associated to (only) a mathematical exercise or to a practical use of Riccati equations. 
Mar 6 
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Matrix equation XAX=B where the solution must be diagonal
@Suvrit, you write too quickly. Firstly, $X$ depends on $A,B$ ; in particular, if $A$ is invertible and $B$ is not, a solution $X$ of $XAX=B$ is certainly not generic. Secondly, if $A,B$ are generic in the PDS set, then $XAX=B$ has $2^n$ symmetric solutions and no diagonal solutions. Thirdly: read my edit. 
Mar 6 
revised 
Matrix equation XAX=B where the solution must be diagonal
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Mar 6 
answered  Matrix equation XAX=B where the solution must be diagonal 
Mar 4 
revised 
Mixing Numerical Range and inner product
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Mar 4 
revised 
Mixing Numerical Range and inner product
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Mar 4 
comment 
Mixing Numerical Range and inner product
The sequel proves nothing ! I give a method to solve your optimization problem: $\min_{u^Tu=1}u^TAU+b^Tu$. This method needs the calculation of $spectrum(A)$ and to find the roots of a polynomial of degree $2N2$. Thus we have $2N2$ candidates and we choose the one that realizes the minimum of $f$. Remark: your problem is not a convex one, even if $b=0$, because $\{u^Tu=1\}$ is not convex and $A$ is not necessarily $\geq 0$ or $\leq 0$. 
Mar 3 
answered  Mixing Numerical Range and inner product 
Feb 21 
awarded  Critic 
Feb 20 
revised 
Regularized Gradient with respect to a matrix (with a specific structure)
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Feb 20 
answered  Regularized Gradient with respect to a matrix (with a specific structure) 
Feb 8 
comment 
Determine the expected size of a lower triangular submatrix of a random matrix?
Anthony, few remarks. i) For your first approximation, you assume $k/n$ is small. ii) The last inequality is $k<4\log(n)/\log(p)+1$. iii) About your "Since there is enough independence around": a choice $(I,J)$ means, in particular, a choice of the orderings of $I,J$. Then if you keep the sets $I,J$ and change only the orderings, you are far from independence, except again, if $k/n$ is very small, that is, if $p$ is small. 
Jan 16 
awarded  Yearling 