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13h
answered Norm Used in Perturbation Matrix Thoery?
14h
comment Find a symmetric matrix B that makes ABC symmetric, A,C known
Since this is still a system of linear equations, if you only need to solve it numerically, just throw the reformulated equation $\pmatrix{C^T\otimes A - (A\otimes C^T)K^{(m,m)}\\ I_{m^2}-K^{(m,m)}}\operatorname{vec}(B)=0$ at some numerical solver like Matlab. Here the symbol "$\otimes$" denotes a Kronecker product, $\operatorname{vec}(\cdot)$ denotes vectorisation and $K^{(m,n)}$ denotes a commutation matrix. See Wikipedia for brief explanations of these terms.
22h
comment Find a symmetric matrix B that makes ABC symmetric, A,C known
Sorry, I've overlooked your requirement that $B$ is also symmetric, but I hope that my comment can still be of some help.
1d
comment Show that $\det(A) > 0$
Uhh, it's a typo. My example $A$ should be $\pmatrix{6&0\\-5&1}$. The definition in your book is equivalent to that $\frac12(A+A^T)$ has a positive spectrum. With $x=(4,9)^T$, we have $x^TAx=-3$.
1d
comment A smart way to bound this function and get rid of covariance matrix
I think there's something wrong with your setting. In general, $\Sigma$ is not necessarily positive semidefinite and $1+h^\ast\Sigma h$ can be negative. Therefore $A(h,\Sigma)$ is not real and your desired inequality does not make sense.
1d
comment Find a symmetric matrix B that makes ABC symmetric, A,C known
If the rank of $A$ is less than $n$, let $v$ be any nonzero vector in the nullspace of $A$ and let every column of $B$ be $v$. Then $ABC=0$ is a symmetric matrix. Similarly when $C$ has deficient rank. If both $A$ and $C$ are full-rank, take $B=A^+SC^+$ for any $n\times n$ real symmetric matrix $S$, where $A^+$ and $C^+$ are the Moore-Penrose pseudoinverses of $A$ and $C$ respectively. Then $ABC=S$.
1d
comment Prove $\det(I + B) = 2(1 + tr(B)).$
The statement in question can be viewed as a generalisation of the identity $\det(I+Q)\equiv 2(1+\operatorname{tr}(Q))$ for $Q\in SO(3;\mathbb R)$. In fact, when the symmetric part of $A$ is positive definite, $B$ is similar to some $Q$ in the special orthogonal group.
2d
revised Find all complex matrices $A$ such that $n\operatorname{Tr}(AB) = \operatorname{Tr}(A)\operatorname{Tr}(B)$ for all $B$.
deleted 43 characters in body; edited tags; edited title
2d
revised Convert from complex exponentials to sinusoids
edited tags
2d
awarded  Necromancer
2d
answered Decomposition for a Sum of Matrix Products
2d
comment Pfaffian of skew symmetric block matrix
If $A=I_m$, can you calculate $\operatorname{Pf}(B)$?
May
19
awarded  Necromancer
May
18
answered How is the perturbation in one column of a symmetric matrix reflected in its eigenvalues?
May
15
answered If ${\sum\limits_{i = 1}^n {\left| {{\lambda _i}} \right|} ^2} = \sum\limits_{i = 1}^n {{\sigma _i}^2} \Rightarrow$A is normal matrix
May
15
comment Calculate the determinant of a matrix given a simple condition
If $a,b,c$ are nonnegative and $a+b+c>0$, it can be shown that $a,b,c$ are actually all positive and $\det(A)\ge0$. However, I'm not sure if anything can be said about $\det(A)$ in general.
May
7
comment If I know $AB$, how can I calculate $BA$?
@MikeF More precisely, $AB$ and $BA$ in general have the same multiset of nonzero eigenvalues.
May
7
comment $X$ is normal matrix and $AX=XB$ and $XA=BX$.why $A{X^*} = {X^*}B$ and ${X^*}A = B{X^*}$?
Closely related: Show that if $T_1, T_2$ are normal operators that commutes then $T_1+T_2$ and $T_1T_2$ are normal.
May
6
comment Give an example of a singular matrix in $M_{3×3}(Q)$ the entries of which are distinct prime positive integers, or show that no such matrix can exist.
Very nice! +1. Did you foresee that examples can be constructed using the first nine primes, or did you just want to try your luck?
May
6
comment Similar matrices NOT over the complex numbers
See the linked (duplicated) question. Put it simply, if both $A$ and $B$ comes from some field $K$ (e.g. $\mathbb Q$) and they are similar over a larger field $F$ (e.g. $\mathbb R$), then they are similar over $K$ too. In other words, if $K$ is a subfield of $F$, the matrices $A,B$ have entries in $K$ and $B=PAP^{-1}$ for some $P\in M_n(F)$, then there exists an invertible matrix $S\in M_n(K)$ such that $B=SAS^{-1}$. (Note that $P$ is not necessarily equal to $S$.)