
xat 333matrix chat
xat 333matrix chat
Characterization of matrices such that $AX=XA^T$
2018年4月14日 · As @Delta-u said, with X =In X = I n, you get A =AT A = A T. By the spectral theorem, there is an orthogonal matrix P P and a diagonal matrix D D such that PTDP = A P T D P = A. Making the change of variables Y = PTXP Y = P T X P, we see that D D must commute with any symmetric matrix Y Y. Write D = diag(d1, …,dn) D = d i a g (d 1, …, d n).
Symmetric, positive semidefinite, and positive definite real …
2001年1月1日 · Given the equations AX = XAT and AX = YB with arbitrary nonzero real matrices A and B of the same size, we seek all real solutions X and Y which are: (1) symmetric, (2) symmetric and positive semidefinite, and (3) symmetric and positive definite.
If A has no special structure we have simply Sij = Jij, that is, the structure matrix is simply the single-entry matrix. Many structures have a representation in singleentry matrices, see Sec. 9.7.6 for more examples of structure matrices.
A particularly attractive feature is that Xm may be written as a low rank matrix, Xm = UmU^ with Um of low column rank, so that only the matrix Um needs to be stored. To the best of our knowledge, no asymptotic convergence analysis of this Galerkin method is available in the literature. The aim of this paper is to fill this gap.
一个关于Lyapunov方程的问题 - 知乎
2022年11月25日 · 设 A,B\in M_ {n} 是正定阵,考虑 Lyapunov 方程 AX+XA=B ,其有唯一解,且可推出唯一解 X 是正定的. 这是一个有趣的问题,为了解决这个问题引入一些概念和结论 (上次有 UU 反应说过程太跳了,这次把要用到的定理先陈述出来,每一个定理都是高代经典问题,具有代表性,没见过的定理可以自己动手做做,如果有问题可以评论区留言,也欢迎私戳我): define:A= (a_ {ij}),B= (b_ {ij})\in M_ {m\times n},定义A,B的Hadamard乘积为A\circ B= (a_ {ij}b_ {ij})\in M_ …
probability - Find the transformation matrix $A$ of a random …
2016年11月15日 · Find the transformation matrix A of a random vector X such that Y = XA T has uncorrelated components. I am trying to solve this problem. I need to find matrix A such that …
《现代控制理论基础》— 4 状态空间的数学模型(二) - 知乎
状态转移矩阵将系统每一刻状态变量的信息与状态变量的初试状态建立关系,状态转移矩阵包含了自由运动的全部信息。 方法有两种, 幂级数法 和 拉普拉斯变换 法。 幂级数法. 设 \dot {x} (t)=Ax (t) 的解是关于 t 的向量幂级数. x (t)=b_ {0}+b_ {1}t+b_ {2}t^ {2}+\cdots+b_ {k}t^ {k}+\cdots. \dot {x} (t)=b_ {1}+2b_ {2}t+\cdots+kb_ {k}t^ {k-1}+\cdots. \Rightarrow A (b_ {0}+b_ {1}t+\cdots+b_ {k}t^ {k}+\cdots )=b_ {1}+2b_ {2}t+\cdots+kb_ {k}t^ {k-1}+\cdots.
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Build advanced, rich communication applications on top of Matrix. Matrix is a rich ecosystem of clients, servers, bots and application services. Find out more in our developer documentation. …
Linear Transformation T (X)=AX-XA and Determinant of Matrix ...
2018年2月12日 · We prove that T (X)=AX-XA is a linear transformation from the vector space of all n by n matrices and show that the determinant of the matrix of T is zero.
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