找到一个与常数或另一个矩阵相乘时得出相同结果的矩阵 [英] Find a matrix that gives same result when multiplied by a constant or another matrix

问题描述

我遇到了类似A*x=lambda*x的问题，其中A的顺序是d*d，x的顺序是d*c，而lambda是一个常数. A和lambda是已知的，而矩阵x是未知的. 有什么办法可以在Matlab中解决这个问题? (与特征值类似，但x是d*c矩阵，而不是矢量).

I have got a problem like A*x=lambda*x, where A is of order d*d, x is of order d*c and lambda is a constant. A and lambda are known and the matrix x is unknown. Is there any way to solve this problem in matlab?? (Like eigen values but x is a d*c matrix instead of being a vector).

推荐答案

如果我对您的理解正确，那么x不一定有任何解决方案.如果A*x=lambda*x，则x的任何列y都满足A*y=lambda*y，因此x的列只是A的特征向量，对应于特征值lambda，并且只有在以下情况下才有任何解lambda实际上是一个特征值.

If I've understood you correctly, there will not necessarily be any solutions for x. If A*x=lambda*x, then any column y of x satisfies A*y=lambda*y, so the columns of x are simply eigenvectors of A corresponding to the eigenvalue lambda, and there will only be any solutions if lambda is in fact an eigenvalue.

从文档:

[V，D] = eig(A)产生特征值(D)和特征向量的矩阵 (V)是矩阵A，因此A * V = V * D.矩阵D是的标准形式 A —对角矩阵，在主对角线上具有A的特征值. 矩阵V是模态矩阵，其列是A的特征向量.

[V,D] = eig(A) produces matrices of eigenvalues (D) and eigenvectors (V) of matrix A, so that A*V = V*D. Matrix D is the canonical form of A — a diagonal matrix with A's eigenvalues on the main diagonal. Matrix V is the modal matrix — its columns are the eigenvectors of A.

您可以使用它来检查lambda是否为特征值，并找到任何对应的特征向量.

You can use this to check if lambda is an eigenvalue, and find any corresponding eigenvectors.

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