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Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? First of all, am I being crazy in thinking that if λ λ is an eigenvalue of AB A B, where A A and B B are both N × N N × N matrices (not necessarily invertible), then λ λ is also an eigenvalue of BA B A?
矩阵A和矩阵B可交换（AB=BA）时，矩阵A和矩阵B之间应该满足什么条件？ - 知乎 在Gilbert Strang 的《Introduction to linear algebra 》的Chapter 6 section 6 1 Introduction to Eigenvalues 的最后部分，有提到当且仅当AB=BA时，A和B 有相同的n个线性无关的特征向量。
证明：矩阵 AB 与 BA 具有相同的非零特征值 - CSDN博客 矩阵 AB 与 BA 具有相同的非零特征值。 可以从两个方面证明该定理，第一种，借助相似矩阵之间拥有相同特征值的结论进行（要求 A,BA,BA,B 是可逆的）；第二种，则从公式 ABx=λxABx=\lambda xABx=λx 着手。
Eigenvalues of - Brown University If 0 is an eigenvalue of AB with algebraic multiplicity k ≥ 0, k ≥ 0, then 0 is an eigenvalues of BA with algebraic multiplicity k+n-m If m=n, then the eigenvalues of AB and BA are the same, with the same algebraic multiplicities
Eigenvalues of AB and BA : r mathematics - Reddit While AB and BA have the same eigenvalue, they have different eigenvectors I'm pretty sure the proof assumes A and B square, but it looks like we can extend to your case
Proofs Homework Set 10 - dept. math. lsa. umich. edu Since B has n distinct eigenvalues, they all have multiplicity 1 which means that all of the eigenspaces of B are one-dimensional (see Theorem 7(b) in Section 5 3)
Eigenvalues of AB and BA - Springer Given below are several proofs of the fact that AB and BA have the same eigenvalues Each proof brings outadif-ferent viewpoint andmay be presented at heappropri-ate time ina linear algebra course
Introduction to Linear Algebra, 5th Edition - MIT Mathematics We will show that det(A − λI) = 0 This section will explain how to compute the x’s and λ’s It can come early in the course because we only need the determinant of a 2 by 2 matrix Let me use det(A − λI) = 0 to find the eigenvalues for this first example, and then derive it properly in equation (3)
linear algebra - Eigenvalues of $AB$ in terms of $A$ and $B . . . Let A and B A and B be two n × n n × n matrix Where a1,a2, ,an a 1, a 2,, a n and b1,b2, ,bn b 1, b 2,, b n be the eigenvalues of A A and B B respectively Now can we describe eigenvalues of AB in terms of eigenvalues of A and B?
On the eigenvalues of A + B and AB - NIST In the case of AB-I we may refer to the well-known theorem stating that AB- I and B- IA have the same eigenvalues, or we may repeat the proof given above using a left-hand eigensolution x* instead of x