Let A and B be two n x n matrices. We recall that two matrices A and B are called similar if and only if there exists an invertible matrix P such that A = P^(-1)BP. Prove that similar matrices have the same eigenvalues.
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Let λ be an eigenvalue of matrix A, and let x be its corresponding eigenvector. So, we have Ax = λx. Show more…
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Show that if $A$ and $B$ are two $n \times n$ matrices, then the matrices $A B$ and $B A$ have the same characteristic polynomial, and thus the same eigenvalues (matrices $A B$ and $B A$ need not be similar though; see Exercise 55 ). Hint: \[ \left[\begin{array}{cc} A B & 0 \\ B & 0 \end{array}\right]\left[\begin{array}{cc} I_{n} & A \\ 0 & I_{n} \end{array}\right]=\left[\begin{array}{cc} I_{n} & A \\ 0 & I_{n} \end{array}\right]\left[\begin{array}{cc} 0 & 0 \\ B & B A \end{array}\right] \].
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a) Prove that similar matrices have the same eigenvalues b) Find two similar matrices that do not have the same eigenvectors
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(a) Show that if the two (n x n) matrices A and B are similar, then det(A) = det(B). (b) Now show that if A and B are similar, then they have the same characteristic polynomial (therefore, they have the same set of eigenvalues, with the same algebraic multiplicities.)
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