How do you show that AB and BA have the same eigenvalues?

How do you show that AB and BA have the same eigenvalues?

Say A is nonsingular. Then BA = A^{-1}AB A. So AB and BA are similar matrices, and they therefore have the same eigenvalues. (If x is an eigenvector of AB with eigenvalue \lambda, then y=A^{-1}x is the eigenvalue of BA with the same eigenvalue.)

Is it true that A and B have the same eigenvectors if a B?

Just write down two generic diagonal matrices and you will see that they must commute. d) First suppose that AB = BA. Then by part b) A and B have the same eigenvectors. Moreover since A has distinct real eigenvalues, it has n linearly independent eigenvectors.

How do you show A and B are similar matrices?

Also, if two matrices have the same distinct eigen values then they are similar. Suppose A and B have the same distinct eigenvalues. Then they are both diagonalizable with the same diagonal 2 Page 3 matrix A. So, both A and B are similar to A, and therefore A is similar to B.

How do you show two matrices have the same eigenvalues?

Since similar matrices A and B have the same characteristic polynomial, they also have the same eigenvalues. If B = PAP−1 and v = 0 is an eigenvector of A (say Av = λv) then B(Pv) = PAP−1(Pv) = PA(P−1P)v = PAv = λPv. Thus Pv (which is non-zero since P is invertible) is an eigenvector for B with eigenvalue λ.

Are the eigenvectors of A and A 2 the same?

Hence, eigenvectors need not match. However, if A is symmetric, then by the spectral theorem for symmetric matrices, indeed A and A2 have exactly the same set of eigenvectors as well.

Is a is similar to B then a 2 is similar to B 2?

A = and B = . Then A2 = B2 and so A2 is similar to B2, but A is not similar to B because nothing but the zero matrix is similar to the zero matrix.

How do you show two matrices similar?

Two square matrices A and B are similar if B = M−1 AM for some matrix M. This allows us to put matrices into families in which all the matrices in a family are similar to each other. Then each family can be represented by a diagonal (or nearly diagonal) matrix. 6 .

When A and B are similar matrices then det A det B?

If A and B are similar, then A and B have the same determinant, rank and charac- teristic polynomial. Corollary 7. If A and B are similar, then they have the same eigenvalues, and A is invertible if and only if B is invertible. = det(P)det(B − λI) 1 det(P) = det(B − λI).

Do similar matrices have the same minimal polynomial?

Similar matrices have the same minimal polynomial . In other words, the sets of annihilating polynomials of the two matrices coincide. As a consequence, also their respective minimal polynomials need to coincide (because they are the lowest-degree monic polynomials in the respective sets of annihilating polynomials).

What is the eigenvalue of BA and AB with same eigenvalues?

So AB and BA are similar matrices, and they therefore have the same eigenvalues. (If x is an eigenvector of AB with eigenvalue \\lambda, then y=A^ {-1}x is the eigenvalue of BA with the same eigenvalue.) Tools for everyone who codes.

How to find the eigenvalues of similar matrices with different eigenvectors?

Take the special case in which either A or B is nonsingular. (This can be fixed later.) Say A is nonsingular. Then BA = A^ {-1}AB A. So AB and BA are similar matrices, and they therefore have the same eigenvalues. (If x is an eigenvector of AB with eigenvalue \\lambda, then y=A^ {-1}x is the eigenvalue of BA with the same eigenvalue.)

Can AB and BA be just any two matrices?

No, AB and BA cannot be just any two matri- ces. They must have the same determinant, where for 2 × 2 matrices the determinant is defined by det. “. a b c d. #. = ad − bc . The determinant function has the remarkable property that det(AB) = det(A)det(B).

Can A and B be both diagonal matrices?

A and B are both diagonal matrices. There exists an invertible matrix P such that P − 1 A P and P − 1 B P are both diagonal. There is actually a sufficient and necessary condition for M n ( C):