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No, in general they do not. They have the same eigenvalues but not the same eigenvectors.

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Q: Do similar matrices have the same eigenvectors?
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Related questions

Do similar matrices have the same eigenvalues?

Yes, similar matrices have the same eigenvalues.


Do commutative matrices have the same eigenvectors?

It is true that diagonalizable matrices A and B commute if and only if they are simultaneously diagonalizable. This result can be found in standard texts (e.g. Horn and Johnson, Matrix Analysis, 1999, Theorem 1.3.12.) One direction of the if and only if proof is straightforward, but the other direction is more technical: If A and B are diagonalizable matrices of the same order, and have the same eigenvectors, then, without loss of generality, we can write their diagonalizations as A = VDV-1 and B = VLV-1, where V is the matrix composed of the basis eigenvectors of A and B, and D and L are diagonal matrices with the corresponding eigenvalues of A and B as their diagonal elements. Since diagonal matrices commute, DL = LD. So, AB = VDV-1VLV-1 = VDLV-1 = VLDV-1 = VLV-1VDV-1 = BA. The reverse is harder to prove, but one online proof is given below as a related link. The proof in Horn and Johnson is clear and concise. Consider the particular case that B is the identity, I. If A = VDV-1 is a diagonalization of A, then I = VIV-1 is a diagonalization of I; i.e., A and I have the same eigenvectors.


How can you prove that similar matrices have the same trace?

you tell me


What is the condition for the addition of matrices?

The matrices must have the same dimensions.


Can matrices of the same dimension be multiplied?

No. The number of columns of the first matrix needs to be the same as the number of rows of the second.So, matrices can only be multiplied is their dimensions are k*l and l*m. If the matrices are of the same dimension then the number of rows are the same so that k = l, and the number of columns are the same so that l = m. And therefore both matrices are l*l square matrices.


Can a Hermitian Matrix possess Complex Eigenvectors?

Yes. Simple example: a=(1 i) (-i 1) The eigenvalues of the Hermitean matrix a are 0 and 2 and the corresponding eigenvectors are (i -1) and (i 1). A Hermitean matrix always has real eigenvalues, but it can have complex eigenvectors.


What has the author Jan R Magnus written?

Jan R. Magnus has written: 'Linear structures' -- subject(s): Matrices 'The bias of forecasts from a first-order autoregression' 'The exact multiperiod mean-square forecast error for the first-order autoregressive model with an intercept' 'On differentiating Eigenvalues and Eigenvectors' 'The exact moments of a ratio of quadratic forms in normal variables' 'Symmetry, 0-1 matrices, and Jacobians'


What is linear combination in matrices?

If X1, X2 , ... , Xn are matrices of the same dimensions and a1, a2, ... an are constants, then Y = a1*X1 + a2*X2 + ... + an,*Xn is a linear combination of the X matrices.


What must be true in order to add matrices?

They must have the same dimensions.


How can you tell whether two matrices can be added?

If both matrices have the same number of columns and rows ex: {1 2 3 4} can not be added with {5 4} b/c they dont have the same amount of numbers


Can the elimnation matrices only be applied to square matrices?

Only square matrices have inverses.


How are a numerator and denominator similar?

They are elements of of a set which may consist of integers, real or complex numbers, polynomial expressions, matrices.