No, in general they do not. They have the same eigenvalues but not the same eigenvectors.
Yes, similar matrices have the same eigenvalues.
The matrices must have the same dimensions.
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.
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.
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.
Yes, similar matrices have the same eigenvalues.
you tell me
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.
The matrices must have the same dimensions.
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.
To calculate eigenvectors in MATLAB, you can use the "eig" function. This function returns both the eigenvalues and eigenvectors of a given matrix. Simply input your matrix as an argument to the "eig" function, and it will output the eigenvectors corresponding to the eigenvalues.
Mathematica can be used to compute and normalize eigenvectors of a given matrix by using the Eigensystem function to find the eigenvectors and eigenvalues of the matrix. Then, the Normalize function can be applied to normalize the eigenvectors.
To find the eigenvalues and eigenvectors of a matrix using the numpy diagonalize function in Python, you can first create a matrix using numpy arrays. Then, use the numpy.linalg.eig function to compute the eigenvalues and eigenvectors. Here's an example code snippet: python import numpy as np Create a matrix A np.array(1, 2, 3, 4) Compute eigenvalues and eigenvectors eigenvalues, eigenvectors np.linalg.eig(A) print("Eigenvalues:", eigenvalues) print("Eigenvectors:", eigenvectors) This code will output the eigenvalues and eigenvectors of the matrix A.
To calculate eigenvalues and eigenvectors in MATLAB using the 'eig' function, the syntax is as follows: eigenvectors, eigenvalues eig(matrix) This command will return the eigenvectors and eigenvalues of the input matrix in a specific order.
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.
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.
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'