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.
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.
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, similar matrices have the same eigenvalues.
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.
you tell me
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.
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'
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.
They must have the same dimensions.
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
Only square matrices have inverses.
They are elements of of a set which may consist of integers, real or complex numbers, polynomial expressions, matrices.