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Do similar matrices have the same eigenvectors?
No, in general they do not. They have the same eigenvalues but not the same eigenvectors.
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.
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.
How do you find eigenvalues of a 3 by 3 matrix?
Call your matrix A, the eigenvalues are defined as the numbers e for which a nonzero vector v exists such that Av = ev. This is equivalent to requiring (A-eI)v=0 to have a non zero solution v, where I is the identity matrix of the same dimensions as A. A matrix A-eI with this property is called singular and has a zero determinant. The determinant of A-eI is a polynomial in e, which has the eigenvalues of A as roots. Often setting this polynomial to zero and solving for e is the easiest way to compute the eigenvalues of A.
Do similar matrices have the same eigenvectors?
No, in general they do not. They have the same eigenvalues but not the same eigenvectors.
How can you prove that similar matrices have the same trace?
you tell me
What has the author Carl Sheldon Park written?
Carl Sheldon Park has written: 'Real eigenvalues of unsymmetric matrices' -- subject(s): Aeronautics
What has the author V L Girko written?
V. L. Girko has written: 'Theory of random determinants' -- subject(s): Determinants, Stochastic matrices 'An introduction to statistical analysis of random arrays' -- subject(s): Eigenvalues, Multivariate analysis, Random matrices
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.
What has the author Doron Gill written?
Doron Gill has written: 'An O(N2) method for computing the Eigensystem of N x N symmetric tridiagonal matrices by the divide and conquer approach' -- subject(s): Eigenvalues
How can I prove that similar matrices have same eigenvalues?
First, we'll start with the definition of an eigenvalue. Let v be a non-zero vector and A be a linear transformation acting on v. k is an eigenvalue of the linear transformation A if the following equation is satisfied:Av = kvMeaning the linear transformation has just scaled the vector, v, not changed its direction, by the value, k.By definition, two matrices, A and B, are similar if B = TAT-1, where T is the change of basis matrix.Let w be some vector that has had its base changed via Tv.Therefore v = T-1wWe want to show that Bw = kvBw = TAT-1w = TAv = Tkv = kTv= kwQ.E.D.
Is every unitary matrix hermitian?
Absolutely not. They are rather quite different: hermitian matrices usually change the norm of vector while unitary ones do not (you can convince yourself by taking the spectral decomposition: eigenvalues of unitary operators are phase factors while an hermitian matrix has real numbers as eigenvalues so they modify the norm of vectors). So unitary matrices are good "maps" whiule hermitian ones are not. If you think about it a little bit you will be able to demonstrate the following: for every Hilbert space except C^2 a unitary matrix cannot be hermitian and vice versa. For the particular case H=C^2 this is not true (e.g. Pauli matrices are hermitian and unitary).
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 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'
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.
