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.
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.
int matrix[][]; // the matrix to find the max in int max = matrix[0][0]; int r,c; for(r = 0; r < 3; ++r) { for(c = 0; c < 3; ++c) { if(matrix[r][c] > max) { max = matrix[r][c]; } } } // max is now the maximum number in matrix
An eigenvector is a vector which, when transformed by a given matrix, is merely multiplied by a scalar constant; its direction isn't changed. An eigenvalue, in this context, is the factor by which the eigenvector is multiplied when transformed.
Yes, similar matrices have the same eigenvalues.
For example, if you have [ -4 1 0 3] as your matrix, it would be negative 4. Whatever negative number is in your matrix is your answer.
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.
int matrix[][]; // the matrix to find the max in int max = matrix[0][0]; int r,c; for(r = 0; r < 3; ++r) { for(c = 0; c < 3; ++c) { if(matrix[r][c] > max) { max = matrix[r][c]; } } } // max is now the maximum number in matrix
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Yes, similar matrices have the same eigenvalues.
An eigenvector is a vector which, when transformed by a given matrix, is merely multiplied by a scalar constant; its direction isn't changed. An eigenvalue, in this context, is the factor by which the eigenvector is multiplied when transformed.
For example, if you have [ -4 1 0 3] as your matrix, it would be negative 4. Whatever negative number is in your matrix is your answer.
This is just one of the ways:Choose the variables couple in question defining the SISO form of the system. Write out the state space matrix commonly denoted as "A" of the synchronous machine. Calculate the eigenvalues of that matrix. Then calculate the residues of the matrix with respect to the selected SISO system (the chosen variables in question define the input matrix B and output matrix C). The eigenvalues are the zeros of the transfer function while the residues are the constants in the fractionally partitioned form of the transfer function.The matrices I was talking about define the linearised system in the form :dx/dt=Ax+Buy=CxFor a more thorough explanation seePower System Stability And Control By Prabha Kundur
The answer is yes, and here's why: Remember that for the eigenvalues (k) and eigenvectors (v) of a matrix (M) the following holds: M.v = k*v, where "." denotes matrix multiplication. This operation is only defined if the number of columns in the first matrix is equal to the number of rows in the second, and the resulting matrix/vector will have as many rows as the first matrix, and as many columns as the second matrix. For example, if you have a 3 x 2 matrix and multiply with a 2 x 4 matrix, the result will be a 3 x 4 matrix. Applying this to the eigenvalue problem, where the second matrix is a vector, we see that if the matrix M is m x n and the vector is n x 1, the result will be an m x 1 vector. Clearly, this can never be a scalar multiple of the original vector.
It is the matrix 1/3It is the matrix 1/3It is the matrix 1/3It is the matrix 1/3
No. Say your matrix is called A, then a number e is an eigenvalue of A exactly when A-eI is singular, where I is the identity matrix of the same dimensions as A. A-eI is singular exactly when (A-eI)T is singular, but (A-eI)T=AT-(eI)T =AT-eI. Therefore we can conclude that e is an eigenvalue of A exactly when it is an eigenvalue of AT.
An eigenvector of a square matrix Ais a non-zero vector v that, when the matrix is multiplied by v, yields a constant multiple of v, the multiplier being commonly denoted by lambda. That is: Av = lambdavThe number lambda is called the eigenvalue of A corresponding to v.
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