DavidWhitbeck said:Trambolin, looking at your first matrix the characteristic equation is
[tex](1-\lambda)(\lambda^2-1) = 0 \Rightarrow \lambda = \pm 1[/tex]
That permits two unique eigenvalues, not one.
Besides if you simply choose the coordinate system aligned with the eigenvectors, in that coordinate system the matrix (call it A) will be proportional to the identity matrix. Then the matrix [tex]A-\lambda I[/tex] will vanish in that coordinate system for some complex number [tex]\lambda[/tex], but then it would have to vanish in all coordinate systems and therefore the matrix is proportional to the identity matrix.
A normal nxn matrix is a square matrix where the commutative property holds, meaning that the order of multiplication of two matrices does not matter.
In the eigenvalue case of a normal nxn matrix, the matrix has a set of complex eigenvalues and corresponding eigenvectors. This means that when multiplied by its eigenvectors, the matrix behaves like a scalar multiple of the eigenvector.
To find the eigenvalues of a normal nxn matrix with complex numbers, you can use the characteristic polynomial method. This involves finding the determinant of the matrix minus a scalar multiple of the identity matrix, and then solving for the roots of the polynomial equation.
Yes, a normal nxn matrix with complex eigenvalues can have real eigenvectors. This is because the eigenvectors represent the direction of the transformation of the matrix, while the eigenvalues represent the magnitude of the transformation. So, even if the eigenvalues are complex, the eigenvectors can still be real.
No, not all normal nxn matrices are diagonalizable with complex numbers. A normal nxn matrix is diagonalizable if and only if it has n linearly independent eigenvectors, which may not always be the case for a normal matrix with complex eigenvalues.