Can a matrix have more than N eigenvalues?

Can a matrix have more than N eigenvalues?

So a square matrix A of order n will not have more than n eigenvalues. So the eigenvalues of D are a, b, c, and d, i.e. the entries on the diagonal. So depending on the values you have on the diagonal, you may have one eigenvalue, two eigenvalues, or more. Anything is possible.

How many eigen values does a matrix have?

two eigenvalues
Since the characteristic polynomial of matrices is always a quadratic polynomial, it follows that matrices have precisely two eigenvalues — including multiplicity — and these can be described as follows.

Does a matrix always have eigenvalues?

If the scalar field is the field of complex numbers, then the answer is YES, every square matrix has an eigenvalue. This stems from the fact that the field of complex numbers is algebraically closed.

Can a 3×3 matrix have only 2 eigenvalues?

This follows from the determinant formula for the eigenvalues of a matrix and the Fundamental Theorem of Algebra. If you take the 3×3 (multiplicative) identity matrix I_{3}, it has the eigenvalue 1 repeated 3 times. If you take the diagonal matrix diag(1,1,2), it has two distinct eigenvalues 1,2, with 1 being repeated.

Can a matrix have infinite eigenvectors?

Since a nonzero subspace is infinite, every eigenvalue has infinitely many eigenvectors. On the other hand, there can be at most n linearly independent eigenvectors of an n × n matrix, since R n has dimension n .

Are there infinite eigenvectors?

Since a nonzero subspace is infinite, every eigenvalue has infinitely many eigenvectors. (For example, multiplying an eigenvector by a nonzero scalar gives another eigenvector.) On the other hand, there can be at most n linearly independent eigenvectors of an n × n matrix, since R n has dimension n .

How do you know if a matrix has eigenvalues?

In order to determine the eigenvectors of a matrix, you must first determine the eigenvalues. Substitute one eigenvalue λ into the equation A x = λ x—or, equivalently, into ( A − λ I) x = 0—and solve for x; the resulting nonzero solutons form the set of eigenvectors of A corresponding to the selectd eigenvalue.

Is a matrix defective?

In linear algebra, a defective matrix is a square matrix that does not have a complete basis of eigenvectors, and is therefore not diagonalizable. In particular, an n × n matrix is defective if and only if it does not have n linearly independent eigenvectors.