Can a matrix have no eigenvalues?

Therefore, any real matrix with odd order has at least one real eigenvalue, whereas a real matrix with even order may not have any real eigenvalues. The eigenvectors associated with these complex eigenvalues are also complex and also appear in complex conjugate pairs.

Are there matrices without eigenvalues?

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.

Do all matrix have eigenvalues?

Theorem EMHE Every Matrix Has an Eigenvalue

Suppose A is a square matrix. Then A has at least one eigenvalue.

Does every complex matrix have n eigenvalues?

As a consequence of the fundamental theorem of algebra as applied to the characteristic polynomial, we see that: Every n × n matrix has exactly n complex eigenvalues, counted with multiplicity.

What does it mean if a matrix has 0 as an eigenvalue?

If 0 is an eigenvalue, then the nullspace is non-trivial and the matrix is not invertible.

Similar matrices and eigenvalues

Is matrix invertible if eigenvalue is zero?

A square matrix is invertible if and only if zero is not an eigenvalue. Solution note: True. Zero is an eigenvalue means that there is a non-zero element in the kernel. For a square matrix, being invertible is the same as having kernel zero.

Does zero count as an eigenvalue?

Fact. Let A be an n × n matrix. The number 0 is an eigenvalue of A if and only if A is not invertible.

Can an eigenvalue have no eigenvectors?

Yes. If it has rank n then its columns span an n-dimensional space, which is all of Rn. By the rank theorem, its nullspace is zero-dimensional, and only contains the zero vector. So there is no nonzero vector v with (A-\lambda I)v=0,orAv=\lambda v$, and do no eigenvector.

Does every Nxn matrix have n eigenvalues?

All N X N square matrices have N eigenvalues; that's just the same as saying that an Nth order polynomial has N roots. While a defective matrix still has N eigenvalues, it does not have N independent eigenvectors.

Can a matrix have less than N eigenvectors?

If an n×n matrix has less than n linearly independent eigenvectors, it is said to be deficient.

Do all matrices have non zero eigenvectors?

Yes it must have. It is a direct consequence of the Jordan Normal Form. As the answers show, even more is true: every square complex matrix has a nonzero eigenvector.

Does the zero matrix have eigenvectors?

All nonzero vectors are eigenvectors, since all vectors v satisfy Mv=0v if M is the zero matrix. Thanks @Hans.

Does every linear transformation have an eigenvalue?

Caution: Not every linear transformation has an eigenvalues! π/2. Since no non-zero vector is taken to a scalar multiple of itself, ρ has no eigenvectors.

Do only square matrices have eigenvalues?

It is not exactly true that non-square matrices can have eigenvalues. Indeed, the definition of an eigenvalue is for square matrices.

Can an orthogonal matrix have no real eigenvalues?

The second statement should say that the determinant of an orthogonal matrix is ±1 and not the eigenvalues themselves. R is an orthogonal matrix, but its eigenvalues are e±i. The eigenvalues of an orthogonal matrix needs to have modulus one. If the eigenvalues happen to be real, then they are forced to be ±1.

Do real matrices have real eigenvalues?

No, a real matrix does not necessarily have real eigenvalues; an example is (01−10).

How many eigenvalues does an Nxn matrix have?

If you take the diagonal matrix diag(1,1,2), it has two distinct eigenvalues 1,2, with 1 being repeated. If you take the diagonal matrix diag(1,2,3), it has 3 distinct eigenvalues 1, 2, 3.

Can an invertible matrix have eigenvalues?

Eigenvalues of an Inverse

An invertible matrix cannot have an eigenvalue equal to zero. Furthermore, the eigenvalues of the inverse matrix are equal to the inverse of the eigenvalues of the original matrix: Ax=λx⟹A−1Ax=λA−1x⟹x=λA−1x⟹A−1x=1λx.

How many eigenvalues exist for a 4x4 matrix?

Its only eigenvalues are 1,2,3,4,5, possibly with multiplicities.

Does every vector have an eigenvector?

Solution note: Every non-zero vector is an eigenvector of the identity map, with eigenvalue 1. Every non-zero vector is an eigenvector of the zero map, with eigenvalue 0. Every non-zero vector is an eigenvector of the “scale by k” map, with eigenvalue k.

What is the purpose of eigenvalues?

Eigenvalues and eigenvectors allow us to "reduce" a linear operation to separate, simpler, problems. For example, if a stress is applied to a "plastic" solid, the deformation can be dissected into "principle directions"- those directions in which the deformation is greatest.

What do eigenvalues tell us?

An eigenvalue is a number, telling you how much variance there is in the data in that direction, in the example above the eigenvalue is a number telling us how spread out the data is on the line. The eigenvector with the highest eigenvalue is therefore the principal component.

What happens when all eigenvalues are zero?

As we know the determinant of a matrix is equal to the products of all eigenvalues. So, if one or more eigenvalues are zero then the determinant is zero and that is a singular matrix. If all eigenvalues are zero then that is a Nilpotent Matrix. And for any such matrix A: A^k = 0 for some specific k.

What happens when an eigenvector is 0?

If we let zero be an eigenvector, we would have to repeatedly say "assume v is a nonzero eigenvector such that..." since we aren't interested in the zero vector. The reason being that v=0 is always a solution to the system Av=λv. An eigenvalue always has at least a one-dimensional space of eigenvectors.

Does a singular matrix have eigenvalues?

Selected Properties of Eigenvalues and Eigenvectors

A matrix with a 0 eigenvalue is singular, and every singular matrix has a 0 eigenvalue.