What does it mean when eigenvalue is complex?

That the two eigenvalues are complex conjugate to each other is no coincidence. If the n × n matrix A has real entries, its complex eigenvalues will always occur in complex conjugate pairs. If λ ∈ C is a complex eigenvalue of A, with a non-zero eigenvector v ∈ Cn, by definition this means: Av = λv, v = 0.

What happens when eigenvalues are repeated?

We say an eigenvalue A1 of A is repeated if it is a multiple root of the char acteristic equation of A; in our case, as this is a quadratic equation, the only possible case is when A1 is a double real root. This gives the solution x1 = eA1tv1 to the system (1).

Is the eigenvalue Decomposition unique?

1, applies to square symmetric matrices and is the basis of the singular value decomposition described in Theorem 18.2. If the eigenvalues are distinct, then this decomposition is unique.

What is the difference between Eigen decomposition and SVD?

In the eigendecomposition, the entries of D can be any complex number – negative, positive, imaginary, whatever. The SVD always exists for any sort of rectangular or square matrix, whereas the eigendecomposition can only exists for square matrices, and even among square matrices sometimes it doesn’t exist.

What is the eigenvalue decomposition of a matrix?

Eigenvalue decomposition is a decomposition of a matrix into a “canonical form”. we want to constrict a diagonal matrix from a given one. a matrix $A$ is diagonalizable if it’s similar to a diagonal matrix (a matrix $A$ is similar to $B$ if there exists an invertible $M$ s.t.

What do you need to know about eigendecomposition?

The eigendecomposition is one form of matrix decomposition. Decomposing a matrix means that we want to find a product of matrices that is equal to the initial matrix. In the case of eigendecomposition, we decompose the initial matrix into the product of its eigenvectors and eigenvalues.

How to reduce the complexity of finding the eigenvalues?

Basically, you can reduce eigenvalue computation to matrix multiplication by computing a symbolic determinant. This gives a running time of O ( n ω + 1 m) to get m bits of the eigenvalues; the best currently known runtime is O ( n 3 + n 2 log 2 b) for an approximation within 2 − b.

Is the complexity of eigenvalue decomposition in IEEE machine arithmetic?

EDIT: this is in the usual numerical linear algerbra model where the basic operations (+,-,*,/) are performed approximately in IEEE machine arithmetic and cost O ( 1) each. If you consider multiple precision and variable complexities depending on the bit length of numbers, that is a completely different beast.