For what conditions on A and B is the resulting matrix A diagonalizable?

for a=(−1), the eigenspaces are linearly dependent. for a=1, the trace of the diagonal form matrix (call it D ) isn’t equal to the trace of the matrix that’s composed of the eigenvectors (call it Q ). (‘ a ‘ must be 1 or (-1) according to the homogeneous equations with which I found the eigenspaces)

What makes something diagonalizable?

A diagonalizable matrix is any square matrix or linear map where it is possible to sum the eigenspaces to create a corresponding diagonal matrix. An n matrix is diagonalizable if the sum of the eigenspace dimensions is equal to n. A matrix that is not diagonalizable is considered “defective.”

Can a matrix be diagonalizable and not invertible?

Diagonalizability does not imply invertibility: Any diagonal matrix with a somewhere on the main diagonal is an example. Most matrices are invertible: Since the determinant is a polynomial in the matrix entries, the set of matrices with determinant equal to is a subvariety of dimension .

How do you check if a transformation is diagonalizable?

A linear transformation, T : Rn → Rn, is diagonalizable if there is a basis B of Rn so that [T]B is diagonal. This means [T] is similar to the diagonal matrix [T]B. Similarly, a matrix A ∈ Rn×n is diagonalizable if it is similar to some diagonal matrix D.

Under what conditions on a B and C is a diagonalizable?

How does a,b,c affects the nullity of the matrices? To be diagonalizable, it has to be has nullity of 2 (the algebraic multiplicity of eigenvalue 1), i.e. the matrix A−I has to be of rank 1.

Which of the following is not a necessary condition for a matrix A to be diagonalizable?

1. Which of the following is not a necessary condition for a matrix, say A, to be diagonalizable? Explanation: The theorem of diagonalization states that, ‘An n×n matrix A is diagonalizable, if and only if, A has n linearly independent eigenvectors.

Why symmetric matrix is diagonalizable?

Symmetric matrices are diagonalizable because there is an explicit algorithm for finding a basis of eigenvectors for them. The key fact is that the unit ball is compact.

How do you prove a map is diagonalizable?

Definition: A linear mapping T on a finite-dimensional vector space V is diagonalizable if there is an ordered basis β for V such that [T]β is a diagonal matrix. proof: ⇒ suppose β={v1,…,vn} is a basis for V for which [T]β is diagonal, then by definition , we have [T]β=(λ1…

Is matrix diagonalizable characteristic polynomial?

diagonalizable matrices are similar to diagonal matrices). In particu- lar, if the characteristic polynomial of a matrix d oesn ‘t split, then it can ‘t be diagonalizable. from an earlier example cannot be diagonalizable over the reals, because its characteristic polynomial does not split over the reals.