## Distinct eigenvalues

Suppose that the eigenvalues  of  are distinct. Then the corresponding eigenvectors  are linearly independent.

We prove it by contradiction. For this we assume that are linearly dependent. Then we can find the largest integer  so that are linearly independent, but are not. Thus, we should be able to find a nontrivial solution  to the homogeneous equation

 (15)

By operating on the above equation, we obtain

which implies a nontrivial solution for the homogeneous equation consisting of  . This is a contradiction! Therefore, must be linearly independent.

As a corollary, we can find that is diagonalizable.

EXAMPLE 2. Determine whether the matrix is diagonalizable or not.

EXERCISE 3. Diagonalize each of the following matrices if possible.

Department of Mathematics
Last modified: 2005-10-21