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

By operating on the above equation, we obtain

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

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