Distinct eigenvalues

Suppose that the eigenvalues  $ \lambda_1,\ldots,\lambda_n$ of $ A$ are distinct. Then the corresponding eigenvectors  $ \mathbf{v}_1,\ldots,\mathbf{v}_n$ are linearly independent.

We prove it by contradiction. For this we assume that $ \mathbf{v}_1,\ldots,\mathbf{v}_n$ are linearly dependent. Then we can find the largest integer $ p$ so that $ \mathbf{v}_1,\ldots,\mathbf{v}_p$ are linearly independent, but $ \mathbf{v}_1,\ldots,\mathbf{v}_p,\mathbf{v}_{p+1}$ are not. Thus, we should be able to find a nontrivial solution  $ x_1,\ldots,x_p,x_{p+1}$ to the homogeneous equation

$\displaystyle x_1 \mathbf{v}_{1} + \cdots + x_p \mathbf{v}_{p} + x_{p+1} \mathbf{v}_{p+1} = \mathbf{0}.$ (15)

By operating $ (A - \lambda_{p+1}I)$ on the above equation, we obtain
    $\displaystyle (\lambda_1 - \lambda_{p+1}) x_1 \mathbf{v}_{1} + \cdots +
(\lambd...
...) x_p \mathbf{v}_{p} +
(\lambda_{p+1} - \lambda_{p+1}) x_{p+1} \mathbf{v}_{p+1}$  
    $\displaystyle = (\lambda_1 - \lambda_{p+1}) x_1 \mathbf{v}_{1} + \cdots +
(\lambda_p - \lambda_{p+1}) x_p \mathbf{v}_{p}
= \mathbf{0},$  

which implies a nontrivial solution for the homogeneous equation consisting of  $ \mathbf{v}_1,\ldots,\mathbf{v}_p$. This is a contradiction! Therefore, $ \mathbf{v}_1,\ldots,\mathbf{v}_n$ must be linearly independent.

As a corollary, we can find that $ A$ is diagonalizable.

EXAMPLE 2. Determine whether the matrix $ A = \begin{bmatrix}
5 &-8 & 1 \\
0 & 0 & 7 \\
0 & 0 &-2
\end{bmatrix}$ is diagonalizable or not.

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

  1. $ A = \begin{bmatrix}
5 & 1 \\
0 & 5
\end{bmatrix}$
  2. $ A = \begin{bmatrix}
4 & 2 & 2 \\
2 & 4 & 2 \\
2 & 2 & 4
\end{bmatrix}$



Department of Mathematics
Last modified: 2005-10-21