Matrix Algebra

Distinct Eigenvalues

Theorem for distinct eigenvalues. Let

be an $n\times n$ square matrix. If $\lambda_1,\ldots,\lambda_k$ are distinct eigenvalues of

then the corresponding eigenvectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$ are linearly independent.

Proof by contradiction. For this we assume that $\mathbf{v}_1, \ldots, \mathbf{v}_k$ are linearly dependent, and draw a contradiction in the end. Then we can find the largest integer $p<k$ 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

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

Proof, continued. By operating $(A - \lambda_{p+1}I)$ on the above homogeneous equation, we can find

$\begin{displaymath} \begin{array}{l} (\lambda_1 - \lambda_{p+1}) x_1 \mathbf{... ...lambda_{p+1}) x_p \mathbf{v}_{p} = \mathbf{0}, \end{array} \end{displaymath}$

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

Application for diagonalization. As a corollary to the previous theorem we can show the following: Suppose that has distinct eigenvalues $\lambda_1,\ldots,\lambda_n$ . Then $P = [\mathbf{v}_1 \cdots \mathbf{v}_n]$ is invertible, and therefore, is diagonalizable.

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

EXERCISE 6. Diagonalize each of the following matrices if possible.

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