2.3 Distinct eigenvalues

We consider, in this section, the IVP

$$x' = Ax, \qquad x (0) = x_{0}$$ (2.4)

for the case when A has distinct eigenvalues.

1.

A has n real and distinct eigenvalues $\lambda_{1}, \ldots , \lambda_{n}$.

Let $v_{1}, \ldots , v_{n}$ be the corresponding eigenvectors of A and let $P = [ v_{1}, \ldots , v_{n} ]$. Then P is nonsingular and

$$P^{-1} AP = {\rm diag} \, [ \lambda_{1} , \ldots , \lambda_{n} ] \, .$$

Thus by Corollary 2.2.2,

$$x(t) = P {\rm diag} \; [e^{\lambda_{1} t}, \ldots , e^{\lambda_{n}t} ] P^{-1} x_{0}$$

is a solution of the IVP (2.4).

2.

A has 2m distinct complex eigenvalues.

Let $\lambda_{j} = \alpha_{j} + i \beta_{j}$ and $\ol{\lambda}_{j} = \alpha_{j} - i\beta_{j}$, $j=1, \ldots , m$ be 2m complex eigenvalues of A, and w_j = u_j + iv_j, $\ol{w}_{j} = u_{j} - iv_{j}$, $j=1, \ldots , m$ be the corresponding complex eigenvectors. Then $[ u_{1}, v_{1}, \ldots , u_{m}, v_{m} ]$ forms a basis for Rⁿ, n = 2m, the matrix

$$P = [u_{1}, v_{1}, \ldots , u_{m}, v_{m} ]$$

is nonsingular and

$$P^{-1} AP = {\rm diag} \; \left\{ \left[ \begin{array}{rl} \a... ... \\ - \beta_{j} & \alpha_{j} \end{array} \right] \right\} \, ,$$

is a real $2m \times 2m$ matrix with $2 \times 2$ blocks along the diagonal.

Thus by Theorem 2.2.1

$$x(t) = P {\rm diag} \; \left\{ e^{\alpha_{j}t} \left[ \begin{... ...t & \cos \beta_{j} t \end{array} \right] \right\} P^{-1} x_{0}$$

is a solution of the IVP (2.4).

Example 2.3.1:    Solve the IVP (2.4) with

$$A = \left[ \begin{array}{lrlr} 1 & -1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 3 & -2 \\ 0 & 0 & 1 & 1 \end{array} \right] \, .$$

A has eigenvalues $\lambda = 1 \pm i$, $2 \pm i$. We only need to consider $\lambda_{1} = 1+i$, $\lambda_{2} = 2 +i$.

The corresponding eigenvectors are $w_{1} = u_{1} + iv_{1} = \left[ \begin{array}{c} i \\ 1 \\ 0 \\ 0 \end{array} \right]$, $w_{2} = u_{2} + iv_{2} = \left[ \begin{array}{c} 0 \\ 0 \\ 1+i \\ 1 \end{array} \right]$.

3.

A has both real and complex eigenvalues and they are distinct.

Let $\lambda_{1}, \ldots , \lambda_{k}$ be the real eigenvalues and $v_{1}, \ldots , v_{k}$ be the corresponding eigenvectors. Let $\lambda_{k+j} = \alpha_{k+j} + i \beta_{k +j}$, $\ol{\lambda}_{k+j} = \alpha_{k+j} - i \beta_{k+j}$, $j = 1, 2, \ldots , m$ be the complex eigenvalues and w_k+j = u_k+j + iv_k+j, $\ol{w}_{k+j} = u_{k+j} - iv_{k+j}$, $j = 1,2, \ldots m$ be the corresponding eigenvectors, where k + 2m = n. Then the matrix

$$P = [v_{1}, \ldots , v_{k}, u_{k+1}, v_{k+1}, \ldots , u_{n}, v_{n} ]$$

is nonsingular and

$$P^{-1} AP = {\rm diag} \; [\lambda_{1}, \ldots , \lambda_{k}, B_{k+1}, \ldots , B_{n}],$$

where $B_{j} = \left[ \begin{array}{rl} \alpha_{j} & \beta_{j} \\ - \beta_{j} & \alpha_{j} \end{array} \right]$ $j = k+1, \ldots , n$.

Thus

$$x(t) = P {\rm diag} \; \Biggl\{ e^{\lambda_{1}t} , \ldots , e... ...beta_{k+1}t & \cos \beta_{k+1}t \end{array}\right] , \ldots ,$$

$$e^{\alpha_{n}t} \left[ \begin{array}{rl} \cos \beta_{n} t & \... ... \cos \beta_{n} t \end{array} \right] \Biggr\} \, P^{-1} x_{0}$$

is a solution of the IVP (2.4).

Example 2.3.2: (Perko, p. 30-31)     Solve the IVP (2.4) with

$$A = \left[ \begin{array}{rlr} -3 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 1 & 1 \end{array} \right] \, .$$