Exercise 765Calculus › Differential equations › Linear systemsL2Added 2 Nov 2026

A diagonalizable system

#diagonalization#eigenvalues#eigenvectors

Statement

Solve the system $X' = AX$ where
$A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \\ -1 & -2 & 1 \end{pmatrix}.$

Hint

Find the eigenvalues and eigenvectors of $A$ . The characteristic polynomial is $X^2(X-6)$ .

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Eigenvalues. $\det(A - \lambda I) = -\lambda^2(\lambda - 6)$ , so the eigenvalues are $0$ (double) and $6$ .

Eigenvectors for $\lambda=0$ . Solve $AX=0$ : the kernel has dimension $2$ , with basis
$u_1 = (1,0,1), \qquad u_2 = (2,-1,0).$

Eigenvector for $\lambda=6$ . Solve $(A-6I)X=0$ : a basis is
$u_3 = (1,2,-1).$

General solution.
$X(t) = \lambda u_1 + \mu u_2 + \gamma e^{6t} u_3, \qquad \lambda,\mu,\gamma \in \mathbb{R}.$