Recall the solution to the IVP $y^{\text{'}}=\lambda y$, $y\left(0\right)=0$ is $y\left(t\right)=e^{\lambda t}{y}_0$, with $y:ℝ\to ℝ$

The solution to the system IVP ${𝒚}^{\text{'}}=𝑨𝒚$, $𝒚\left(0\right)={𝒚}_0$ is similar, $𝒚\left(t\right)=e^{𝑨t}{𝒚}_0$, with:

• ${𝒚}_0\in {ℝ}^n,𝒚:ℝ\to {ℝ}^n$

• $𝑨\in {ℝ}^{n\times n}$

• $e^{𝑨t}\in {ℝ}^{n\times n}$ is known as the matrix exponential defined by

  ${\displaystyle e^{𝑨t}=𝑰+𝑨\frac{t}{1!}+{𝑨}^2\frac{t^2}{2!}+\cdots +{𝑨}^k\frac{t^k}{k!}+\cdots }$

If $𝑨$ is diagonalizable, i.e., the eigenvector matrix $𝑿\in {ℝ}^{n\times n}$

arising in the eigenvalue problem $𝑨𝑿=𝑿𝚲$ (or $𝑨{𝒙}_k={\lambda }_k{𝒙}_k$, $k=1,\dots ,n$) is invertible, then

Diagonalizable matrices include:

• matrices with distinct eigenvalues, $j\ne k\Rightarrow {\lambda }_j\ne {\lambda }_k$

• symmetric matrices, ${𝑨}^T=𝑨$, the eigenvalues of which are purely real

• skew-symmetric matrices, ${𝑨}^T=-𝑨$, the eigenvalues of which are purely imaginary

Assume henceforth that $𝑨$ is diagonalizable

Rewrite solution as

and observe that $\{{𝒚}_1,\dots ,{𝒚}_n\}=\{e^{{\lambda }_{1}t}{𝒙}_1,\dots ,e^{{\lambda }_{n}t}{𝒙}_n\}$ are independent and form a basis for solutions of ${𝒚}^{\text{'}}=𝑨𝒚$