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For $z\ne0$, the polar form of $z$ is

where $\theta$ is an argument of $z$. The expressions $a+bi$ and $re^{i\theta}$ are two writings of the same complex number.

Write a characteristic root in polar form as

Then

Thus $\rho_j<1$ gives a decaying mode, $\rho_j>1$ gives an unbounded mode and $\rho_j=1$ gives a non-decaying mode.

For a characteristic root $r=\alpha+i\beta$,

Therefore $\alpha<0$ gives decay, $\alpha>0$ gives an unbounded mode and, for a simple root, $\alpha=0$ gives a non-decaying oscillatory or constant mode. A repeated root on the imaginary axis introduces a term $t e^{rt}$, whose magnitude is unbounded.

Solve the characteristic equation and choose the real homogeneous modes according to the roots:

• If $r_1\neq r_2$ are distinct real roots, use $$y_h=Ae^{r_1x}+Be^{r_2x}.$$
• If $r$ is a repeated real root, use $$y_h=(A+Bx)e^{rx}.$$
• If the roots are $r=\alpha\pm i\beta$ with $\beta\neq0$, use $$y_h=e^{\alpha x} \bigl(A\cos(\beta x)+B\sin(\beta x)\bigr).$$

Suppose $A=PDP^{-1}$ is diagonalisable, so it has an eigenbasis. A mode contribution $c\lambda^kv$ with $c\ne0$ decays to zero when $|\lambda|<1$, has constant magnitude when $|\lambda|=1$ and grows when $|\lambda|>1$. Without an eigenbasis, a Jordan block can produce additional factors such as $k\lambda^{k-1}$, so the diagonalisation hypothesis is what makes this scalar classification complete (especially at the unit circle).

For a constant matrix $A$ and a non-zero vector $\mathbf u$, the function

is a solution of $\mathbf{x}'=A\mathbf{x}$ if and only if $\lambda$ is an eigenvalue of $A$ and $\mathbf{u}$ is a corresponding eigenvector. The zero solution is handled separately by taking $\mathbf u=\mathbf0$.

Suppose that a transition matrix $A$ satisfies:

1. $1$ is an eigenvalue with algebraic multiplicity $1$;
2. every other eigenvalue $\lambda_j$ has $|\lambda_j|<1$;
3. there is a basis of eigenvectors for $\mathbb R^n$.

Then $\mathbf{x}(k+1)=A\mathbf{x}(k)$ has a limit $\mathbf{p}$ in the eigenspace for $\lambda=1$. It is the unique vector in that eigenspace whose component sum equals the component sum of $\mathbf{x}(0)$; it may be zero when that sum is zero. If $\mathbf{x}(0)$ is a probability vector, then $\mathbf{p}$ is the stationary distribution and does not depend on the initial probabilities.