Diagonalization and Discrete Dynamics

Separate independent modes

If an $n\times n$ matrix has $n$ independent eigenvectors, put them in columns of $V$ and their eigenvalues in the same order on diagonal $D$ . The equations $Av_j=\lambda_jv_j$ combine into $AV=VD$ , hence $A=VDV^{-1}$ .

Then $A^k=VD^kV^{-1}$ for nonnegative integers $k$ . A state evolving by $x_{k+1}=Ax_k$ has solution $x_k=A^kx_0$ . Each eigenmode is multiplied by $\lambda^k$ . Modes with $|\lambda|<1$ decay; those with $|\lambda|>1$ grow in magnitude when their initial coefficient is nonzero.

Visual guide

VISUAL GUIDEAn eigenbasis separates the modes

First eigenline
Second eigenline

For A = [[2, 1], [1, 2]], (1, 1) has eigenvalue 3 and (1, −1) has eigenvalue 1. In this tilted basis, the matrix simply stretches each coordinate independently; diagonalization changes the coordinates used to describe that action.

Worked example: two decaying modes

Let $A=\begin{pmatrix}0.75&0.25\\0.25&0.75\end{pmatrix}$ . The vectors $(1,1)^T$ and $(1,-1)^T$ have eigenvalues $1$ and $1/2$ . Starting from $x_0=(1,0)^T$ ,

x_k=\frac12(1,1)^T+\frac12(1/2)^k(1,-1)^T.

The state tends to $(1/2,1/2)^T$ : the difference mode decays while the total remains constant. The eigenvalue-one mode does not decay.

Worked example: why a repeated root is not enough

The matrix $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has repeated eigenvalue $1$ , but its eigenspace contains only multiples of $(1,0)^T$ . There are too few independent eigenvectors to diagonalize it.

Writing $A=I+N$ with $N^2=0$ gives $A^k=I+kN=\begin{pmatrix}1&k\\0&1\end{pmatrix}$ . Some states grow linearly despite every eigenvalue having magnitude one. Boundary stability claims require attention to defective eigenvalues, not just magnitudes.

Distinct eigenvalues guarantee independent eigenvectors, so $n$ distinct eigenvalues guarantee diagonalizability over the field containing them. Repeated eigenvalues may still be diagonalizable; the identity matrix is a simple example.

Continuous-time bridge

For the constant-coefficient system $x'(t)=Ax(t)$ , the solution is $x(t)=e^{tA}x(0)$ , where $e^{tA}=\sum_{k=0}^{\infty}t^kA^k/k!$ . When $A=VDV^{-1}$ , this becomes $Ve^{tD}V^{-1}$ : each diagonal entry is $e^{t\lambda_i}$ . For $A=\operatorname{diag}(-1,2)$ and $x(0)=(3,1)^T$ , the solution is $(3e^{-t},e^{2t})^T$ . Continuous decay depends on negative real parts of eigenvalues, while discrete decay depends on magnitudes below one. The differential-equations course develops this distinction.

Practice

Compute $\operatorname{diag}(2,3)^4$ .
Does $A=2I$ diagonalize despite its repeated eigenvalue?
For a diagonalizable matrix with all $|\lambda|<1$ , what happens to $A^kx_0$ ?

Show worked solutions

$\operatorname{diag}(16,81)$ .
Yes. Every basis is an eigenvector basis.
Every mode decays, so the state tends to zero for every fixed initial vector.