Why the Eigenmode Formula Holds

From a State Update to Eigenmodes

A discrete-time linear system updates its state by applying the same matrix at every step. If the state at step \(k\) is \(\mathbf{x}_k\in\mathbb{R}^n\), and the transition matrix is \(\mathbf{A}\in\mathbb{R}^{n\times n}\), the update rule is \(\mathbf{x}_{k+1}=\mathbf{A}\mathbf{x}_k\). Repeated updates give \(\mathbf{x}_k=\mathbf{A}^k\mathbf{x}_0\), where \(\mathbf{x}_0\) is the initial state and \(\mathbf{A}^0=\mathbf{I}_n\). The central question here is how to understand this repeated action when \(\mathbf{A}\) is diagonalizable.

An eigenvector is a nonzero vector \(\mathbf{v}\) for which \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for some scalar \(\lambda\), called its eigenvalue. Such a vector has a particularly simple response to one update: it keeps the same direction and is scaled by \(\lambda\). A component of the initial state that lies along an eigenvector is called an eigenmode. We will show that after \(k\) updates this component is scaled by \(\lambda^k\). This describes the exact evolution of the components; it does not, by itself, make a claim about what happens as \(k\) becomes large.

Here the coefficients \(c_i\) are coordinates in the eigenvector basis, not necessarily the ordinary entries of \(\mathbf{x}_0\). A basis spans the space, so such coefficients exist. They are unique because if two coefficient lists gave the same vector, subtracting the expressions would give a linear combination of the basis vectors equal to \(\mathbf{0}\); linear independence forces every coefficient in that difference to be zero.

Why One Eigenvector Picks Up a Power

Proof. For \(k=0\), \(\mathbf{A}^0\mathbf{v}=\mathbf{I}_n\mathbf{v}=\mathbf{v}=1\mathbf{v}=\lambda^0\mathbf{v}\). Suppose the formula holds for an integer \(k\geq0\). Then, using the definition of a matrix power, the induction hypothesis, and the eigenvector equation,

Thus induction proves the claim for every \(k\geq0\). The convention \(\lambda^0=1\) makes the initial step valid even for \(\lambda=0\); for \(k\geq1\), a zero eigenvalue gives \(\mathbf{A}^k\mathbf{v}=\mathbf{0}\). \(\square\)

The theorem concerns one eigenvector at a time. To apply it to every initial state, we need the eigenvectors to form a basis, so that every state is a sum of eigenvector components. The next result combines the individual formula with that decomposition.

Proof. Fix an integer \(k\geq0\). Since \(\mathbf{x}_0=\sum_{i=1}^{n}c_i\mathbf{v}_i\), linearity of matrix multiplication gives \(\mathbf{A}^k\mathbf{x}_0=\sum_{i=1}^{n}c_i\mathbf{A}^k\mathbf{v}_i\). The eigenvector iteration theorem applies to each \(\mathbf{v}_i\), so \(\mathbf{A}^k\mathbf{v}_i=\lambda_i^k\mathbf{v}_i\). Substituting these equalities yields \(\mathbf{A}^k\mathbf{x}_0=\sum_{i=1}^{n}c_i\lambda_i^k\mathbf{v}_i\), as claimed. At \(k=0\), every factor \(\lambda_i^0\) is \(1\), so the right side is the original decomposition of \(\mathbf{x}_0\). \(\square\)

This proof also explains the role of diagonalizability. The eigenvector iteration theorem always applies to an eigenvector, but a basis of eigenvectors is what lets us use it to describe an arbitrary initial state. The formula does not require the eigenvalues to be distinct: separate basis vectors may have the same eigenvalue, in which case their components are each multiplied by the same power.

Worked Examples: Computing the Components

Reading the Formula Correctly

The eigenmode formula separates the state into contributions that can be computed independently and then added. For each basis vector \(\mathbf{v}_i\), the initial coefficient \(c_i\) remains in the expression, while the repeated updates contribute the factor \(\lambda_i^k\). The factor is a power because each update applies one additional multiplication by \(\lambda_i\): after one step it is \(\lambda_i\), after two it is \(\lambda_i\lambda_i\), and the induction proof covers every nonnegative integer number of steps.

In matrix notation, if \(\mathbf{P}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_n\end{bmatrix}\) and \(\mathbf{D}\) is the diagonal matrix with entries \(\lambda_1,\dots,\lambda_n\), then \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). The columns of \(\mathbf{P}\) are a basis, so \(\mathbf{P}^{-1}\mathbf{x}_0\) is the column of eigenbasis coefficients. Accordingly, \(\mathbf{A}^k\mathbf{x}_0=\mathbf{P}\mathbf{D}^k\mathbf{P}^{-1}\mathbf{x}_0\): the diagonal entries of \(\mathbf{D}^k\) are \(\lambda_i^k\). This is the same component-by-component statement as the eigenmode formula.

The formula requires an eigenvector basis to represent every possible initial state in this way. If a matrix has an eigenvector that does not belong to any basis of eigenvectors, the individual eigenvector iteration result still holds, but this particular formula does not give a basis decomposition for every state. Therefore, before using the formula for a system, check that the stated diagonalization is valid and use the eigenbasis coordinates of \(\mathbf{x}_0\).

Common Pitfall: Using the Eigenvalue Instead of Its Power

A frequent mistake is to apply the one-step rule once and then treat it as the rule for all \(k\). For the first worked example, an incorrect calculation at \(k=2\) would replace each component by just one factor of its eigenvalue and write \(2\cdot2\mathbf{v}_1+2\cdot3\mathbf{v}_2=4\mathbf{v}_1+6\mathbf{v}_2=\begin{bmatrix}10\\6\end{bmatrix}\). That is the state after one update, not two.

The correction is to use the exponent \(k\) on each eigenvalue. At \(k=2\), the first component is \(2\cdot2^2\mathbf{v}_1=8\mathbf{v}_1\), and the second is \(2\cdot3^2\mathbf{v}_2=18\mathbf{v}_2\). Their sum is \(\begin{bmatrix}26\\18\end{bmatrix}\), which agrees with the direct calculation \(\mathbf{A}^2\mathbf{x}_0=\begin{bmatrix}26\\18\end{bmatrix}\). A reliable check is to compare a claimed iterate with direct multiplication for a small value of \(k\).