From Repeated Updates to a Power
A discrete-time linear system updates its state by applying the same matrix at each step. If the state at step \(k\) is \(\mathbf{x}_k\in\mathbb{R}^n\) and the state matrix is \(\mathbf{A}\in\mathbb{R}^{n\times n}\), the update rule is \(\mathbf{x}_{k+1}=\mathbf{A}\mathbf{x}_k\). Starting from \(\mathbf{x}_0\), the first few states are \(\mathbf{x}_1=\mathbf{A}\mathbf{x}_0\), \(\mathbf{x}_2=\mathbf{A}^2\mathbf{x}_0\), and \(\mathbf{x}_3=\mathbf{A}^3\mathbf{x}_0\). Thus the state after \(k\) steps is determined by the matrix power \(\mathbf{A}^k\).
Calculating \(\mathbf{A}^k\) by multiplying \(\mathbf{A}\) by itself repeatedly can become cumbersome when \(k\) is large. If \(\mathbf{A}\) is diagonalizable over \(\mathbb{R}\), its powers can instead be computed using powers of a diagonal matrix. A diagonal matrix is especially convenient to raise to a power: each diagonal entry is raised to that power separately.
The diagonal entries of \(\mathbf{D}\) may be repeated, and they may be zero. To use this method, the essential requirement is the displayed factorization with \(\mathbf{P}\) invertible; having real eigenvalues alone is not the definition of diagonalizability. Once the factorization is known, the power formula applies for every nonnegative integer \(k\), including \(k=0\).
Proof. For \(k=0\), the definition of the zeroth power gives \(\mathbf{A}^0=\mathbf{I}_n\), and \(\mathbf{P}\mathbf{D}^0\mathbf{P}^{-1}=\mathbf{P}\mathbf{I}_n\mathbf{P}^{-1}=\mathbf{P}\mathbf{P}^{-1}=\mathbf{I}_n\). Now suppose the formula holds for some \(k\geq0\). Using \(\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}_n\), we obtain $$ \mathbf{A}^{k+1}=\mathbf{A}^k\mathbf{A} =(\mathbf{P}\mathbf{D}^k\mathbf{P}^{-1})(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}) =\mathbf{P}\mathbf{D}^k\mathbf{D}\mathbf{P}^{-1} =\mathbf{P}\mathbf{D}^{k+1}\mathbf{P}^{-1}. $$ Induction proves the formula for every \(k\geq0\). The rule for \(\mathbf{D}^k\) follows from multiplying diagonal matrices: the diagonal entries multiply with one another, and every off-diagonal entry remains zero. \(\square\)
A Practical Calculation
To calculate a state from a diagonalization, first express the initial state in the eigenvector basis. If \(\mathbf{P}\) has the eigenvectors as its columns, this means computing \(\mathbf{c}=\mathbf{P}^{-1}\mathbf{x}_0\). Then \(\mathbf{D}^k\) scales each coordinate by the corresponding eigenvalue raised to \(k\), and multiplying by \(\mathbf{P}\) returns the result to the original coordinates. In symbols, \(\mathbf{x}_k=\mathbf{P}\mathbf{D}^k\mathbf{c}\). This is the power formula applied to the initial state, with \(\mathbf{P}^{-1}\mathbf{x}_0=\mathbf{c}\).
Place the basis eigenvectors, in a fixed order, as the columns of \(\mathbf{P}\), and place the matching eigenvalues in the same order along the diagonal of \(\mathbf{D}\).
Confirm that \(\mathbf{P}\) is invertible and that \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), or equivalently that \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\).
Compute \(\mathbf{c}=\mathbf{P}^{-1}\mathbf{x}_0\). These entries give the initial state’s coordinates in the eigenvector basis.
Compute \(\mathbf{x}_k=\mathbf{P}\mathbf{D}^k\mathbf{c}\), then check the result at \(k=0\) and against the update rule.
Worked Examples
Worked Example: A Symmetric Two-State System
Let $$ \mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}, \qquad \mathbf{x}_0=\begin{bmatrix}2\\0\end{bmatrix}. $$ Find \(\mathbf{x}_k\) for any integer \(k\geq0\).
The vector \(\begin{bmatrix}1\\1\end{bmatrix}\) is an eigenvector with eigenvalue \(3\), because $$ \mathbf{A}\begin{bmatrix}1\\1\end{bmatrix} =\begin{bmatrix}3\\3\end{bmatrix} =3\begin{bmatrix}1\\1\end{bmatrix}. $$ The vector \(\begin{bmatrix}1\\-1\end{bmatrix}\) is an eigenvector with eigenvalue \(1\), because $$ \mathbf{A}\begin{bmatrix}1\\-1\end{bmatrix} =\begin{bmatrix}1\\-1\end{bmatrix}. $$ Use these vectors as the columns of \(\mathbf{P}\), in that order, and match their eigenvalues in \(\mathbf{D}\):
Indeed, direct multiplication gives \(\mathbf{P}\mathbf{P}^{-1}=\mathbf{I}_2\), and $$ \mathbf{P}\mathbf{D}\mathbf{P}^{-1} =\frac12\begin{bmatrix}3&1\\3&-1\end{bmatrix} \begin{bmatrix}1&1\\1&-1\end{bmatrix} =\begin{bmatrix}2&1\\1&2\end{bmatrix} =\mathbf{A}. $$ The initial coordinates are $$ \mathbf{c}=\mathbf{P}^{-1}\mathbf{x}_0 =\frac12\begin{bmatrix}1&1\\1&-1\end{bmatrix} \begin{bmatrix}2\\0\end{bmatrix} =\begin{bmatrix}1\\1\end{bmatrix}. $$ Therefore $$ \mathbf{x}_k=\mathbf{P}\mathbf{D}^k\mathbf{c} =\begin{bmatrix}1&1\\1&-1\end{bmatrix} \begin{bmatrix}3^k&0\\0&1\end{bmatrix} \begin{bmatrix}1\\1\end{bmatrix} =\begin{bmatrix}3^k+1\\3^k-1\end{bmatrix}. $$ This also gives \(\mathbf{x}_0=\begin{bmatrix}2\\0\end{bmatrix}\) when \(k=0\). To check the update, multiply the general state by \(\mathbf{A}\): $$ \mathbf{A}\mathbf{x}_k =\begin{bmatrix}2&1\\1&2\end{bmatrix} \begin{bmatrix}3^k+1\\3^k-1\end{bmatrix} =\begin{bmatrix}3^{k+1}+1\\3^{k+1}-1\end{bmatrix} =\mathbf{x}_{k+1}. $$ So the formula satisfies the initial condition and the system’s recurrence.
Worked Example: A Nonsymmetric Matrix with Distinct Eigenvalues
Consider $$ \mathbf{A}=\begin{bmatrix}4&1\\0&2\end{bmatrix}, \qquad \mathbf{x}_0=\begin{bmatrix}0\\1\end{bmatrix}. $$ Find the state after \(k\) steps. The vector \(\begin{bmatrix}1\\0\end{bmatrix}\) has eigenvalue \(4\), and \(\begin{bmatrix}1\\-2\end{bmatrix}\) has eigenvalue \(2\): multiplying by \(\mathbf{A}\) gives, respectively, \(\begin{bmatrix}4\\0\end{bmatrix}\) and \(\begin{bmatrix}2\\-4\end{bmatrix}\).
Set $$ \mathbf{P}=\begin{bmatrix}1&1\\0&-2\end{bmatrix}, \qquad \mathbf{D}=\begin{bmatrix}4&0\\0&2\end{bmatrix}, \qquad \mathbf{P}^{-1}=\begin{bmatrix}1&\frac12\\0&-\frac12\end{bmatrix}. $$ The inverse is verified by multiplying: $$ \mathbf{P}\mathbf{P}^{-1} =\begin{bmatrix}1&\frac12-\frac12\\0&1\end{bmatrix} =\mathbf{I}_2. $$ The diagonalization also checks directly: $$ \mathbf{P}\mathbf{D}\mathbf{P}^{-1} =\begin{bmatrix}4&2\\0&-4\end{bmatrix} \begin{bmatrix}1&\frac12\\0&-\frac12\end{bmatrix} =\begin{bmatrix}4&1\\0&2\end{bmatrix} =\mathbf{A}. $$ Convert the initial state to eigenvector coordinates: $$ \mathbf{c}=\mathbf{P}^{-1}\mathbf{x}_0 =\begin{bmatrix}\frac12\\-\frac12\end{bmatrix}. $$ Consequently, $$ \mathbf{x}_k =\mathbf{P}\mathbf{D}^k\mathbf{c} =\begin{bmatrix}1&1\\0&-2\end{bmatrix} \begin{bmatrix}4^k&0\\0&2^k\end{bmatrix} \begin{bmatrix}\frac12\\-\frac12\end{bmatrix} =\begin{bmatrix}\frac12(4^k-2^k)\\2^k\end{bmatrix}. $$ At \(k=0\), this is \(\begin{bmatrix}0\\1\end{bmatrix}\), as required. For the recurrence check, $$ \mathbf{A}\mathbf{x}_k =\begin{bmatrix}2(4^k-2^k)+2^k\\2\cdot2^k\end{bmatrix} =\begin{bmatrix}2\cdot4^k-2^k\\2^{k+1}\end{bmatrix} =\begin{bmatrix}\frac12(4^{k+1}-2^{k+1})\\2^{k+1}\end{bmatrix} =\mathbf{x}_{k+1}. $$ The expression therefore gives the state at every nonnegative integer step.
Worked Example: Three Eigenmodes in Three Coordinates
Let $$ \mathbf{A}=\begin{bmatrix}2&1&0\\0&3&1\\0&0&4\end{bmatrix}, \qquad \mathbf{x}_0=\begin{bmatrix}1\\0\\1\end{bmatrix}. $$ Use a diagonalization to find \(\mathbf{x}_k\). The eigenvalues \(2,3,4\) have respective eigenvectors \(\begin{bmatrix}1\\0\\0\end{bmatrix}\), \(\begin{bmatrix}1\\1\\0\end{bmatrix}\), and \(\begin{bmatrix}1\\2\\2\end{bmatrix}\). For example, multiplying the third vector by \(\mathbf{A}\) gives \(\begin{bmatrix}4\\8\\8\end{bmatrix}=4\begin{bmatrix}1\\2\\2\end{bmatrix}\). The same multiplication gives \(2\begin{bmatrix}1\\0\\0\end{bmatrix}\) and \(3\begin{bmatrix}1\\1\\0\end{bmatrix}\) for the first two.
Place these eigenvectors in the columns of \(\mathbf{P}\), and use their matching eigenvalues to form \(\mathbf{D}\): $$ \mathbf{P}=\begin{bmatrix}1&1&1\\0&1&2\\0&0&2\end{bmatrix}, \qquad \mathbf{D}=\begin{bmatrix}2&0&0\\0&3&0\\0&0&4\end{bmatrix}, \qquad \mathbf{P}^{-1}=\begin{bmatrix}1&-1&\frac12\\0&1&-1\\0&0&\frac12\end{bmatrix}. $$ Multiplying \(\mathbf{P}\mathbf{P}^{-1}\) gives diagonal entries \(1\) and off-diagonal entries \(0\), so the displayed inverse is correct. To check the diagonalization, first calculate \(\mathbf{P}\mathbf{D}=\begin{bmatrix}2&3&4\\0&3&8\\0&0&8\end{bmatrix}\). Then $$ \mathbf{P}\mathbf{D}\mathbf{P}^{-1} =\begin{bmatrix}2&3&4\\0&3&8\\0&0&8\end{bmatrix} \begin{bmatrix}1&-1&\frac12\\0&1&-1\\0&0&\frac12\end{bmatrix} =\begin{bmatrix}2&1&0\\0&3&1\\0&0&4\end{bmatrix} =\mathbf{A}. $$ The initial coordinates are $$ \mathbf{c}=\mathbf{P}^{-1}\mathbf{x}_0 =\begin{bmatrix}\frac32\\-1\\\frac12\end{bmatrix}. $$ Thus $$ \mathbf{x}_k =\mathbf{P}\mathbf{D}^k\mathbf{c} =\frac32\,2^k\begin{bmatrix}1\\0\\0\end{bmatrix} -3^k\begin{bmatrix}1\\1\\0\end{bmatrix} +\frac12\,4^k\begin{bmatrix}1\\2\\2\end{bmatrix} =\begin{bmatrix}\frac32\,2^k-3^k+\frac12\,4^k\\-3^k+4^k\\4^k\end{bmatrix}. $$ At \(k=0\), the coordinates give \(\begin{bmatrix}1\\0\\1\end{bmatrix}\). For a recurrence check, multiply the general expression by \(\mathbf{A}\). Its first entry becomes \(2(\frac32\,2^k-3^k+\frac12\,4^k)+(-3^k+4^k)=\frac32\,2^{k+1}-3^{k+1}+\frac12\,4^{k+1}\). Its second entry becomes \(3(-3^k+4^k)+4\cdot4^k=-3^{k+1}+4^{k+1}\), and its third entry is \(4\cdot4^k=4^{k+1}\). These are exactly the entries of \(\mathbf{x}_{k+1}\).
Common Pitfall: Leaving Out the Inverse
The formula is \(\mathbf{A}^k=\mathbf{P}\mathbf{D}^k\mathbf{P}^{-1}\), not \(\mathbf{P}\mathbf{D}^k\). The inverse is needed to convert from the original coordinates to eigenvector coordinates before applying the diagonal scaling; the factor \(\mathbf{P}\) then converts back. A concrete incorrect calculation using the first example would give $$ \mathbf{P}\mathbf{D} =\begin{bmatrix}1&1\\1&-1\end{bmatrix} \begin{bmatrix}3&0\\0&1\end{bmatrix} =\begin{bmatrix}3&1\\3&-1\end{bmatrix}, $$ which is not \(\mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\). The correction is to include \(\mathbf{P}^{-1}\): $$ \mathbf{P}\mathbf{D}\mathbf{P}^{-1} =\frac12\begin{bmatrix}3&1\\3&-1\end{bmatrix} \begin{bmatrix}1&1\\1&-1\end{bmatrix} =\begin{bmatrix}2&1\\1&2\end{bmatrix}. $$ A second check is to substitute \(k=0\): the correct power formula gives \(\mathbf{P}\mathbf{I}\mathbf{P}^{-1}=\mathbf{I}\), whereas omitting the inverse would give \(\mathbf{P}\), which is generally not the identity.
When the Method Applies
The method calculates powers once a diagonalization is available; it does not guarantee that every real matrix has one over \(\mathbb{R}\). A proposed eigenvector matrix must be invertible, and the matching eigenvalues must be in the same order in \(\mathbf{D}\) as their eigenvectors are in \(\mathbf{P}\). In particular, an eigenvalue repeated on the diagonal must be paired with the correct eigenvector columns. If a real diagonalization has been established, the power formula needs no further restrictions on \(k\) beyond \(k\geq0\).
For a discrete system, this gives a direct route from an initial state to a state many steps later: compute the initial eigenvector coordinates once, raise the diagonal entries to the desired power, and convert back. The recurrence check \(\mathbf{A}\mathbf{x}_k=\mathbf{x}_{k+1}\) is a useful way to catch arithmetic or ordering errors. The long-term meaning of the eigenvalue magnitudes is a separate question; here the main task is exact computation of the state at a specified step.
Check Your Understanding
For each question, keep the order of eigenvectors and diagonal entries consistent, and verify any proposed state formula against the initial condition.
- If \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), write the formula for \(\mathbf{A}^k\) and explain why it holds at \(k=0\).
- For \(\mathbf{D}=\begin{bmatrix}5&0\\0&-2\end{bmatrix}\), write \(\mathbf{D}^k\) for an integer \(k\geq1\).
- Suppose \(\mathbf{P}^{-1}\mathbf{x}_0=\begin{bmatrix}2\\-1\end{bmatrix}\) and \(\mathbf{D}=\begin{bmatrix}3&0\\0&1\end{bmatrix}\). Express \(\mathbf{x}_k\) in terms of the columns of \(\mathbf{P}\).
- For the matrix in the second worked example, calculate \(\mathbf{x}_2\) using its closed-form expression and verify that it equals \(\mathbf{A}\mathbf{x}_1\).
- What condition on \(\mathbf{P}\) is required for the factorization \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) to be a diagonalization?
