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Lesson 115 — Diagonalization

Decomposition A = PDP⁻¹. Conditions of diagonalizability, construction algorithm, matrix powers, matrix exponential and applications in dynamical systems.

Used in: 3.º year advanced HS · Equiv. Lineare Algebra LK German · Equiv. Math III Japanese · Equiv. H2 Mathematics Singaporean

A = PDP^{-1}

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Rigorous notation, full derivation, hypotheses

Spectral decomposition — definition and theory

Fundamental definition

"A matrix $A$ is diagonalizable if it is similar to a diagonal matrix — there exists an invertible $P$ such that $P^{-1}AP$ is diagonal." — Beezer, A First Course in Linear Algebra, §SD

Equivalent conditions

Theorem· Criterion of diagonalizability

The following statements are equivalent for $A \in \mathcal{M}_n(\mathbb{K})$ :

$A$ is diagonalizable.
$A$ possesses $n$ linearly independent eigenvectors.
The sum of the dimensions of the eigenspaces equals $n$ : $\sum_\lambda \dim E_\lambda = n$ .
For each eigenvalue $\lambda$ , the geometric multiplicity equals the algebraic multiplicity: $m_g(\lambda) = m_a(\lambda)$ .
$\mathbb{K}^n = \bigoplus_\lambda E_\lambda$ — the space is a direct sum of eigenspaces.

Proof of equivalence 2 $\Leftrightarrow$ 1. If $\{v_1, \ldots, v_n\}$ are $n$ linearly independent eigenvectors with eigenvalues $\lambda_1, \ldots, \lambda_n$ , form $P = [v_1 \mid \cdots \mid v_n]$ . Since the columns are LI, $P$ is invertible. Calculate:

$AP = A[v_1 \mid \cdots \mid v_n] = [\lambda_1 v_1 \mid \cdots \mid \lambda_n v_n] = [v_1 \mid \cdots \mid v_n]\operatorname{diag}(\lambda_i) = PD$

Thus $P^{-1}AP = D$ . The converse is immediate: if $A = PDP^{-1}$ , the columns of $P$ are linearly independent eigenvectors. $\square$

"An $n \times n$ matrix $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors." — Beezer, A First Course in Linear Algebra, §SD Theorem DED

Cases guaranteeing diagonalizability

Sufficient conditions for diagonalizability. Real symmetric: P orthogonal (Spectral Theorem, L116). Normal: P unitary.

Diagonalization algorithm

Compute the characteristic polynomial $p_A(\lambda) = \det(A - \lambda I)$ and find the roots $\lambda_1, \ldots, \lambda_k$ with algebraic multiplicities $m_a(\lambda_i)$ .
For each $\lambda_i$ , solve $(A - \lambda_i I)v = 0$ and find a basis for $E_{\lambda_i} = \ker(A - \lambda_i I)$ . Verify $m_g(\lambda_i) = \dim E_{\lambda_i}$ .
If $\sum m_g(\lambda_i) = n$ : form $P$ with the eigenvectors as columns and $D = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)$ (respecting column order).
If $\sum m_g(\lambda_i) < n$ : $A$ is not diagonalizable — resort to Jordan form.

Immediate applications

A^k = P D^k P^{-1}, \quad D^k = \operatorname{diag}(\lambda_1^k, \ldots, \lambda_n^k)

what this means · Matrix power via diagonalization: D^k has the eigenvalues raised to k on the diagonal.

e^{At} = P e^{Dt} P^{-1}, \quad e^{Dt} = \operatorname{diag}(e^{\lambda_1 t}, \ldots, e^{\lambda_n t})

what this means · Matrix exponential: each eigenvalue lambda_i generates e^{lambda_i t} on the diagonal.

For any analytic function $f$ : $f(A) = Pf(D)P^{-1}$ with $f(D) = \operatorname{diag}(f(\lambda_i))$ .

Worked examples

Example— 1· Diagonalizing a 2×2 matrix (basic)

Problem: Diagonalize $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ .

Strategy: Find eigenvalues and eigenvectors, form $P$ and $D$ , verify $AP = PD$ .

Solution:

Characteristic polynomial: $p_A(\lambda) = \det(A - \lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda - 5)(\lambda - 2)$ Eigenvalues: $\lambda_1 = 5$ , $\lambda_2 = 2$ .
Eigenvectors:

For $\lambda_1 = 5$ : $(A - 5I)v = 0 \Rightarrow \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}v = 0$ . Basis: $v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .

For $\lambda_2 = 2$ : $(A - 2I)v = 0 \Rightarrow \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}v = 0$ . Basis: $v_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ .

Assembly: $P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}, \quad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$

Verification: $AP = \begin{pmatrix} 5 & 2 \\ 5 & -4 \end{pmatrix}$ and $PD = \begin{pmatrix} 5 & 2 \\ 5 & -4 \end{pmatrix}$ . Equal. $\checkmark$

Source. Beezer, A First Course in Linear Algebra §SD, Example DMS22 — GNU FDL license.

Example— 2· Non-diagonalizable matrix (comprehension)

Problem: Determine if $A = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$ is diagonalizable. Justify.

Strategy: Compute algebraic and geometric multiplicities of the sole eigenvalue.

Solution:

Eigenvalues: $p_A(\lambda) = (3-\lambda)^2 = 0$ , so $\lambda = 3$ with $m_a = 2$ .
Eigenspace: $(A - 3I) = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ . The system $(A - 3I)v = 0$ requires $v_2 = 0$ , with $v_1$ free. Basis: $\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix}\right\}$ . Therefore $m_g(3) = 1$ .
Conclusion: $m_g(3) = 1 < 2 = m_a(3)$ . There are not two linearly independent eigenvectors. $A$ is not diagonalizable.

Verification: If there existed $P$ with $P^{-1}AP = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = 3I$ , we would have $A = P(3I)P^{-1} = 3I$ . But $A \neq 3I$ . Contradiction. $\checkmark$

Source. Beezer, A First Course in Linear Algebra §SD, Example NDMS3 — GNU FDL license.

Example— 3· Matrix power via diagonalization (intermediate)

Problem: Compute $A^{10}$ for $A = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}$ .

Strategy: Diagonalize $A$ , raise $D$ to power 10 (trivial), compose.

Solution:

Eigenvalues: $p_A(\lambda) = (2-\lambda)(1-\lambda) = 0$ , so $\lambda_1 = 2$ , $\lambda_2 = 1$ .
Eigenvectors:

$\lambda_1 = 2$ : $(A - 2I)v = 0 \Rightarrow \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}v = 0$ . $v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ .

$\lambda_2 = 1$ : $(A - I)v = 0 \Rightarrow \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}v = 0$ . $v_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ .

Decomposition: $P = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$ , $D = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ .

$P^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$ (verify: $\det P = -1$ , so $P^{-1} = \frac{1}{-1}\begin{pmatrix} -1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$ ).

Power: $D^{10} = \begin{pmatrix} 2^{10} & 0 \\ 0 & 1^{10} \end{pmatrix} = \begin{pmatrix} 1024 & 0 \\ 0 & 1 \end{pmatrix}$ .

$A^{10} = PD^{10}P^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1024 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1024 & 1023 \\ 0 & 1 \end{pmatrix}$

Verification: $A^2 = \begin{pmatrix} 4 & 3 \\ 0 & 1 \end{pmatrix}$ . By the formula: $A^2 = \begin{pmatrix} 2^2 & 2^2-1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 3 \\ 0 & 1 \end{pmatrix}$ . $\checkmark$

Source. Hefferon, Linear Algebra, ch. 5 §II — CC-BY-SA license.

Example— 4· Binet's formula for Fibonacci (advanced)

Problem: Derive the closed formula of Binet $F_n = \dfrac{\phi^n - \psi^n}{\sqrt{5}}$ , where $\phi = \frac{1+\sqrt{5}}{2}$ and $\psi = \frac{1-\sqrt{5}}{2}$ .

Strategy: Rewrite the recurrence as matrix iteration, diagonalize the transition matrix.

Solution:

Matrix form: $\begin{pmatrix} F_{n+1} \\ F_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} F_n \\ F_{n-1} \end{pmatrix}$ . Let $M = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ .
Eigenvalues of $M$ : $\lambda^2 - \lambda - 1 = 0 \Rightarrow \lambda = \frac{1 \pm \sqrt{5}}{2}$ , so $\lambda_1 = \phi$ , $\lambda_2 = \psi$ .
Eigenvectors: $v_1 = \begin{pmatrix} \phi \\ 1 \end{pmatrix}$ , $v_2 = \begin{pmatrix} \psi \\ 1 \end{pmatrix}$ .
Decomposition: $\begin{pmatrix} F_{n+1} \\ F_n \end{pmatrix} = M^n \begin{pmatrix} 1 \\ 0 \end{pmatrix} = PD^nP^{-1}\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ .

$P = \begin{pmatrix} \phi & \psi \\ 1 & 1 \end{pmatrix}$ , $\det P = \phi - \psi = \sqrt{5}$ . $P^{-1} = \frac{1}{\sqrt{5}}\begin{pmatrix} 1 & -\psi \\ -1 & \phi \end{pmatrix}$ .

Second component: $F_n = \frac{\phi^n(\phi - \psi + \psi \cdot \text{...}) - \psi^n(\ldots)}{\sqrt{5}}$ . Computing explicitly the second row of $PD^nP^{-1}\begin{pmatrix}1\\0\end{pmatrix}$ :

$F_n = \frac{1}{\sqrt{5}}\left(\phi^n \cdot 1 \cdot \frac{1}{\sqrt{5}}\sqrt{5} - \psi^n \cdot 1 \cdot \frac{1}{\sqrt{5}}\sqrt{5}\right) = \frac{\phi^n - \psi^n}{\sqrt{5}}$

Verification: $F_1 = \frac{\phi - \psi}{\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{5}} = 1$ . $F_2 = \frac{\phi^2 - \psi^2}{\sqrt{5}} = \frac{(\phi-\psi)(\phi+\psi)}{\sqrt{5}} = \frac{\sqrt{5} \cdot 1}{\sqrt{5}} = 1$ . $\checkmark$

Source. Beezer, A First Course in Linear Algebra §SD — GNU FDL license.

Example— 5· Stability of dynamical system (modeling)

Problem: The system $\dot{x} = Ax$ with $A = \begin{pmatrix} -3 & 1 \\ 1 & -3 \end{pmatrix}$ and $x(0) = \begin{pmatrix} 2 \\ 0 \end{pmatrix}$ . Determine $x(t)$ and classify the stability.

Strategy: Diagonalize $A$ , solve the decoupled system, compose the solution.

Solution:

Eigenvalues: $p_A(\lambda) = (-3-\lambda)^2 - 1 = \lambda^2 + 6\lambda + 8 = (\lambda+4)(\lambda+2) = 0$ . $\lambda_1 = -4$ , $\lambda_2 = -2$ .
Eigenvectors:

$\lambda_1 = -4$ : $(A + 4I) = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ . $v_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ .

$\lambda_2 = -2$ : $(A + 2I) = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}$ . $v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .

General solution: $x(t) = c_1 e^{-4t}\begin{pmatrix} 1 \\ -1 \end{pmatrix} + c_2 e^{-2t}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .
Initial condition: $x(0) = \begin{pmatrix} 2 \\ 0 \end{pmatrix} \Rightarrow c_1 + c_2 = 2$ and $-c_1 + c_2 = 0$ , so $c_1 = c_2 = 1$ .

$x(t) = e^{-4t}\begin{pmatrix} 1 \\ -1 \end{pmatrix} + e^{-2t}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$

Stability: Both eigenvalues are negative ( $-4 < 0$ , $-2 < 0$ ), so $x(t) \to 0$ as $t \to \infty$ . The system is asymptotically stable.

Verification: $x(0) = \begin{pmatrix} 1 \\ -1 \end{pmatrix} + \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix}$ . $\checkmark$

Source. Hefferon, Linear Algebra, ch. 5 §II — CC-BY-SA license.

Exercise list

45 exercises · 11 with worked solution (25%)

Application 18Understanding 6Modeling 9Challenge 7Proof 5

Ex. 115.1Application
Diagonalize $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ .
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Ex. 115.2Application
Diagonalize $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ .
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Ex. 115.3ApplicationAnswer key
Is $A = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ diagonalizable? Justify.
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Ex. 115.4Understanding
Is $A = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$ diagonalizable?
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Ex. 115.5Application
Check whether $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is diagonalizable over $\mathbb{R}$ and over $\mathbb{C}$ .
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Ex. 115.6Application
Diagonalize $A = \begin{pmatrix} 5 & -1 \\ 4 & 2 \end{pmatrix}$ .
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Ex. 115.7Application
Diagonalize the symmetric matrix $A = \begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix}$ and verify that $P$ is orthogonal.
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Ex. 115.8Application
Diagonalize $A = \begin{pmatrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ .
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Ex. 115.9Application
Determine whether $A = \operatorname{diag}(1, 2, 3)$ is diagonalizable.
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Ex. 115.10UnderstandingAnswer key
Is the projection $P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ diagonalizable?
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Ex. 115.11UnderstandingAnswer key
For which values of $a, b \in \mathbb{R}$ is the matrix $\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}$ diagonalizable?
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Ex. 115.12ChallengeAnswer key
Determine the eigenvalues of $C = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}$ and decide: is it diagonalizable over $\mathbb{R}$ ? Over $\mathbb{C}$ ?
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Ex. 115.13ApplicationAnswer key
Use diagonalization to compute $A^{10}$ , with $A = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}$ .
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Ex. 115.14Application
Compute $A^{100}$ for $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ (Fibonacci matrix) via eigenvalues.
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Ex. 115.15Proof
Prove by induction that $A^k = PD^kP^{-1}$ for all $k \in \mathbb{N}$ .
Ex. 115.16Application
Compute $e^{At}$ for $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ . Interpret geometrically.
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Ex. 115.17Application
Compute $\sqrt{A}$ for $A = \begin{pmatrix} 4 & 0 \\ 0 & 9 \end{pmatrix}$ .
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Ex. 115.18Application
Compute $\cos A$ for $A = \begin{pmatrix} 0 & \pi \\ -\pi & 0 \end{pmatrix}$ .
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Ex. 115.19ApplicationAnswer key
Verify that $A^k \to 0$ for $A = \begin{pmatrix} 0{,}5 & 0 \\ 0 & 0{,}3 \end{pmatrix}$ .
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Ex. 115.20Application
With $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ (eigenvalues 3 and 1), compute $A^k \begin{pmatrix} 2 \\ 0 \end{pmatrix}$ in terms of $k$ .
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Ex. 115.21ApplicationAnswer key
Solve $\dot{x} = Ax$ with $A = \begin{pmatrix} -1 & 0 \\ 0 & -2 \end{pmatrix}$ , $x(0) = (1, 1)^T$ .
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Ex. 115.22Application
Solve $\dot{x} = Ax$ with $A = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}$ .
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Ex. 115.23Application
Show that $F_n = \frac{1}{\sqrt{5}}\left(\phi^n - \psi^n\right)$ for the Fibonacci sequence, where $\phi = \frac{1+\sqrt{5}}{2}$ .
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Ex. 115.24Challenge
If $A$ is diagonalizable and $f$ is a polynomial, show that $f(A) = Pf(D)P^{-1}$ with $f(D) = \operatorname{diag}(f(\lambda_1), \ldots, f(\lambda_n))$ .
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Ex. 115.25Modeling
Markov chain of weather: $M = \begin{pmatrix} 0{,}7 & 0{,}3 \\ 0{,}4 & 0{,}6 \end{pmatrix}$ . Compute $M^{10}$ via diagonalization and find the stationary distribution.
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Ex. 115.26Modeling
Coupled mass-spring system of 2 masses with stiffness matrix $K = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$ (unit masses). Find the normal modes and the natural frequencies of vibration.
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Ex. 115.27ModelingAnswer key
Leslie matrix of population of 2 age groups: $L = \begin{pmatrix} 0 & 1{,}2 \\ 0{,}4 & 0 \end{pmatrix}$ . Compute the dominant eigenvalue and interpret as population growth rate.
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Ex. 115.28Modeling
Simplified PageRank: 4 pages with transition matrix $P = \frac{1}{4}\begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix}$ . Find the stationary distribution (eigenvector of $\lambda = 1$ ).
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Ex. 115.29Modeling
Covariance matrix of 2 assets: $\Sigma = \begin{pmatrix} 4 & 2 \\ 2 & 3 \end{pmatrix}$ . Diagonalize $\Sigma$ and interpret the eigenvectors as principal risk directions.
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Ex. 115.30Modeling
Discrete control system $x_{k+1} = Ax_k$ with $A = \begin{pmatrix} 0{,}5 & 0{,}3 \\ 0{,}1 & 0{,}4 \end{pmatrix}$ . Determine if the system is stable by checking the spectral radius $\rho(A) = \max|\lambda_i|$ .
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Ex. 115.31Modeling
In recurrent neural networks, exploding/vanishing gradients occur when the spectral radius $\rho(J)$ of the layer's Jacobian is $> 1$ or $< 1$ . Explain the mechanism via diagonalization and suggest an architectural solution.
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Ex. 115.32Modeling
Model of drainage of two coupled tanks: $A = \begin{pmatrix} -2 & 1 \\ 0 & -1 \end{pmatrix}$ . Solve $\dot{x} = Ax$ with $x(0) = (1, 2)^T$ and determine when $\|x(t)\| < 0{,}01$ .
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Ex. 115.33Modeling
In an information diffusion network, the discrete dynamics is $x_{k+1} = Wx_k$ where $W$ is symmetric with eigenvalues $1, 0{,}8, 0{,}2$ . Interpret what happens to $x_k$ for large $k$ .
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Ex. 115.34Understanding
Why is an $n \times n$ matrix with $n$ distinct eigenvalues (over $\mathbb{C}$ ) always diagonalizable?
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Ex. 115.35ProofAnswer key
Prove that eigenvectors corresponding to distinct eigenvalues are linearly independent.
Ex. 115.36Proof
Is every $2 \times 2$ matrix with $\det A = 0$ and $\operatorname{tr} A \neq 0$ diagonalizable? Justify.
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Ex. 115.37Challenge
Find a $2 \times 2$ non-diagonalizable matrix with eigenvalue 5 of algebraic multiplicity 2.
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Ex. 115.38Proof
Prove that similar matrices have the same characteristic polynomial (and thus the same eigenvalues).
Ex. 115.39ChallengeAnswer key
Show that if $A$ is diagonalizable and $f$ is a polynomial, then $f(A)$ is diagonalizable with eigenvalues $f(\lambda_i)$ .
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Ex. 115.40Challenge
If $A$ is diagonalizable, prove that $A^T$ is also diagonalizable (with the same eigenvalues).
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Ex. 115.41Understanding
If $A = QDQ^T$ with $Q$ orthogonal and $D$ real diagonal, prove that $A$ is symmetric.
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Ex. 115.42Understanding
If $A$ is diagonalizable with eigenvalues $\lambda_1, \ldots, \lambda_n$ , what is the relation between $\operatorname{tr} A$ , $\det A$ and the eigenvalues?
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Ex. 115.43Challenge
Show that $AB$ and $BA$ have the same nonzero eigenvalues (even though $AB \neq BA$ ).
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Ex. 115.44ProofAnswer key
If $A$ is diagonalizable and invertible, prove that $A^{-1}$ is also diagonalizable with eigenvalues $1/\lambda_i$ .
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Ex. 115.45Challenge
System of chemical reactions $A \rightleftharpoons B$ with equations $\dot{c}_A = -k_1 c_A + k_{-1}c_B$ , $\dot{c}_B = k_1 c_A - k_{-1}c_B$ . Solve via diagonalization and find the equilibrium.
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Sources

A First Course in Linear Algebra — Robert A. Beezer · 2022 · EN · GNU FDL. Primary reference: §SD (Similar Matrices and Diagonalization) with rigorous definitions and numbered exercises.
Linear Algebra Done Right (4th ed) — Sheldon Axler · 2024 · EN · CC-BY-NC. Ch. 5C–5D: diagonalizable operators, polynomials and functions of operators.
Linear Algebra — Jim Hefferon · 2022 · EN · CC-BY-SA. Ch. 5 §II: diagonalization, introductory Jordan, examples of dynamical systems.