Chapter 25 Quiz — Diagonalization
Eleven conceptual checks. Try each before opening the answer; every answer includes a one-line reason so you can repair a wrong mental model on the spot.
Q1. In the factorization $A = PDP^{-1}$, what are the columns of $P$ and the diagonal entries of $D$?
Answer
The columns of $P$ are the **eigenvectors** of $A$, and the diagonal entries of $D$ are the corresponding **eigenvalues**, in matching order. The factorization is just "change to the eigenbasis": $D = P^{-1}AP$ is $A$ rewritten in the basis of its own eigenvectors.Q2. Read $A = PDP^{-1}$ right to left. What does each factor do to a vector?
Answer
$P^{-1}$ converts the vector into **eigenbasis coordinates**; $D$ **stretches each eigen-axis independently** by its eigenvalue; $P$ converts the stretched result **back to standard coordinates**. The real action is the diagonal $D$ in the middle — pure independent scaling — sandwiched by the round trip into and out of the good coordinate system.Q3. State the exact condition for an $n \times n$ matrix to be diagonalizable.
Answer
It must have **$n$ linearly independent eigenvectors** (equivalently, an eigenbasis of $\mathbb{R}^n$ or $\mathbb{C}^n$). That is exactly when the eigenvector matrix $P$ is invertible, which is exactly when $A = PDP^{-1}$ is possible.Q4. Having $n$ distinct eigenvalues guarantees diagonalizability. Is the converse true — must a diagonalizable matrix have distinct eigenvalues?
Answer
No. Distinct eigenvalues are **sufficient but not necessary**. The identity matrix has the single eigenvalue $1$ repeated $n$ times yet is perfectly diagonalizable (it is already diagonal). A repeated eigenvalue only *risks* failure; it fails only if that eigenvalue is short on eigenvectors.Q5. A $2 \times 2$ matrix has the repeated eigenvalue $7$ and a one-dimensional eigenspace. Is it diagonalizable?
Answer
**No** — it is *defective*. The algebraic multiplicity of $7$ is $2$ but its geometric multiplicity is only $1$, so it supplies just one independent eigenvector, not the two a $2\times 2$ eigenbasis needs. There is no invertible $P$ of eigenvectors. (Its Jordan form, Chapter 36, is $\begin{bmatrix}7&1\\0&7\end{bmatrix}$.)Q6. Why is computing $A^{100}$ via $PD^{100}P^{-1}$ so much cheaper than multiplying $A$ by itself $100$ times?
Answer
Because $D^{100}$ costs only $n$ **scalar** exponentiations (raise each eigenvalue to the $100$th power), then two matrix multiplications sandwich it. The expensive part — the eigendecomposition — is done **once**, and the exponent $100$ never multiplies the matrix cost. The repeated-multiplication route is $99$ matrix products.Q7. A diagonalizable matrix has eigenvalues $3, -0.5, 0.2$. What happens to $A^k\mathbf{x}_0$ as $k \to \infty$ (for a generic $\mathbf{x}_0$)?
Answer
It **grows without bound in the direction of the eigenvector for $\lambda = 3$** (the dominant eigenvalue, $|3| > 1$). The $(-0.5)^k$ and $0.2^k$ modes decay to zero, so for large $k$, $A^k\mathbf{x}_0 \approx c_1 3^k \mathbf{v}_1$. The largest-magnitude eigenvalue sets the growth rate; its eigenvector sets the limiting direction.Q8. In the Fibonacci diagonalization, the eigenvalues are the golden ratio $\varphi \approx 1.618$ and $\psi \approx -0.618$. Why do the Fibonacci numbers grow at rate $\varphi$ and not $\psi$?
Answer
Because $\varphi$ is the **dominant** (larger-magnitude) eigenvalue: $\varphi^n$ grows while $|\psi|^n < 1$ shrinks to nothing. In the closed form $F_n = (\varphi^n - \psi^n)/\sqrt5$, the $\psi^n$ term vanishes, leaving $F_n \approx \varphi^n/\sqrt5$ — so the ratio $F_{n+1}/F_n \to \varphi$.Q9. Every column-stochastic matrix (columns summing to $1$) has $1$ as an eigenvalue. Why does that matter for a Markov chain's long-run behavior?
Answer
The eigenvalue $1$ gives a mode that is **neither amplified nor decayed** — the steady-state distribution. Every *other* eigenvalue has magnitude $\le 1$ (typically $< 1$), so those modes decay under iteration, funneling any starting distribution onto the $\lambda = 1$ eigenvector. That eigenvector, normalized to sum $1$, *is* the steady state.Q10. You diagonalize $A$ and write $P = [\mathbf{v}_1 \mid \mathbf{v}_2]$ but list $D = \operatorname{diag}(\lambda_2, \lambda_1)$ — eigenvalues in the swapped order. Is $A = PDP^{-1}$ still correct?
Answer
**No.** The $i$-th column of $P$ must pair with the $i$-th diagonal entry of $D$. Swapping the eigenvalue order without swapping the columns of $P$ produces a *different* matrix, not $A$. You may reorder eigenpairs freely, but each eigenvector must stay married to its own eigenvalue.Q11. What is the eigenbasis expansion $A^k\mathbf{x}_0 = c_1\lambda_1^k\mathbf{v}_1 + \dots + c_n\lambda_n^k\mathbf{v}_n$ telling you conceptually?