Chapter 36 Quiz — Jordan Normal Form
Eleven conceptual checks. Try each before opening the answer. These test understanding — whether you see that a defective matrix is shearing, not broken, and that the Jordan form is the honest record of that shear.
1. What exactly makes a matrix defective, stated in terms of the two multiplicities?
Answer
A matrix is defective when, for at least one eigenvalue $\lambda$, the **geometric multiplicity** $\dim N(A-\lambda I)$ (number of independent eigenvectors) is **strictly less than** the **algebraic multiplicity** (the eigenvalue's multiplicity as a root of the characteristic polynomial). Equivalently: the eigenvalue is "owed" more eigenvectors than it actually has, so the eigenvectors cannot form a basis and the matrix cannot be diagonalized. The shortfall $d_\lambda = \text{algebraic} - \text{geometric}$ is the number of generalized eigenvectors you must manufacture.2. A matrix has eigenvalue $\lambda = 7$ with algebraic multiplicity $1$. Can it be defective at $\lambda = 7$?
Answer
No. The multiplicity inequality says $1 \le \text{geometric} \le \text{algebraic} = 1$, which forces geometric multiplicity $= 1 = $ algebraic. A *simple* (multiplicity-one) eigenvalue is never defective. Defectiveness requires a **repeated** eigenvalue that fails to supply its full quota of eigenvectors — so a matrix with all distinct eigenvalues is automatically diagonalizable.3. Geometrically, what does the $2\times 2$ Jordan block $\begin{bmatrix}\lambda&1\\0&\lambda\end{bmatrix}$ do to the plane, and what does the off-diagonal $1$ represent?
Answer
It *scales* the plane by $\lambda$ and *shears* it: it fixes the eigen-direction (the $x$-axis), scaling along it, while sliding every other direction toward that axis in proportion to its perpendicular distance. The off-diagonal $1$ *is* the shear — the "leftover" motion that the single eigenvector cannot capture. With $\lambda = 1$ it is a pure shear; a general $\lambda$ shears and then scales by $\lambda$. Diagonal would mean no shear; the $1$ is exactly the matrix's distance from diagonalizable.4. How do you find a generalized eigenvector, and how is the equation different from the one for an ordinary eigenvector?
Answer
Solve the *inhomogeneous* system $(A-\lambda I)\mathbf{w} = \mathbf{v}$, where $\mathbf{v}$ is a genuine eigenvector — the right-hand side is the eigenvector, not zero. An ordinary eigenvector solves the *homogeneous* $(A-\lambda I)\mathbf{v} = \mathbf{0}$. The generalized equation is solvable only when $\mathbf{v}$ lies in the **column space** of $A-\lambda I$; when an eigenvalue has several eigenvectors, you must start the chain from one in that column space (the §36.4 trap).5. State the Jordan canonical form theorem, including the field over which it holds and the sense in which it is unique.
Answer
Every square matrix $A$ **over $\mathbb{C}$** can be written $A = PJP^{-1}$ with $P$ invertible and $J$ block-diagonal, each block a Jordan block (eigenvalue on the diagonal, $1$'s on the superdiagonal). $J$ is **unique up to the order of the blocks**. The complex field is essential: existence relies on the characteristic polynomial splitting into linear factors, guaranteed over $\mathbb{C}$ (Fundamental Theorem of Algebra) but *not* over $\mathbb{R}$ — a real matrix has a real Jordan form only if all its eigenvalues are real.6. For an eigenvalue $\lambda$, what determines the number of Jordan blocks, and what determines their total size?
Answer
The **number of blocks** for $\lambda$ equals its **geometric multiplicity** $\dim N(A-\lambda I)$ — one block per independent eigenvector (each block is built on one eigenvector at its bottom). The **total size** of those blocks equals the **algebraic multiplicity**. The individual sizes, however, require the ranks of higher powers $(A-\lambda I)^2, (A-\lambda I)^3, \dots$; the two multiplicities alone do not always determine them.7. Two matrices have identical eigenvalues with identical algebraic and geometric multiplicities. Must they be similar?
Answer
Not necessarily. The eigenvalues and both multiplicities fix the *number* and *total size* of the Jordan blocks, but not their individual sizes when there is ambiguity. For example, an eigenvalue with algebraic multiplicity $4$ and geometric multiplicity $2$ could have blocks $3+1$ or $2+2$ — same multiplicities, *different* Jordan forms, hence **not similar**. Two matrices are similar iff they have the same Jordan form up to block order; the block sizes are genuine extra information beyond the multiplicities.8. Why does the Jordan form make $A^k$ computable for a defective matrix, and what distinctive term appears that a diagonalizable matrix never produces?
Answer
$A = PJP^{-1}$ gives $A^k = PJ^kP^{-1}$, and a Jordan block powers cleanly via $J = \lambda I + N$ with $N$ nilpotent: $(\lambda I + N)^k$ is a finite binomial sum. For a $2\times 2$ block, $J^k = \begin{bmatrix}\lambda^k & k\lambda^{k-1}\\0&\lambda^k\end{bmatrix}$. The distinctive term is the **$k\lambda^{k-1}$** — a *polynomial in $k$ times a geometric factor*. A diagonalizable matrix's powers are pure $\lambda^k$ with no polynomial factor; the extra $k$ is the signature of defectiveness (the nilpotent part made visible).9. What is the matrix exponential of a $2\times 2$ Jordan block $J = \begin{bmatrix}\lambda&1\\0&\lambda\end{bmatrix}$, and why does it matter for differential equations?
Answer
$e^{Jt} = e^{\lambda t}\begin{bmatrix}1&t\\0&1\end{bmatrix}$, because $J = \lambda I + N$ with $N^2 = 0$ gives $e^{Nt} = I + tN$ (a finite series). It matters because Chapter 37 solves $\mathbf{x}' = A\mathbf{x}$ with $\mathbf{x}(t) = e^{At}\mathbf{x}(0)$. For a defective $A$, the solution carries a $t\,e^{\lambda t}$ term — the slow "creep" of a critically-damped system — which a diagonalizable matrix (pure $e^{\lambda t}$) would never produce.10. Why do numerical libraries like numpy not compute the Jordan form, and what do they use instead?
Answer
Because the Jordan form is **catastrophically unstable**: an arbitrarily small perturbation splits a repeated eigenvalue and makes a defective matrix diagonalizable, with the eigenvalue gap scaling like $\sqrt{\epsilon}$ for a perturbation of size $\epsilon$, and the eigenvector matrix becoming ill-conditioned. In finite precision you cannot reliably tell whether a matrix is exactly defective. Robust software uses the **Schur decomposition** ($A = QTQ^*$, $Q$ unitary, $T$ triangular — always stable) and the **SVD** instead. `sympy` *can* compute the Jordan form because it uses exact rational arithmetic, never rounding.11. A student claims, "Since the identity matrix $I$ has eigenvalue $1$ with multiplicity $n$, it must be defective." Where is the error?