LINEAR-ALGEBRA · Interactive Practice | Unit 27 · Video 4

When Diagonalization Fails: Jordan's Universal Blueprint

IKey Formulas & Facts

Concept	Statement
Diagonalization	$S^{-1} A S = \Lambda$ — requires $n$ independent eigenvectors
Jordan block $J_k(\lambda)$	$k\times k$ matrix with $\lambda$ on diagonal, $1$ 's on superdiagonal
Block / eigenvector count	Each Jordan block contributes exactly one eigenvector
Jordan form	Every square matrix $A$ satisfies $A = M J M^{-1}$ where $J$ is block-diagonal of Jordan blocks
Similarity invariant	Two matrices are similar iff their Jordan block sizes match (up to reordering)

IIVisualization 1 — Anatomy of a Jordan Block

Why does a Jordan block have exactly one eigenvector — no matter how large the block $k$ ?

Eigenvalue λ 2 Block size k 3

IIIVisualization 2 — The Twin Matrices Puzzle

Two nilpotent $4\times 4$ matrices agree on every invariant but one — so why aren't they similar?

IVVisualization 3 — Numerical Instability of the Jordan Form

Nudge the diagonal by $\varepsilon$ and the eigenvalues split — the block structure jumps discontinuously.

Perturbation ε 0.00

💡 Because this jump is discontinuous, numerical linear algebra relies on the SVD — the Jordan form lives in exact arithmetic only.

VQuiz: Test Your Understanding

Question 1

Consider a single $5 \times 5$ Jordan block $J_5(\lambda)$ with eigenvalue $\lambda$ repeated five times on the diagonal and $1$ 's on the superdiagonal.

How many independent eigenvectors does this block have?

✅ Correct! Every Jordan block, no matter how large, contributes exactly one independent eigenvector.

❌ Not quite. The diagonal $\lambda$ 's give algebraic multiplicity 5, but the superdiagonal $1$ 's force most coordinates to zero — only one direction is free.

Show solution

Solution:

Compute $J_5(\lambda) - \lambda I$ . The diagonal becomes zero, and we are left with a matrix whose only nonzero entries are $1$ 's on the superdiagonal.

Solving $(J - \lambda I)v = 0$ with $v = (v_1, v_2, v_3, v_4, v_5)^T$ gives:

$v_2 = 0, \quad v_3 = 0, \quad v_4 = 0, \quad v_5 = 0$

Only $v_1$ is free. So the null space is one-dimensional, spanned by $e_1 = (1, 0, 0, 0, 0)^T$ .

Key fact: every Jordan block — regardless of size — contributes exactly one eigenvector.

Question 2

True or False: If two $n \times n$ matrices have the same eigenvalues (with the same algebraic multiplicities) and the same number of independent eigenvectors, then they must be similar.

✅ Correct! Block sizes are the missing invariant — the twin matrices in the video show two matrices that agree on eigenvalues and eigenvector count but are still not similar.

❌ Not quite. Remember the twin matrices: same eigenvalues, same eigenvector count, but different block signatures $\{2,2\}$ vs $\{3,1\}$ — and therefore not similar.

Show solution

Solution: False.

The "twin matrices" from the video are a direct counterexample. Both are $4 \times 4$ , both have eigenvalue $0$ with multiplicity 4, and both have exactly 2 independent eigenvectors. Yet:

Matrix 1 has Jordan signature $\{2, 2\}$ (two 2×2 blocks)
Matrix 2 has Jordan signature $\{3, 1\}$ (one 3×3 block + one 1×1 block)

Two matrices are similar if and only if their Jordan block sizes match (up to reordering). The block-size signature is the complete invariant — eigenvalues and eigenvector counts alone are not enough.

Question 3

A $6 \times 6$ matrix $A$ has all eigenvalues equal to $\lambda = 7$ and has exactly 3 independent eigenvectors.

Which of the following is a possible Jordan block signature for $A$ ?

✅ Correct! 3 eigenvectors means 3 blocks, and $4 + 1 + 1 = 6$ matches the matrix size.

❌ Not quite. Recall: (# of independent eigenvectors) = (# of Jordan blocks). With 3 eigenvectors, you need exactly 3 blocks whose sizes sum to 6.

Show solution

Solution:

The number of Jordan blocks equals the number of independent eigenvectors. Since $A$ has 3 eigenvectors, it must have exactly 3 blocks. The block sizes must sum to $6$ (the matrix size).

Check each option:

$\{3, 3\}$ : 2 blocks, sum = 6. Eigenvector count = 2. ❌
$\{4, 1, 1\}$ : 3 blocks, sum = 6. Eigenvector count = 3. Correct!
$\{5, 1\}$ : 2 blocks, sum = 6. Eigenvector count = 2. ❌
$\{6\}$ : 1 block, sum = 6. Eigenvector count = 1. ❌

Only $\{4, 1, 1\}$ provides 3 blocks summing to 6 — matching both the eigenvector count and the matrix size.

Question 4

Classical diagonalization $S^{-1} A S = \Lambda$ corresponds to which special case of Jordan form?

✅ Correct! A 1×1 Jordan block is just an eigenvalue on the diagonal, so when every block is 1×1, $J$ collapses to the diagonal matrix $\Lambda$ .

❌ Not quite. Think about what a $1 \times 1$ Jordan block looks like — there's no superdiagonal, just a single eigenvalue. What if every block in $J$ were that small?

Show solution

Solution:

A $1 \times 1$ Jordan block is just a single eigenvalue on the diagonal — it has no superdiagonal $1$ 's. If every block in $J$ is $1 \times 1$ , then $J$ has nothing but eigenvalues on its diagonal and zeros elsewhere. That is precisely a diagonal matrix:

$J = \Lambda = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)$

This happens exactly when the number of blocks equals $n$ , which means the number of independent eigenvectors equals $n$ — the standard diagonalizability condition.

Diagonalization is the special case of Jordan form where every block has size 1×1.

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