LINEAR-ALGEBRA · Interactive Practice | Unit 21 · Video 3

The Hidden Reason Some Matrices Refuse to Diagonalize

IKey Concepts & Formulas

Concept	Formula / Definition	Meaning
Diagonalization	$S^{-1} A S = \Lambda$	Columns of $S$ are eigenvectors; $\Lambda$ is diagonal of eigenvalues
Eigenspace	$\text{null}(A - \lambda I)$	All eigenvectors for eigenvalue $\lambda$ (plus zero)
Algebraic multiplicity	Number of times $\lambda$ appears as a root of $\det(A - \lambda I) = 0$	How often eigenvalue repeats
Geometric multiplicity	$\dim \text{null}(A - \lambda I)$	Dimension of eigenspace
Master criterion	Diagonalizable $\iff$ geometric = algebraic for every $\lambda$	Gap = "defect"; nonzero defect means defective

IIVisualization 1 — Eigenspace Collapse

A single off-diagonal entry decides whether the repeated eigenvalue $\lambda = 2$ owns the whole plane or just one line.

Off-diagonal c 0

IIIVisualization 2 — Distinct vs. Repeated Eigenvalues

When the eigenvalues $a$ and $d$ differ the matrix has two eigenvector directions; forcing them equal can leave only one.

a (top-left) 1 d (bottom-right) 3

IVVisualization 3 — The Multiplicity Table

Diagonalizability is decided eigenvalue by eigenvalue: geometric multiplicity must match algebraic for every one.

Matrix

Eigenvalue λ	Algebraic	Geometric	Independent eigenvectors

💡 An $n \times n$ Jordan block always has geometric multiplicity 1, so its defect grows to $n - 1$ .

VPractice Quiz

Question 1

Consider $A = \begin{pmatrix} 4 & 7 \\ 0 & 9 \end{pmatrix}$ . Without computing eigenvectors explicitly, can you conclude that $A$ is diagonalizable?

✅ Correct! Distinct eigenvalues always guarantee diagonalizability.

❌ Not quite. Remember: the danger zone is repeated eigenvalues. Here the diagonal entries differ.

Show solution

Solution:

Since $A$ is upper triangular, its eigenvalues are the diagonal entries: $\lambda_1 = 4$ and $\lambda_2 = 9$ .

These are distinct, and the theorem from the video says: if an $n \times n$ matrix has $n$ distinct eigenvalues, then eigenvectors from different eigenvalues are automatically linearly independent, so $A$ is diagonalizable.

The off-diagonal entry does not cause trouble here — it only does so when the diagonal entries (and hence eigenvalues) are equal.

Question 2

For $A = \begin{pmatrix} 5 & 2 \\ 0 & 5 \end{pmatrix}$ , what are the algebraic and geometric multiplicities of $\lambda = 5$ ?

✅ Correct! This is the classic defective matrix from the video.

❌ Not quite. Write out $A - 5I$ and find the dimension of its null space.

Show solution

Solution:

Algebraic multiplicity: The characteristic polynomial is $(5-\lambda)^2 = 0$ , so $\lambda = 5$ appears as a root twice. Algebraic multiplicity = 2.

Geometric multiplicity: Compute $A - 5I = \begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}$ The null space requires $2y = 0$ , so $y = 0$ with $x$ free. The eigenspace is spanned by $(1,0)^T$ , a one-dimensional line.

So geometric multiplicity = 1, and the defect is $2 - 1 = 1$ . The matrix is defective.

Question 3

True or False: If an eigenvalue $\lambda$ has algebraic multiplicity 3, then its geometric multiplicity must also be 3.

✅ Correct! Geometric multiplicity can be anywhere from 1 up to the algebraic multiplicity.

❌ Not quite. Recall: geometric multiplicity $\le$ algebraic multiplicity, but equality is not forced.

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Solution:

False. The fundamental inequality is $1 \le \text{geometric multiplicity} \le \text{algebraic multiplicity}.$

So with algebraic multiplicity 3, the geometric multiplicity can be 1, 2, or 3. Any value less than 3 produces a defect and makes the matrix defective.

For example, the $3\times 3$ Jordan block $\begin{pmatrix} 3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{pmatrix}$ has algebraic multiplicity 3 but geometric multiplicity 1.

Question 4

A $4 \times 4$ matrix $A$ has characteristic polynomial $(\lambda - 2)^2 (\lambda - 7)^2$ . For which combination of geometric multiplicities is $A$ diagonalizable?

✅ Correct! Every eigenvalue must individually pass the geom = alg test.

❌ Not quite. The criterion must hold for every eigenvalue simultaneously, not just one.

Show solution

Solution:

The master criterion says $A$ is diagonalizable if and only if geometric multiplicity equals algebraic multiplicity for every eigenvalue.

Here both eigenvalues have algebraic multiplicity 2, so we need: $\text{geom}(2) = 2 \quad \text{AND} \quad \text{geom}(7) = 2.$

This produces a full set of $2 + 2 = 4$ linearly independent eigenvectors, enough to fill the columns of $S$ .

Any other combination leaves a gap (defect) for at least one eigenvalue, making $A$ defective.

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