LINEAR-ALGEBRA

Two Matrices, Same Eigenvalues, Different Worlds

IKey Formulas & Facts

Concept	Formula / Statement
Similarity	$B = M^{-1} A M$ for some invertible $M$
Scalar matrix is fixed under conjugation	$M^{-1}(\lambda I) M = \lambda I$ for every invertible $M$
Eigenvector equation	$(A - \lambda I)\mathbf{x} = \mathbf{0}$
Geometric multiplicity	$\dim \ker(A - \lambda I) =$ number of independent eigenvectors at $\lambda$
Similarity invariant	Geometric multiplicity is preserved under similarity

IIVisualization 1 — Same Eigenvalue, Different Eigenvectors

Both matrices have eigenvalues $4,4$ — yet $4I$ turns every direction into an eigenvector while the Jordan block turns only one.

Direction θ 30°

💡 The number of eigendirections is the geometric multiplicity: $2$ for $4I$ , $1$ for the Jordan block.

IIIVisualization 2 — Conjugation Can't Move 4I

Conjugating by any invertible $M$ leaves $4I$ exactly where it is, but slides the Jordan block to a different-looking twin.

M₁₁ 1 M₁₂ 2 M₂₁ 1 M₂₂ 1

💡 The set of all $M^{-1}BM$ — every twin the right panel can become — is exactly Family 2, the defective similarity class.

IVVisualization 3 — The Invariant That Splits the Families

Geometric multiplicity — the number of independent eigendirections — is $2$ only for $4I$ , and $1$ for every other twin.

a = A₁₁ 4 b = A₁₂ 1

VQuiz Questions

Question 1

Consider the matrix $A = \begin{bmatrix} 4 & 1 \\ 0 & 4 \end{bmatrix}.$

What is the geometric multiplicity of the eigenvalue $\lambda = 4$ (i.e., the number of linearly independent eigenvectors)?

✅ Correct! The Jordan block has exactly one eigendirection, $(1, 0)$ .

❌ Not quite. You may be thinking of the algebraic multiplicity (which is 2). The geometric multiplicity counts independent eigenvectors, and the Jordan block has only one.

❌ Not quite. Solve $(A - 4I)\mathbf{x} = \mathbf{0}$ and count the dimension of the solution space.

Show solution

Solution:

We solve $(A - 4I)\mathbf{x} = \mathbf{0}$ : $A - 4I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.$

This forces the second coordinate $x_2 = 0$ , with $x_1$ free. The kernel is spanned by $(1, 0)$ alone — a 1-dimensional eigenspace.

So the geometric multiplicity is $\boxed{1}$ . (The algebraic multiplicity is 2, since $(\lambda - 4)^2$ is the characteristic polynomial. The mismatch is what makes the matrix defective.)

Question 2

True or False: If two matrices $A$ and $B$ have the same trace, the same determinant, and the same eigenvalues, then they must be similar.

✅ Correct! Eigenvalues alone don't classify matrices up to similarity once eigenvalues repeat. You also need geometric multiplicities.

❌ Not quite. Remember $4I$ and the Jordan block share trace, det, and eigenvalues but are not similar.

Show solution

Solution:

The video's central counterexample shows this is false.

Take $A = 4I = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 1 \\ 0 & 4 \end{bmatrix}$ .

Both have trace $= 8$
Both have determinant $= 16$
Both have eigenvalues $4, 4$

But they are not similar:

$A$ has 2 independent eigenvectors (geometric multiplicity 2)
$B$ has only 1 independent eigenvector (geometric multiplicity 1)

Since geometric multiplicity is a similarity invariant, $A \not\sim B$ .

Question 3

Consider the matrix $C = \begin{bmatrix} 5 & 1 \\ -1 & 3 \end{bmatrix}.$

Which family does $C$ belong to (within the universe of $2 \times 2$ matrices with eigenvalues $4, 4$ )?

✅ Correct! $C - 4I$ has rank 1, so geometric multiplicity is 1 — Family 2.

❌ Not quite. Check: $\operatorname{tr}(C) = 8$ , $\det(C) = 16$ , so eigenvalues are $4, 4$ . Then compute the rank of $C - 4I$ to find the geometric multiplicity.

Show solution

Solution:

Step 1 — Verify the eigenvalues:

$\operatorname{tr}(C) = 5 + 3 = 8$
$\det(C) = (5)(3) - (1)(-1) = 15 + 1 = 16$
Characteristic polynomial: $\lambda^2 - 8\lambda + 16 = (\lambda - 4)^2$

So eigenvalues are $4, 4$ . ✓

Step 2 — Compute the geometric multiplicity: $C - 4I = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}.$

This matrix has rank 1 (the second row is $-1$ times the first), so its kernel is 1-dimensional. Geometric multiplicity = 1.

Step 3 — Decide the family: Since $C \neq 4I$ (in fact, $C$ has only one eigenvector), $C$ lives in Family 2: it is similar to the Jordan block $\begin{bmatrix} 4 & 1 \\ 0 & 4 \end{bmatrix}$ .

Question 4

Which of the following is the fundamental reason that $4I$ forms a similarity family all by itself (a singleton)?

✅ Correct! Commutativity of scalars with everything is exactly what pins $4I$ in place under conjugation.

❌ Not quite. Trace, determinant, and repeated eigenvalues are all shared with the Jordan block, so they can't be what isolates $4I$ . Think about what makes the conjugation $M^{-1}(4I)M$ collapse.

Show solution

Solution:

The key calculation is: $M^{-1}(4I)M = 4 \cdot M^{-1} I M = 4 \cdot M^{-1} M = 4I.$

Scalars commute with every matrix, so the $4$ slides past $M^{-1}$ , the identity drops out, and $M^{-1}M = I$ leaves us with $4I$ no matter what $M$ we picked. That is why $4I$ is stranded by itself.

The other options are red herrings:

❌ Determinant alone doesn't determine the family (the Jordan block also has det 16).
❌ Having repeated eigenvalues doesn't automatically make a matrix a singleton — the Jordan block has repeated eigenvalues and lives in a huge family.
❌ Trace 8 is shared by infinitely many similarity classes.

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