LINEAR-ALGEBRA · Unit 27 · Video 2 · Interactive Practice

The Hidden Family: Why Infinitely Many Matrices Share the Same Eigenvalues

IKey Formulas

Formula	Name	Meaning
$B = M^{-1} A M$	Similarity relation	$A$ and $B$ are similar via invertible $M$
$A = S \Lambda S^{-1}$	Diagonalization	Special similarity to a diagonal $\Lambda$
$\text{tr}(A) = \lambda_1 + \lambda_2$	Trace = sum of eigenvalues	Preserved under similarity
$\det(A) = \lambda_1 \lambda_2$	Determinant = product of eigenvalues	Preserved under similarity

IIVisualization 1 — The Invariant Characteristic Parabola

Fix $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ ; every invertible $M$ gives a different $B = M^{-1}AM$ — so which numbers refuse to move?

M₁₁ 1 M₁₂ 4 M₂₁ 0 M₂₂ 1

💡 Trace and determinant are similarity invariants, so the whole characteristic polynomial $\lambda^2 - 4\lambda + 3 = (\lambda-1)(\lambda-3)$ — and therefore its roots — is frozen for every invertible $M$ .

IIIVisualization 2 — Eigenvalues Stay, Eigenlines Move

A shear basis sends each eigenvector $x$ of $A$ to $M^{-1}x$ for $B$ — does its eigenvalue come along for the ride?

Shear s 1

A — eigenlines fixed

B = M⁻¹AM

IVQuiz

Question 1

Suppose $A$ is a $2 \times 2$ matrix with $\text{tr}(A) = 7$ and $\det(A) = 10$ . You compute $B = M^{-1} A M$ for some invertible $M$ of your choosing. What are the eigenvalues of $B$ ?

Correct! Since $\text{tr}(B) = 7$ and $\det(B) = 10$ , the eigenvalues must satisfy $\lambda_1 + \lambda_2 = 7$ and $\lambda_1 \lambda_2 = 10$ , giving $\{2, 5\}$ .

Not quite. Remember that eigenvalues are similarity invariants, and they are completely determined (for a $2 \times 2$ matrix) by the trace and determinant.

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Solution:

Trace and determinant are invariants of similarity, so $B$ has the same trace and determinant as $A$ : $\lambda_1 + \lambda_2 = 7, \qquad \lambda_1 \lambda_2 = 10.$ The two numbers that sum to $7$ and multiply to $10$ are $2$ and $5$ . So the eigenvalues of $B$ are $\{2, 5\}$ — regardless of which invertible $M$ you choose, and regardless of the specific entries of $A$ (as long as its trace and determinant match).

Question 2

True or False: If $A$ and $B$ are similar matrices, then they have the same eigenvectors.

Correct! Eigenvalues are invariant, but eigenvectors transform: if $Ax = \lambda x$ then $B(M^{-1}x) = \lambda (M^{-1}x)$ .

Not quite. This is a classic trap. Re-examine the derivation: the eigenvector picks up an $M^{-1}$ factor.

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Solution:

The statement is false. The proof in the video shows that if $A x = \lambda x$ , then $B (M^{-1} x) = \lambda (M^{-1} x).$ So the eigenvalue $\lambda$ is preserved, but the eigenvector becomes $M^{-1} x$ , not $x$ .

Eigenvalues: identical between similar matrices.
Eigenvectors: related by $x \mapsto M^{-1} x$ — generally different.

This makes geometric sense: a similarity transformation is a change of basis, so the same physical direction has different coordinates in the new basis.

Question 3

Let $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ and $M = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}$ , so $M^{-1} = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix}$ .

What is $B = M^{-1} A M$ ?

Correct! $B = M^{-1}AM = \begin{pmatrix} -2 & -15 \\ 1 & 6 \end{pmatrix}$ . It looks nothing like $A$ , yet shares its eigenvalues $1$ and $3$ — similarity changes a matrix's appearance, never its spectrum.

Not quite. That is the diagonal cousin $\Lambda$ , which equals $S^{-1}AS$ for the eigenvector matrix $S$ . Here $M$ is a different (non-eigenvector) shear, so $B$ won't be diagonal.

Not quite. Carefully compute $AM$ first, then left-multiply by $M^{-1}$ .

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Solution:

Compute right-to-left.

Step 1: $AM$ . $AM = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 9 \\ 1 & 6 \end{pmatrix}.$

Step 2: $M^{-1}(AM)$ . $M^{-1}(AM) = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 9 \\ 1 & 6 \end{pmatrix} = \begin{pmatrix} 2 - 4 & 9 - 24 \\ 1 & 6 \end{pmatrix} = \begin{pmatrix} -2 & -15 \\ 1 & 6 \end{pmatrix}.$

Check invariants: $\text{tr}(B) = -2 + 6 = 4 = \text{tr}(A)$ . $\det(B) = -12 - (-15) = 3 = \det(A)$ . Eigenvalues are again $3$ and $1$ .

Question 4

Which statement best captures why similar matrices are considered descriptions of the "same" underlying linear transformation?

Correct! $B = M^{-1}AM$ is the change-of-basis formula. $M$ is the translator between two coordinate systems for the same underlying transformation.

Not quite. Similar matrices can have completely different entries (as Visualization 1 shows) and different eigenvectors (as Question 2 established). The deeper reason lies in the geometry of basis changes.

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Solution:

The relation $B = M^{-1} A M$ is precisely the change-of-basis formula. If $A$ describes a linear transformation in one basis, then $B$ describes the same transformation in a basis related to the original by $M$ .

The transformation is the geometric reality (a stretching, rotation, shear, etc.).
The matrix is just a coordinate description of that reality.
$M$ is the translator between two coordinate systems.

That is why eigenvalues — the actual stretching factors — are invariant: they belong to the transformation, not to any particular description of it. Entries, in contrast, can change dramatically (as the sliders in Visualization 1 demonstrate). Eigenvectors also change, because directions get re-expressed in the new coordinates.

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