LINEAR-ALGEBRA · Interactive Practice | Unit 26 · Video 4

From Numbers to Shapes: How a 3×3 Matrix Becomes an Ellipsoid

IKey Formulas

Formula	Name	Description
$A = Q\Lambda Q^T$	Principal Axis Theorem	Symmetric matrix factorization
$\mathbf{x}^T A \mathbf{x} = 1$	Ellipsoid equation	Level set of quadratic form
$\text{length}_i = \dfrac{1}{\sqrt{\lambda_i}}$	Half-axis length	Direction $\mathbf{q}_i$ , eigenvalue $\lambda_i$
$\det(A) = \lambda_1 \lambda_2 \lambda_3$	Determinant identity	And $\text{trace}(A) = \sum \lambda_i$

Our matrix from the video: $A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}, \quad \lambda_1 = 2-\sqrt{2},\ \lambda_2 = 2,\ \lambda_3 = 2+\sqrt{2}$

IIVisualization 1 — Eigenvalues Set the Axis Lengths

In the eigenbasis the principal cross-section is an axis-aligned ellipse with semi-axes $1/\sqrt{\lambda_i}$ .

λ₁ 0.50 λ₂ 2.00 λ₃ 3.50

IIIVisualization 2 — Level Sets Grow With √c

Raising the level $c$ in $\mathbf{x}^T A\mathbf{x}=c$ scales the ellipse like $\sqrt{c}$ while its shape holds fixed.

Level c 1.00

This is the $x_3=0$ slice of the video's $A$ ; the off-diagonal $-1$ tilts the ellipse $45\degree$ off the axes.

IVVisualization 3 — Two Tests for Positive Definiteness

For $A(t)=\begin{bmatrix} 2 & t & 0 \\ t & 2 & t \\ 0 & t & 2 \end{bmatrix}$ , the eigenvalue test and the leading-minor test flag positive definiteness together.

t −1.0

Positive definiteness holds exactly while $|t|<\sqrt{2}$ ; past it the smallest eigenvalue and $\det_3$ turn negative together.

VQuiz Questions

Question 1

For the matrix from the video, $A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix},$ the three eigenvalues are $\lambda_1 = 2 - \sqrt{2}$ , $\lambda_2 = 2$ , $\lambda_3 = 2 + \sqrt{2}$ .

Which eigenvalue corresponds to the longest half-axis of the ellipsoid $\mathbf{x}^T A \mathbf{x} = 1$ ?

✅ Correct! Smallest eigenvalue → longest axis (because length = 1/√λ).

❌ Not quite. Remember: length $= 1/\sqrt{\lambda}$ , so larger $\lambda$ gives a shorter axis.

Show solution

Solution:

The half-axis length along eigenvector $\mathbf{q}_i$ is $\text{length}_i = \frac{1}{\sqrt{\lambda_i}}.$

This is a decreasing function of $\lambda$ : smaller eigenvalues produce longer axes.

$\lambda = 2 - \sqrt{2} \approx 0.586 \Rightarrow$ length $\approx 1/\sqrt{0.586} \approx 1.31$ (longest)
$\lambda = 2 \Rightarrow$ length $= 1/\sqrt{2} \approx 0.71$
$\lambda = 2 + \sqrt{2} \approx 3.414 \Rightarrow$ length $\approx 0.54$ (shortest)

The smallest eigenvalue $2 - \sqrt{2}$ gives the longest axis.

Question 2

True or False: Since the matrix $A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$ has all diagonal entries equal to 2, the ellipsoid $\mathbf{x}^T A \mathbf{x} = 1$ is a sphere.

✅ Correct! Diagonal entries do not determine the axis lengths — eigenvalues do, and they are all different.

❌ Not quite. The eigenvalues, not the diagonal, set the axis lengths. Our matrix has three distinct eigenvalues.

Show solution

Solution: False.

A common trap! The diagonal entries of $A$ do not directly give the axis lengths. The geometry is set by the eigenvalues (lengths) and eigenvectors (directions), not by the matrix entries.

For our $A$ , the diagonal is $2, 2, 2$ , but the eigenvalues are three different numbers: $2 - \sqrt{2}$ , $2$ , $2 + \sqrt{2}$ . So the ellipsoid is a lopsided one with three distinct axis lengths, not a sphere.

A sphere requires all three eigenvalues to be equal. The diagonal entries being equal does not force that.

Question 3

For the same matrix $A$ , the three pivots from elimination are $2$ , $\tfrac{3}{2}$ , $\tfrac{4}{3}$ . What is the product of these pivots, and what does it equal?

✅ Correct! The product of pivots equals the determinant (= product of eigenvalues = 4).

❌ Not quite. Pivots multiply to give the determinant (not the trace). And $\det(A) =$ product of eigenvalues.

Show solution

Solution:

Multiply the pivots: $2 \cdot \frac{3}{2} \cdot \frac{4}{3} = \frac{2 \cdot 3 \cdot 4}{2 \cdot 3} = 4.$

In LU decomposition, the determinant of $A$ equals the product of the pivots (assuming no row swaps). And indeed: $\det(A) = \lambda_1 \lambda_2 \lambda_3 = (2 - \sqrt{2}) \cdot 2 \cdot (2 + \sqrt{2}) = 2(4 - 2) = 4.$

All three views — pivots, leading determinants, and eigenvalues — agree. The trace (sum of diagonal) is $6$ , which is the sum of eigenvalues, not the product.

Question 4

A symmetric $3 \times 3$ matrix $B$ has eigenvalues $\lambda_1 = 4$ , $\lambda_2 = 4$ , $\lambda_3 = 1$ . What kind of shape is the level set $\mathbf{x}^T B \mathbf{x} = 1$ ?

✅ Correct! Two equal eigenvalues plus one different gives a rugby-ball-shaped spheroid.

❌ Not quite. Count distinct eigenvalues: two are equal ( $4,4$ ) and one is different ( $1$ ). That pattern names the shape.

Show solution

Solution:

All three eigenvalues are positive, so $B$ is positive definite and the level set is a bounded ellipsoid.

The half-axis lengths are:

Along $\mathbf{q}_1$ : $1/\sqrt{4} = 1/2$
Along $\mathbf{q}_2$ : $1/\sqrt{4} = 1/2$
Along $\mathbf{q}_3$ : $1/\sqrt{1} = 1$

Two axes have equal length and one is different. Geometrically this is a rugby ball (also called a spheroid): a circular cross-section in the $\mathbf{q}_1, \mathbf{q}_2$ plane, stretched along $\mathbf{q}_3$ .

A sphere would require all three eigenvalues equal. A lopsided ellipsoid would require three distinct eigenvalues.

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