Linear-Algebra-Done-Right · Unit 23 · Video 1 · Interactive Practice

Invertible Operators Turn Balls into Ellipsoids

IKey Formulas

Formula	Name	Where it comes from
$B = \{v \in V : \\|v\\| < 1\}$	The ball	Strict inequality: the sphere $\\|v\\| = 1$ is left out
$E(s_1 f_1, \dots, s_n f_n) = \Big\{v : \dfrac{\|\langle v, f_1\rangle\|^2}{s_1^2} + \cdots + \dfrac{\|\langle v, f_n\rangle\|^2}{s_n^2} < 1\Big\}$	Ellipsoid with principal axes $s_k f_k$	$f_1, \dots, f_n$ orthonormal basis, $s_k > 0$ ; all $s_k = 1$ gives $B$ (Parseval)
$T(B) = E(s_1 f_1, \dots, s_n f_n)$	Ball to ellipsoid	$T$ invertible with SVD $Tv = \sum_k s_k \langle v, e_k\rangle f_k$ ; key identity $\langle Tv, f_k\rangle = s_k\langle v, e_k\rangle$
$T(E) = T(S(B)) = (TS)(B)$	Ellipsoid to ellipsoid	$E = S(B)$ for the stretch $Sv = \sum_k s_k\langle v, f_k\rangle f_k$

Key Insight: $T$ sends $e_k$ to $s_k f_k$ , so the orthonormal directions $e_1, \dots, e_n$ of the ball land on the principal axes of the ellipsoid, each stretched by its singular value.

IIAn Ellipsoid Is a Stretched Ball

$E(s_1 f_1, s_2 f_2)$ reaches out $s_k$ along $f_k$ , and a point $w$ belongs to it exactly when the weighted sum is below 1.

s₂ 1

IIIThe Ball Lands Exactly on the Ellipsoid

For every $v$ , the ellipsoid's sum at $Tv$ equals $\|v\|^2$ , so $Tv \in E(s_1 f_1, s_2 f_2)$ exactly when $v \in B$ .

Operator

IVQuiz Questions

Problem 1 · Inside the Ellipsoid

Given: $f_1, f_2$ the standard basis of $\mathbf{R}^2$ — which point lies in the ellipsoid $E(3f_1, f_2)$ ?

✅ Correct!

\tfrac{2^2}{9} + 0.5^2 = \tfrac{4}{9} + \tfrac14 = \tfrac{25}{36} < 1

❌ On the boundary.

(3, 0) = 3f_1

gives

\tfrac{9}{9} + 0 = 1

, and the defining inequality is strict, so the tip of a principal axis is not in

E

❌ Wrong axis. The stretch by 3 is along

f_1

(the

x

-direction):

\tfrac{0.5^2}{9} + 2^2 > 1

❌ Not quite.

\tfrac{4}{9} + 0.64 = \tfrac{4}{9} + \tfrac{16}{25} > 1

: this point lies just outside.

Show solution

With $f_1, f_2$ standard, $\langle (x, y), f_1\rangle = x$ and $\langle (x, y), f_2\rangle = y$ , so

E(3f_1, f_2) = \Big\{(x, y) : \frac{x^2}{9} + y^2 < 1\Big\}.

Test each point: $(3, 0)$ gives $1$ (boundary, excluded); $(0.5, 2)$ gives $\tfrac{1}{36} + 4 > 1$ ; $(2, 0.5)$ gives $\tfrac{4}{9} + \tfrac{1}{4} = \tfrac{25}{36} < 1$ ; $(2, 0.8)$ gives $\tfrac{4}{9} + \tfrac{16}{25} = \tfrac{244}{225} > 1$ .

Only $(2, 0.5)$ lies in $E(3f_1, f_2)$ .

Problem 2 · The Axes Follow the f's

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (2y, -3x)$ — which inequality describes $T(B)$ ?

✅ Correct!

Te_1 = 3f_1

with

f_1 = (0, -1)

and

Te_2 = 2f_2

with

f_2 = (1, 0)

: the ellipse reaches 2 horizontally and 3 vertically.

❌ Those are the input directions.

T

stretches

e_1 = (1, 0)

by 3, but sends it to the vertical direction

(0, -3)

. The principal axes point along the

f

's, not the

e

's.

❌ That is the inverse picture. Points of

T(B)

have

|x|

up to 2 and

|y|

up to 3, so the semi-axis lengths go in the denominators.

❌ Not quite.

T

changes lengths:

\|T(1, 0)\| = 3

Show solution

Take $e_1 = (1, 0)$ , $e_2 = (0, 1)$ . Then $Te_1 = (0, -3) = 3\,(0, -1)$ and $Te_2 = (2, 0) = 2\,(1, 0)$ . With $f_1 = (0, -1)$ , $f_2 = (1, 0)$ (an orthonormal basis),

Tv = 3\langle v, e_1\rangle f_1 + 2\langle v, e_2\rangle f_2

is a singular value decomposition with $s_1 = 3$ , $s_2 = 2$ . Hence $T(B) = E(3f_1, 2f_2)$ . Since $\langle (x, y), f_1\rangle = -y$ and $\langle (x, y), f_2\rangle = x$ ,

T(B) = \Big\{(x, y) : \frac{y^2}{9} + \frac{x^2}{4} < 1\Big\}.

Problem 3 · The Longest Semi-Axis

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (x + 2y,\ 2x - 2y)$ — find the longest principal axis of the ellipse $T(B)$ .

✅ Correct!

s_1 = 3

and

f_1 = \pm\tfrac{1}{\sqrt5}(1, -2)

❌ Right length, wrong direction.

(2, 1)

is an eigenvector for the eigenvalue

2

, which gives the shorter axis

s_2 = 2

❌ Take the square root.

9

is an eigenvalue of

T^*T

; the singular value is

\sqrt9 = 3

❌ That is the shorter axis. The eigenvalue

-3

has the larger absolute value, so the longest axis has length

3

Show solution

The matrix $A = \begin{pmatrix} 1 & 2 \\ 2 & -2 \end{pmatrix}$ is symmetric, so $T^*T = T^2$ . Its characteristic polynomial is $\lambda^2 + \lambda - 6 = (\lambda - 2)(\lambda + 3)$ , with eigenvectors $(2, 1)$ for $2$ and $(1, -2)$ for $-3$ .

Then $T^*T$ has eigenvalues $9$ and $4$ with the same eigenvectors, so $s_1 = 3$ , $e_1 = \tfrac{1}{\sqrt5}(1, -2)$ , and

f_1 = \frac{Te_1}{s_1} = \frac{-3e_1}{3} = -e_1 .

The longest semi-axis $s_1 f_1$ has length $3$ and lies along the line through $(1, -2)$ ; the other has length $s_2 = 2$ along $(2, 1)$ .

Problem 4 · Rotating an Ellipsoid Back

Given: $f_1 = \big(\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2}\big)$ , $f_2 = \big(-\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2}\big)$ , $E = E(2f_1, f_2)$ , and $T(x, y) = \big(\tfrac{x + y}{\sqrt2}, \tfrac{y - x}{\sqrt2}\big)$ — which inequality describes $T(E)$ ?

✅ Correct!

T

is the rotation by

-45^\circ

Tf_1 = (1, 0)

and

Tf_2 = (0, 1)

, so

T(E) = E(2Tf_1, Tf_2)

is axis-aligned.

❌ That is $E$ itself.

T

is not the identity; it moves the axes

f_1, f_2

Tf_1, Tf_2

❌ Check where $f_1$ goes.

Tf_1 = \big(\tfrac{2/\sqrt2}{\sqrt2}, 0\big) = (1, 0)

, so the long axis (length 2) is horizontal.

❌ $T$ preserves lengths, but $E$ was never a ball. Its semi-axes have lengths 2 and 1, and so do those of

T(E)

Show solution

Write $E = S(B)$ with $Sv = 2\langle v, f_1\rangle f_1 + \langle v, f_2\rangle f_2$ . Then $T(E) = (TS)(B)$ and

(TS)v = 2\langle v, f_1\rangle\, Tf_1 + \langle v, f_2\rangle\, Tf_2 .

Compute $Tf_1 = \big(\tfrac{1}{\sqrt2}\cdot\tfrac{2}{\sqrt2}, 0\big) = (1, 0)$ and $Tf_2 = \big(0, \tfrac{1}{\sqrt2}\cdot\tfrac{2}{\sqrt2}\big) = (0, 1)$ , an orthonormal basis. So this is a singular value decomposition of $TS$ with singular values $2, 1$ , and the ball theorem gives

T(E) = E\big(2(1, 0),\ (0, 1)\big) = \Big\{(x, y) : \frac{x^2}{4} + y^2 < 1\Big\}.

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