Linear-Algebra-Done-Right · Unit 23 · Video 2 · Interactive Practice

Boxes to Boxes: Parallelepipeds and the SVD

IKey Formulas

Formula	Name	Where it comes from
$u + P(v_1, \dots, v_n)$ , $\ P(v_1, \dots, v_n) = \{a_1 v_1 + \cdots + a_n v_n : a_k \in (0, 1)\}$	Parallelepiped with edges $v_1, \dots, v_n$	$v_1, \dots, v_n$ a basis of $V$ ; open interval, so the boundary is excluded
$T\big(u + P(v_1, \dots, v_n)\big) = Tu + P(Tv_1, \dots, Tv_n)$	Image of a parallelepiped	Linearity keeps the coefficients; invertibility keeps $Tv_1, \dots, Tv_n$ a basis
$u + P(r_1 e_1, \dots, r_n e_n)$	Box	$e_1, \dots, e_n$ orthonormal basis, $r_k > 0$ the edge lengths
$T\big(u + P(r_1 e_1, \dots, r_n e_n)\big) = Tu + P(r_1 s_1 f_1, \dots, r_n s_n f_n)$	Boxes along the $e$ 's stay boxes	SVD of invertible $T$ : $Te_k = s_k f_k$ , every $s_k > 0$

Key Insight: A box whose edges point along the first basis $e_1, \dots, e_n$ of a singular value decomposition comes out as a box along the second basis $f_1, \dots, f_n$ , with edge $r_k e_k$ becoming $r_k s_k f_k$ .

IISame Coefficients In, Same Coefficients Out

A point $u + a_1 v_1 + a_2 v_2$ goes to $Tu + a_1 Tv_1 + a_2 Tv_2$ , with the coefficients unchanged.

Operator

IIIWhich Boxes Stay Boxes?

Under $T(x, y) = (3x + 3y,\ y - x)$ , a box keeps its right angles only when its edges lie along $e_1 = \tfrac{1}{\sqrt2}(1, 1)$ and $e_2 = \tfrac{1}{\sqrt2}(-1, 1)$ .

Tilt θ of e₁ 0° r₁ 1.00 r₂ 1.00

IVQuiz Questions

Problem 1 · A Point of a Parallelogram

Given: the parallelogram $(1, 1) + P\big((2, 1), (0, 2)\big)$ in $\mathbf{R}^2$ — which point belongs to it?

✅ Correct!

(2, 2.5) = (1, 1) + \tfrac12 (2, 1) + \tfrac12 (0, 2)

, both coefficients in

(0, 1)

❌ On the boundary.

(3, 2) = (1, 1) + 1\cdot(2, 1) + 0\cdot(0, 2)

: the coefficients

1

and

0

are not in the open interval

(0, 1)

❌ That is the corner $u$ . It needs

a_1 = a_2 = 0

, which the open interval excludes.

❌ Not quite. Solve

(1, 1) + a_1(2, 1) + a_2(0, 2) = (2, 4)

a_1 = \tfrac12

a_2 = \tfrac54 > 1

Show solution

A point $(x, y)$ is in the set when $(x, y) = (1 + 2a_1,\ 1 + a_1 + 2a_2)$ with $a_1, a_2 \in (0, 1)$ . Solve: $a_1 = \tfrac{x - 1}{2}$ , $a_2 = \tfrac{y - 1 - a_1}{2}$ .

$(3, 2)$ : $a_1 = 1$ , $a_2 = 0$ — boundary, excluded.
$(2, 4)$ : $a_1 = \tfrac12$ , $a_2 = \tfrac54$ — outside.
$(1, 1)$ : $a_1 = a_2 = 0$ — the corner, excluded.
$(2, 2.5)$ : $a_1 = \tfrac12$ , $a_2 = \tfrac12$ — inside.

Problem 2 · A Square Under a Shear

Given: $T(x, y) = (x + y,\ y)$ on $\mathbf{R}^2$ — find the image of the box $P\big((1, 0), (0, 1)\big)$ .

✅ Correct!

T(1, 0) = (1, 0)

and

T(0, 1) = (1, 1)

, and

\langle (1, 0), (1, 1)\rangle = 1 \ne 0

❌ The edge $(0, 1)$ moves.

T(0, 1) = (0 + 1, 1) = (1, 1)

❌ Apply $T$ to each edge.

T(1, 0) = (1, 0)

, not

(1, 1)

: the first coordinate of

T(x, y)

x + y

❌ The corner stays put. The corner is

u = 0

and

T0 = 0

Show solution

The box has corner $u = 0$ and edges $(1, 0), (0, 1)$ . By $T\big(u + P(v_1, v_2)\big) = Tu + P(Tv_1, Tv_2)$ :

T\Big(P\big((1, 0), (0, 1)\big)\Big) = P\big((1, 0), (1, 1)\big).

Its edges have inner product $1 \cdot 1 + 0 \cdot 1 = 1 \ne 0$ , so they are not orthogonal and the image is a parallelogram that is not a box.

Problem 3 · Edge Lengths of the Image Box

Given: $T(x, y) = (3x + 3y,\ y - x)$ , with $s_1 = 3\sqrt2$ , $s_2 = \sqrt2$ , $e_1 = \tfrac{1}{\sqrt2}(1, 1)$ , $e_2 = \tfrac{1}{\sqrt2}(-1, 1)$ — find the edge lengths of $T\big(P(2e_1, e_2)\big)$ .

✅ Correct!

r_1 s_1 = 2 \cdot 3\sqrt2 = 6\sqrt2

and

r_2 s_2 = 1 \cdot \sqrt2 = \sqrt2

❌ Those belong to the diamond. The video's box had

r_1 = r_2 = \sqrt2

; here

r_1 = 2

and

r_2 = 1

❌ The singular values are already square roots. Multiply

r_k

s_k

, not by

s_k^2

❌ $T$ stretches the edges. The edge

r_k e_k

becomes

r_k s_k f_k

Show solution

The box $P(2e_1, e_2)$ has edges along the $e$ 's with $r_1 = 2$ , $r_2 = 1$ , so

T\big(P(2e_1, e_2)\big) = P(2 s_1 f_1,\ s_2 f_2) = P\big(6\sqrt2\, f_1,\ \sqrt2\, f_2\big).

Check directly: $T(2e_1) = \sqrt2\, T(1, 1) = \sqrt2\,(6, 0) = (6\sqrt2, 0)$ and $T(e_2) = \tfrac{1}{\sqrt2} T(-1, 1) = \tfrac{1}{\sqrt2}(0, 2) = (0, \sqrt2)$ . The edges are orthogonal, with lengths $6\sqrt2$ and $\sqrt2$ .

Problem 4 · When Every Box Stays a Box

Given: $T(x, y) = (x - y,\ x + y)$ on $\mathbf{R}^2$ — which statement is true?

✅ Correct!

T = \sqrt2 \times

(rotation by

45^\circ

), so

\langle Tu, Tw\rangle = 2\langle u, w\rangle

: right angles survive in every direction.

❌ The theorem guarantees boxes, it does not exclude others. Here both singular values equal

\sqrt2

, so every orthonormal basis can serve as

e_1, e_2

❌ Axis-aligned boxes do stay boxes, but not only those.

T

rotates and scales uniformly, so a tilted box stays a box too.

❌ Boxes depend on singular values, not eigenvalues.

T

indeed has no real eigenvalues, yet

T(1, 0) = (1, 1)

and

T(0, 1) = (-1, 1)

are orthogonal.

Show solution

The matrix $\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$ gives $T^*T = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = 2I$ . So $\langle Tu, Tw\rangle = \langle T^*Tu, w\rangle = 2\langle u, w\rangle$ for all $u, w$ .

If $u + P(r_1 e_1, r_2 e_2)$ is any box, the image edges $r_1 Te_1$ , $r_2 Te_2$ satisfy $\langle Te_1, Te_2\rangle = 0$ and $\|Te_k\| = \sqrt2$ . So the image is the box $Tu + P\big(\sqrt2 r_1 f_1, \sqrt2 r_2 f_2\big)$ with $f_k = Te_k/\sqrt2$ , for every orthonormal basis $e_1, e_2$ .

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