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Summary Mathematics 3 : Advanced Linear Algebra | UvA | 2026/27

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Voorbeeld 5 van de 25 pagina's

Complete course summary for Mathematics 3: Advanced Linear Algebra at Universiteit van Amsterdam, covering orthogonality, least squares, diagonalization, complex eigenvalues, the spectral theorem, quadratic forms, and singular value decomposition. The document includes detailed lecture notes with step-by-step solution guides, worked exam exercises, and quick-reference guides for all major topics. Ideal for exam preparation and consolidating understanding of advanced linear algebra concepts—organized by topic with practice questions included.

Voorbeeld van de inhoud

Mathematics 3
Advanced Linear Algebra — Complete Course Summary


Orthogonality & Least Squares, Diagonalization, Complex Eigenvalues,
the Spectral Theorem, Quadratic Forms, and Singular Value Decomposition




Lecture Notes, Step-by-Step Solution Guides &
Worked Exam Exercises




This document is an independent study summary and is not affiliated with, endorsed by, or a substitute for
any university’s official course materials.

,Mathematics 3 — Advanced Linear Algebra Summary 1



Contents


I Lecture Notes & Worked Exercises 2

1 Orthogonality and Least Squares 2
1.1 Orthogonal Projections and Orthonormal Bases . . . . . . . . . . . . . . . . . . . . . 2
1.2 Gram–Schmidt Process and QR Factorization . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Orthogonal Transformations and Orthogonal Matrices . . . . . . . . . . . . . . . . . 4
1.4 Least Squares and Data Fitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Diagonalization 5
2.1 Eigenvalues, Eigenvectors, and Diagonalization . . . . . . . . . . . . . . . . . . . . . 5
2.2 Finding Eigenvalues: the Characteristic Equation . . . . . . . . . . . . . . . . . . . . 5
2.3 Algebraic and Geometric Multiplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.4 How Eigenvalues Transform (a Useful Reference List) . . . . . . . . . . . . . . . . . 6

3 Complex Numbers and Complex Eigenvalues 7
3.1 Complex Number Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3.2 Polynomials and the Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . 7
3.3 Complex Eigenvalues of Real Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4 Schur Decomposition and the Spectral Theorem 8
4.1 Real Schur Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4.2 Complex Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4.3 The Spectral Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.4 Geometric Classification of Orthogonal Matrices . . . . . . . . . . . . . . . . . . . . 9

5 Quadratic Forms and Positive Definite Matrices 10
5.1 Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
5.2 Definiteness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
5.3 Constrained Optimization of Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . 12

6 Singular Value Decomposition 12
6.1 Singular Values and the SVD Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 12


II Quick-Reference Step-by-Step Guides 14

7 Guide: Finding a Basis (Kernel, Image, Orthogonal Complement) 14

8 Guide: Orthogonal Projections, Distance, and Gram–Schmidt 14

9 Guide: Diagonalization 15

10 Guide: Complex Eigenvalues and Rotation-Scaling Matrices 15

11 Guide: Spectral Theorem, Orthogonal Diagonalization 16

12 Guide: Rotation and Reflection Matrices 16

13 Guide: Quadratic Forms 17

14 Guide: Singular Value Decomposition 17

,Mathematics 3 — Advanced Linear Algebra Summary 2


III Worked Exam Exercises 18

15 Orthogonality, Gram–Schmidt, and Least Squares 18

16 Diagonalization and Complex Eigenvalues 19

17 Spectral Theorem and Rotations/Reflections 20

18 Quadratic Forms and Positive Definite Matrices 21

19 Singular Value Decomposition 22


IV Practice Questions 23

, Mathematics 3 — Advanced Linear Algebra Summary 3




Part I Lecture Notes & Worked Ex-
ercises
1 Orthogonality and Least Squares
1.1 Orthogonal Projections and Orthonormal Bases
√
Two vectors ⃗v , w ⃗ ∈ Rn are orthogonal if ⃗v · w
⃗ = 0. The norm of ⃗v is ∥⃗v ∥ = ⃗v · ⃗v , and ⃗v is a unit
vector if ∥⃗v ∥ = 1; any nonzero ⃗v can be normalized via û = ∥⃗v1∥ ⃗v .
A vector ⃗x ∈ Rn is orthogonal to a subspace V if it is orthogonal to every vector in V .
Vectors ⃗u1 , . . . , ⃗um are orthonormal if they are all unit vectors and pairwise orthogonal: ⃗ui ·⃗uj = 1
if i = j, and 0 otherwise.

Orthonormal vectors are linearly independent, and if there are n of them in Rn , they
form a basis of Rn .

Orthogonal decomposition. For a subspace V of Rn and ⃗x ∈ Rn , there is a unique decomposi-
tion
⃗x = ⃗xq + ⃗x⊥ , ⃗xq ∈ V, ⃗x⊥ ∈ V ⊥ ,
where ⃗xq = projV (⃗x) is the orthogonal projection of ⃗x onto V . If ⃗u1 , . . . , ⃗um is an orthonormal
basis of V ,

projV (⃗x) = (⃗u1 · ⃗x)⃗u1 + · · · + (⃗um · ⃗x)⃗um .
If ⃗u1 , . . . , ⃗un is an orthonormal basis of all of Rn : ⃗x = (⃗u1 · ⃗x)⃗u1 + · · · + (⃗un · ⃗x)⃗un for every ⃗x.

The transformation T (⃗x) = projV (⃗x) is linear, and its matrix is P = QQT , where Q =
[⃗u1 · · · ⃗um ] has the orthonormal basis of V as columns; P is symmetric since P T = (QQT )T =
QQT = P .

Orthogonal complement. V ⊥ = {⃗x ∈ Rn : ⃗v · ⃗x = 0 ∀⃗v ∈ V } is the kernel of the orthogonal
projection onto V .

For a subspace V of Rn : (1) V ⊥ is a subspace; (2) V ∩ V ⊥ = {⃗0}; (3) dim(V ) + dim(V ⊥ ) = n;
(4) (V ⊥ )⊥ = V .


Key inequalities and identities.
Pythagorean theorem: ∥⃗x + ⃗y ∥2 = ∥⃗x∥2 + ∥⃗y ∥2 holds iff ⃗x ⊥ ⃗y .
Projection is a contraction: ∥projV (⃗x)∥ ≤ ∥⃗x∥, with equality iff ⃗x ∈ V .
Cauchy–Schwarz: |⃗x · ⃗y | ≤ ∥⃗x∥ ∥⃗y ∥, with equality iff ⃗x, ⃗y are parallel.
⃗x · ⃗y
Angle between vectors: θ = arccos ∈ [0, π].
∥⃗x∥∥⃗y ∥

,Mathematics 3 — Advanced Linear Algebra Summary 4


Proof of the triangle inequality via Cauchy–Schwarz (a proof pattern that recurs throughout the
course):
C-S
⃗ 2 = ∥⃗v ∥2 + ∥w∥
∥⃗v + w∥ ⃗ 2 + 2(⃗v · w)
⃗ ≤ ∥⃗v ∥2 + ∥w∥
⃗ 2 + 2∥⃗v ∥∥w∥ ⃗ 2,
⃗ = (∥⃗v ∥ + ∥w∥)

and taking square roots gives ∥⃗v + w∥
⃗ ≤ ∥⃗v ∥ + ∥w∥.
⃗
Worked example: distance to a hyperplane via the normal vector

Compute the distance from ⃗b = (1, 2, 3, 4) to the subspace V = {⃗x ∈ R4 : 4x1 −6x2 +2x3 +5x4 =
0}.
Solution. The equation defining V says ⃗n · ⃗x = 0 where ⃗n = (4, −6, 2, 5), so V ⊥ = span{⃗n}.
Since V ⊥ is one-dimensional, the projection onto V ⊥ is easy:

⃗n · ⃗b
projV ⊥ (⃗b) = ⃗n,
⃗n · ⃗n

and d(⃗b, V ) = ∥⃗b − projV (⃗b)∥ = ∥projV ⊥ (⃗b)∥ — this is much faster than projecting onto V itself
when V is a hyperplane (codimension 1).

Worked example: minimizing a linear functional on the unit sphere
Among all unit vectors ⃗u = (x, y, z) ∈ R3 , find the one minimizing x + 2y + 3z.
Solution. Write x + 2y + 3z = ⃗u · ⃗v with ⃗v = (1, 2, 3). By Cauchy–Schwarz, −∥⃗u∥∥⃗v ∥ ≤ ⃗u √ · ⃗v ≤
∥⃗u∥∥⃗v ∥, with√equality iff ⃗u, ⃗v are parallel. Since ∥⃗u∥ = 1, write ⃗u = t⃗v with 1 = ∥t⃗v√∥ = |t| 14,
so t = ±1/ 14. Since ⃗u · ⃗v = t∥⃗v ∥2 = 14t, the minimum occurs at t = −1/ 14, giving
⃗u = − √114 (1, 2, 3).


1.2 Gram–Schmidt Process and QR Factorization

Gram–Schmidt. Given a basis ⃗v1 , . . . , ⃗vm of a subspace V of Rn , resolve each ⃗vj (j = 2, . . . , m)
into components parallel and perpendicular to span(⃗v1 , . . . , ⃗vj−1 ):

1 ⊥
⃗vj⊥ = ⃗vj − (⃗u1 · ⃗vj )⃗u1 − · · · − (⃗uj−1 · ⃗vj )⃗uj−1 , ⃗uj = ⃗v ,
∥⃗vj⊥ ∥ j

with ⃗u1 = ⃗v1 /∥⃗v1 ∥. Then ⃗u1 , . . . , ⃗um is an orthonormal basis of V .

QR factorization. Any n × m matrix M with linearly independent columns ⃗v1 , . . . , ⃗vm can
be written M = QR, where Q is n × m with orthonormal columns ⃗u1 , . . . , ⃗um and R is m × m
upper triangular with positive diagonal entries. This representation is unique, with

rjj = ∥⃗vj⊥ ∥ (r11 = ∥⃗v1 ∥), rij = ⃗ui · ⃗vj for i < j, rij = 0 for i > j.

Equivalently: apply Gram–Schmidt to the columns of M to get Q; then R = QT M (since
QT Q= Im ).
Worked example: QR factorization
 
2 2
Find the QR factorization of M =  1 7 .
−2 −8
Solution. r11 = ∥⃗v1 ∥ = 3, ⃗u1 = 3 (2, 1, −2). Then r12 = ⃗u1 · ⃗v2 = 9, so ⃗v2⊥ = ⃗v2 − 9⃗u1 =
1

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