Math Notes

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Math

Linear Algebra Basics

Linear algebra is the branch of mathematics concerning linear equations, linear maps, and their representations in vector spaces and through matrices.

1. Vector Spaces

Definition 1.1(Vector Space)
A vector space over a field FF is a set VV together with two operations:
  • Vector addition: V×V→VV \times V \to V
  • Scalar multiplication: F×V→VF \times V \to V
satisfying the eight axioms of associativity, commutativity, identity, inverse, distributivity, and compatibility.
Example 1.1
The space Rn\mathbb{R}^n of all nn-tuples of real numbers is a vector space. For example, in R3\mathbb{R}^3:
v⃗=(123),w⃗=(456)\vec{v} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \quad \vec{w} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}
v⃗+w⃗=(579),2v⃗=(246)\vec{v} + \vec{w} = \begin{pmatrix} 5 \\ 7 \\ 9 \end{pmatrix}, \quad 2\vec{v} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}

2. Linear Transformations

Definition 2.1(Linear Transformation)
A function T:V→WT: V \to W between vector spaces is called alinear transformation if for all vectors u⃗,v⃗∈V\vec{u}, \vec{v} \in V and scalars c∈Fc \in F:
T(u⃗+v⃗)=T(u⃗)+T(v⃗)T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})
T(cv⃗)=cT(v⃗)T(c\vec{v}) = cT(\vec{v})

3. Eigenvalues and Eigenvectors

Definition 3.1(Eigenvalue and Eigenvector)
Let AA be an n×nn \times n matrix. A scalar λ\lambda is an eigenvalue of AA if there exists a non-zero vector v⃗\vec{v} such that:
Av⃗=λv⃗A\vec{v} = \lambda\vec{v}
The vector v⃗\vec{v} is called an eigenvector corresponding to λ\lambda.
Lemma 3.1
The eigenvalues of a matrix AA are the roots of its characteristic polynomial:
det⁡(A−λI)=0\det(A - \lambda I) = 0
Theorem 3.1(Spectral Theorem)
If AA is a real symmetric matrix, then:
  1. All eigenvalues of AA are real.
  2. Eigenvectors corresponding to distinct eigenvalues are orthogonal.
  3. AA can be diagonalized by an orthogonal matrix.
Example 3.1
Find the eigenvalues of A=(4123)A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}.

Solution: We solve det⁡(A−λI)=0\det(A - \lambda I) = 0:

det⁡(4−λ123−λ)=(4−λ)(3−λ)−2=0\det\begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix} = (4-\lambda)(3-\lambda) - 2 = 0
λ2−7λ+10=0  ⟹  (λ−5)(λ−2)=0\lambda^2 - 7\lambda + 10 = 0 \implies (\lambda - 5)(\lambda - 2) = 0

Thus, the eigenvalues are λ1=5\lambda_1 = 5 and λ2=2\lambda_2 = 2.