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Linear Algebra --- Lecture 12—13 Study Notes

Topics Covered

  • Vector Norms
  • Frobenius Norm
  • Determinants
  • Eigenvalues
  • Eigenvectors
  • Characteristic Polynomial
  • Finding Eigenvalues
  • Finding Eigenvectors
  • Practice Problems

Vector Norms

Definition

A norm measures the length or magnitude of a vector.

Key Points

L1 Norm (Manhattan Norm)

∥x∥1=∑i∣xi∣\|x\|_1=\sum_i |x_i|

L2 Norm (Euclidean Norm)

∥x∥2=∑ixi2\|x\|_2=\sqrt{\sum_i x_i^2}

The L2 norm represents the shortest distance from the origin.

Example

Let

x=[−142]x=\begin{bmatrix} -1\\ 4\\ 2 \end{bmatrix}

L1 Norm

∥x∥1=∣−1∣+∣4∣+∣2∣=7\|x\|_1=|-1|+|4|+|2|=7

L2 Norm

∥x∥2=(−1)2+42+22=21\|x\|_2=\sqrt{(-1)^2+4^2+2^2} =\sqrt{21}

Frobenius Norm

Definition

The Frobenius norm is the matrix version of the L2 norm.

∥A∥F=∑i∑jaij2\|A\|_F= \sqrt{\sum_i\sum_j a_{ij}^2}

Example

A=[120010003]A= \begin{bmatrix} 1&2&0\\ 0&1&0\\ 0&0&3 \end{bmatrix} ∥A∥F=12+22+12+32=15\|A\|_F= \sqrt{1^2+2^2+1^2+3^2} = \sqrt{15}

Determinants

2×2 Matrix

∣abcd∣=ad−bc\begin{vmatrix} a&b\\ c&d \end{vmatrix} =ad-bc

3×3 Matrix

Expand along any row or column using cofactors.

Important Facts

If

∣A∣=0|A|=0

then

  • Matrix is singular.
  • A−1A^{-1} does not exist.
  • Columns are linearly dependent.
  • No unique solution exists.

Eigenvalues and Eigenvectors

Definition

If

Av=λvAv=\lambda v

then

  • vv is an eigenvector.
  • λ\lambda is the corresponding eigenvalue.

Finding Eigenvalues

Move all terms to one side.

Av=λvAv=\lambda v Av−λIv=0Av-\lambda Iv=0 (A−λI)v=0(A-\lambda I)v=0

For non-zero solutions,

det⁡(A−λI)=0\det(A-\lambda I)=0

This equation is called the Characteristic Polynomial.

Example

Let

A=[2112]A= \begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}

Characteristic equation

∣2−λ112−λ∣=0\left| \begin{matrix} 2-\lambda&1\\ 1&2-\lambda \end{matrix} \right| =0 (2−λ)2−1=0(2-\lambda)^2-1=0 λ2−4λ+3=0\lambda^2-4\lambda+3=0 (λ−3)(λ−1)=0(\lambda-3)(\lambda-1)=0

Eigenvalues

λ=3,  1\lambda=3,\;1

Finding Eigenvectors

For each eigenvalue solve

(A−λI)v=0(A-\lambda I)v=0

For

λ=3\lambda=3

the eigenvector is

[11]\begin{bmatrix} 1\\ 1 \end{bmatrix}

For

λ=1\lambda=1

the eigenvector is

[−11]\begin{bmatrix} -1\\ 1 \end{bmatrix}

General Procedure

  1. Compute A−λIA-\lambda I
  2. Find ∣A−λI∣|A-\lambda I|
  3. Solve characteristic polynomial
  4. Find eigenvalues
  5. Substitute each eigenvalue into (A−λI)v=0(A-\lambda I)v=0
  6. Solve for eigenvectors

Common Mistakes

  • Forgetting the identity matrix.
  • Incorrect determinant calculations.
  • Mixing eigenvalues with eigenvectors.
  • Forgetting eigenvectors are any non-zero scalar multiple.

Short Exam Notes

  • Av=λvAv=\lambda v
  • Characteristic equation:
∣A−λI∣=0|A-\lambda I|=0
  • Solve determinant first.
  • Then compute eigenvectors.
  • If ∣A∣=0|A|=0, inverse does not exist.