Week 10 - Vector Spaces, Orthogonality, and Linear Least-Squares

• Solving underdetermined systems
• Orthogonal Vectors & Orthogonal Spaces
  • Normal Vector
  • Orthogonal Spaces
  • Fundamental Spaces
• Approximating a Solution
• Refers
• Words

1. Solving underdetermined systems

• Important attributes of a linear system $Ax = b$ and associated matrix A:
  • (example:$\left(\begin{array}{c c c c}1 & 3 & 1 & 2 \\ 2 & 6 & 4 & 8 \\ 0 & 0 & 2 & 4\end{array}\right)\left(\begin{array}{c} \chi_0 \\ \chi_1 \\ \chi_2 \\ \chi_3\end{array}\right) = \left(\begin{array}{c} 1 \\ 3 \\ 1\end{array}\right)$)
  • The row-echelon form of the system.
    • 
  • The pivots.
    • the first nonzero entry in each row: 1, 2.
  • The free variables.
    • the columns that has no pivots: $\chi_1, \chi_3$
  • The dependent variables.
    • the columns that has pivots: $\chi_0, \chi_2$
  • A specific solution.
    • Often called a particular solution.
    • The most straightforward way is to set the free variables equal to zero
      • => $\chi_1 = \chi_3 = 0$
      • => $\left(\begin{array}{c} 1/2 \\ 0 \\ 1/2 \\ 0\end{array}\right)$
  • A basis for the null space.
    • Often called the kernel of the matrix.
    • $\chi_0 + 3\chi_1 + \chi_2 + 2\chi_3 = 0, 2\chi_2 + 4\chi_3 = 0$ => $\chi_2 = -2\chi_3, \chi_0 = -3\chi_1$
    • $\begin{bmatrix} \chi_0 \\ \chi_1 \\ \chi_2 \\ \chi_3 \end{bmatrix} = \chi_1 \begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \end{bmatrix} + \chi_3\begin{bmatrix} 0 \\ 0 \\ -2 \\ 1\end{bmatrix}$
    • So the basic for $\mathcal{N}(A) = \text{Span }(\begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ -2 \\ 1\end{bmatrix})$
  • A general solution.
    • Often called a complete solution.
    • given by:
      • $\begin{bmatrix} 1/2 \\ 0 \\ 1/2 \\ 0 \end{bmatrix} + \beta_0 \begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \end{bmatrix} + \beta_1 \begin{bmatrix} 0 \\ 0 \\ -2 \\ 1\end{bmatrix}$
  • A basis for the column space, $\mathcal{C}(A)$.
    • Often called the range of the matrix.
    • equal to the number of dependent variables.
    • The columns that have pivots in them are linearly independent. The corresponding columns in the original matrix are also linearly independent:
      • 
      • 
  • A basis for the row space, $\mathcal{R}(A) = \mathcal{C}(A^T)$.
    • The row space is the subspace of all vectors that can be created by taking linear combinations of the rows of a matrix.
    • List the rows that have pivots in the row echelon form as column vectors:
      • 
      • Notice these are the first and second row of A.
  • The dimension of the row and column space.
    • = number of pivots
    • = 2
  • The rank of the matrix.
    • = number of pivots
    • = 2
  • The dimension of the null space.
    • = the number of non-pivots columns
    • = 2

2. Orthogonal Vectors & Orthogonal Spaces

• Vectors x and y are considered to be orthogonal (perpendicular) if they meet at a right angle: $x^T y = 0$

2.1. Normal Vector

• The normal vector, often simply called the "normal," to a surface is a vector which is perpendicular to the surface at a given point.
• For example:
  • 
  • Define the plane as $Ax + By + Cz = D$
    • Vector $\vec{n} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$ is normal to the plane.
    • Vector $\vec{x_0} = \begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix}$ is pointing to the plane.
    • Vector $\vec{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is pointing to the plane.
  • Then $\vec{x} - \vec{x_0}$ should be on the plane and perpendicular to $\vec{n}$
  • Then $(\vec{x} - \vec{x_0})^T \vec{n} = 0$
  • $\begin{bmatrix} x - x_0 \\ y - y_0 \\ z - z_0 \end{bmatrix}^T \begin{bmatrix} a \\ b \\ c \end{bmatrix} = 0$
  • $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$
  • So we can use $ax + by + cz = ax_0 + by_0 + cz_0$ to represent the plane.

Cross Product

• the cross product or vector product is a binary operation on two vectors in three-dimensional space ($\mathbb{R}^3$) and is denoted by the symbol ×.
  • $\begin{bmatrix} a_0 \\ a_1 \\ a_2 \end{bmatrix} \times \begin{bmatrix} b_0 \\ b_1 \\ b_2 \end{bmatrix} = \begin{bmatrix} a_1 b_2 - a_2 b_1 \\ a_2 b_0 - a_0 b_2 \\ a_0 b_1 - a_1 b_0 \end{bmatrix}$
• Given two linearly independent vectors a and b, the cross product, a × b, is a vector that is perpendicular to both a and b and thus normal to the plane containing them.
  • Because $\begin{bmatrix} a_1 b_2 - a_2 b_1 \\ a_2 b_0 - a_0 b_2 \\ a_0 b_1 - a_1 b_0 \end{bmatrix}^T \begin{bmatrix} a_0 \\ a_1 \\ a_2 \end{bmatrix} = 0$ and $\begin{bmatrix} a_1 b_2 - a_2 b_1 \\ a_2 b_0 - a_0 b_2 \\ a_0 b_1 - a_1 b_0 \end{bmatrix}^T \begin{bmatrix} b_0 \\ b_1 \\ b_2 \end{bmatrix} = 0$
• 
• So we can use vectors a and b to get n. ($n = a \times b$)

Visualizing a column space as a plane in R3

• For example:
• $A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \end{bmatrix}$. $\text{rref }(A) = \begin{bmatrix} 1 & 0 & 3 & 2 \\ 0 & 1 & -2 & -1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$
  • rref: reduced row-echelon form.
• $\mathcal{C}(A) = \text{Span }(\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}, \begin{bmatrix}1 \\ 1 \\ 4\end{bmatrix} )$
• Define $n$ is the normal vector to $\mathcal{C}(A)$, And vector $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ is point to the surface. Then:
  • Use cross product, we get $n = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix} \times \begin{bmatrix}1 \\ 1 \\ 4\end{bmatrix} = \begin{bmatrix}5 \\ -1 \\ -1\end{bmatrix}$
  • $n \cdot (\begin{bmatrix}x \\ y \\ z\end{bmatrix} - \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}) = 0$
  • $5x - y - z = 0$ <=> $\mathcal{C}(A)$

2.2. Orthogonal Spaces

• Definition: Let $V, W \subset \mathbb{R}^n$ be subspaces. Then $V$ and $W$ are said to be orthogonal iff $v \in V$ and $w \in W$ implies $v^T w = 0$. Denoted by $V \perp W$
• Definition: Given subspace $V \subset \mathbb{R}^n$, the set of all vectors in $\mathbb{R}^n$ that are orthogonal to $V$ is denoted by $V^{\perp}$ (pronounced as “V-perp”).

2.3. Fundamental Spaces

• Recall some definitions. Let $A \in \mathbb{R}^{m \times n}$ and have k pivots. Then:
  • Column space: $\mathcal{C}(A) = \{y|y = Ax\}\subset \mathbb{R}^m$.
    • dimension: k
  • Null space: $\mathcal{N}(A) = \{x|Ax = 0\} \subset \mathbb{R}^n$.
    • dimension: n - k
    • $0$ is vector $\in \mathbb{R}^n$
  • Row space: $\mathcal{R}(A) = \mathcal{C}(A^T) =\{y|y = A^T x\} \subset \mathbb{R}^n$.
    • dimension: k
  • Left null space: $\mathcal{N}(A^T) = \{x|x^T A = 0^T\} \subset \mathbb{R}^m$.
    • dimension: m - k
    • $0$ is vector $\in \mathbb{R}^m$

• Theorem: Let $A \in \mathbb{R}^{m \times n}$. Then:

  • $\mathcal{R}(A) \perp \mathcal{N}(A)$.
  • every $x \in \mathbb{R}^n$ can be written as $x = x_r + x_n$ where $x_r \in \mathcal{R}(A)$ and $x_n \in \mathcal{N}(A)$.
  • $A$ is a one-to-one, onto mapping from $\mathcal{R}(A)$ to $\mathcal{C}(A)$.
  • $\mathcal{N}(A^T)$ is orthogonal to $\mathcal{C}(A)$ and the dimension of $\mathcal{N}(A^T)$ equals $m-r$, where $r$ is the dimension of $\mathcal{C}(A)$.

• 

• For example: $A = \begin{bmatrix}2 & -1 & -3 \\ -4 & 2 & 6\end{bmatrix}$
  • $T(x) = Ax$, $T: \mathbb{R}^3 \Rightarrow \mathbb{R}^2$
  • $\mathcal{C}(A) = \text{Span }(\begin{bmatrix} 2 \\ -4 \end{bmatrix}) \subseteq \mathbb{R}^2$
  • $\mathcal{N}(A^T) = \text{Span }(\begin{bmatrix} 2 \\ 1 \end{bmatrix}) \subseteq \mathbb{R}^2$
  • $\mathcal{N}(A) = \text{Span }(\begin{bmatrix} \frac{1}{2} \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} \frac{3}{2} \\ 0 \\ 1 \end{bmatrix}) \subseteq \mathbb{R}^3$
  • $\mathcal{R}(A) = \mathcal{C}(A^T) = \text{Span }(\begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix}) \subseteq \mathbb{R}^3$
    • => $\mathcal{R}(A) \perp \mathcal{N}(A)$
    • => $\mathcal{C}(A) \perp \mathcal{N}(A^T)$

3. Approximating a Solution

• Find a line $y = \gamma_0 + \gamma_1 x$ to interpolate these points:
  • $x = \left(\begin{array}{c} \chi_0 \\ \chi_1 \\ \chi_2 \\ \chi_3 \end{array}\right) = \left(\begin{array}{c}1 \\ 2 \\ 3 \\ 4\end{array}\right) \text{ and } y = \left(\begin{array}{c} \psi_0 \\ \psi_1 \\ \psi_2 \\ \psi_3 \end{array}\right) = \left(\begin{array}{c}1.97 \\ 6.97 \\ 8.89 \\ 10.01\end{array}\right)$
  • 
• Clearly, there is no line could go through all these points, then what is the best approximation?
• Set $A = \begin{bmatrix}1 & 1 \\ 1 & 2 \\ 1 & 3 \\ 1 & 4\end{bmatrix}, b = \begin{bmatrix}1.97 \\ 6.97 \\ 8.89 \\ 10.01\end{bmatrix}$
• We've learned before that $Ax=b$ has a solution iff $b \in \mathcal{C}(A)$. In other words, b is in the plane of $\text{Span}(a_1, a_2,\ldots, a_n)$.
• So, here we are solving $Ax \approx b$.

• Set the projection of b = $z$, $A\hat{x} = z$

  • 

• We can get

  • $b = z + w$ where $w^T v = 0$ for all $v \in \mathcal{C}(A)$.
• Also $w \subset \mathcal{C}(A)^{\perp}$ => $w \subset \mathcal{N}(A^T)$. So, $A^Tw = 0$(same as $w^T A = 0^T$), which means
  • $0 = A^Tw = A^T(b - z) = A^T(b - A \hat{x})$
  • Rewrite it, we get $A^TA \hat{x} = A^T b$.
  • This is known as the normal equation associated with the problem $A\hat{x} \approx b$.
• Although $A^TA$ is nonsingular, then
  • $\hat{x} = (A^T A)^{-1} A^T b$
• And the vector $z \in \mathcal{C}(A)$ closest to $b$ is given by
  • $z = A \hat{x} = A (A^T A)^{-1} A^T b$
• This shows that if A has linearly independent columns, then $z = A \hat{x} = A (A^T A)^{-1} A^T b$ is the vector in the columns space closest to b. This is the projection of b onto the column space of A.
• And the “best” solutionis known as the “linear least-squares” solution.

4. Refers

• https://en.wikipedia.org/wiki/Row_and_column_spaces
• https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/dot-cross-products/v/defining-a-plane-in-r3-with-a-point-and-normal-vector
• https://en.wikipedia.org/wiki/Normal_(geometry))
• http://mathworld.wolfram.com/NormalVector.html
• https://en.wikipedia.org/wiki/Cross_product

5. Words

• orthogonality [,ɔ:θɔɡə'næləti] n. [数] 正交性；相互垂直