Linear Algebra
Table of Contents
1. Linear Equations¶
- Set of linear equations (two equations, two unknowns)
$$ \begin{align*} 4x_{1} − 5x_{2} &= −13\\ −2x_{1} + 3x_{2} &= 9 \end{align*} $$
1.1. Solving Linear Equations¶
- Two linear equations
$$ \begin{align*} 4x_{1} − 5x_{2} &= −13\\ −2x_{1} + 3x_{2} &= 9 \end{align*} $$
- In vector form, $Ax = b$, with
$$A = \begin{bmatrix} 4 & -5 \\ -2 & 3 \end{bmatrix} , \quad x = \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} , \quad b = \begin{bmatrix} -13 \\ 9 \end{bmatrix} $$
- Solution using inverse
$$ \begin{align*} Ax &= b \\ A^{-1}Ax &= A^{-1}b \\ x &= A^{-1}b \end{align*} $$
Don’t worry here about how to compute inverse, but it’s very siminp.linargr to the standard method for solving linear equations
We will use a
numpyto compute
import numpy as np
A = np.array([[4, -5],[-2, 3]])
print(A)
b = np.array([[-13],[9]])
print(b)
$A^{-1} b$
x = np.linalg.inv(A).dot(b)
print(x)
A = np.asmatrix(A)
b = np.asmatrix(b)
x = A.I*b
print(x)
A = np.array([[4, -5],
[-2, 3]])
b = np.array([[-13],
[9]])
x = np.linalg.inv(A).dot(b)
print(x)
A = np.asmatrix(A)
b = np.asmatrix(b)
x = A.I*b
print(x)
1.2. System of Linear Equations¶
- Consider system of linear equations
$$ \begin{align*} y_1 &= a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} \\ y_2 &= a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} \\ &\, \vdots \\ y_m &= a_{m1}x_{1} + a_{m2}x_{2} + \cdots + a_{mn}x_{n} \end{align*} $$
- Can be written in a matrix form as $y = Ax$, where
$$ y= \begin{bmatrix} y_{1} \\ y_{2} \\ \vdots \\ y_{m} \end{bmatrix} \qquad A = \begin{bmatrix} a_{11}&a_{12}&\cdots&a_{1n} \\ a_{21}&a_{22}&\cdots&a_{2n} \\ \vdots&\vdots&\ddots&\vdots\\ a_{m1}&a_{m2}&\cdots&a_{mn} \\ \end{bmatrix} \qquad x= \begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{bmatrix} $$
1.3. Elements of a Matrix¶
- Can write a matrix in terms of its columns
$$A = \begin{bmatrix} \mid&\mid&&\mid\\ a_{1} & a_{2} & \cdots & a_{n}\\ \mid&\mid&&\mid\\ \end{bmatrix} $$
- Careful, $a_{i}$ here corresponds to an entire vector $a_{i} \in \mathbb{R}^{m}$, not an element of a vector
- Can write a matrix in terms of rows
$$A = \begin{bmatrix} - & b_{1}^T& - \\ - & b_{2}^T& - \\ &\vdots& \\ - & b_{m}^T& - \end{bmatrix} $$
- $b_{i} \in \mathbb{R}^{n}$
1.4. Vector-Vector Products¶
- Inner product: $x, y \in \mathbb{R}^{n}$
$$x^{T}y = \sum\limits_{i=1}^{n}x_{i}\,y_{i} \quad \in \mathbb{R} $$
x = np.array([[1],
[1]])
y = np.array([[2],
[3]])
print(x.T.dot(y))
x = np.asmatrix(x)
y = np.asmatrix(y)
print(x.T*y)
z = x.T*y
print(z.A)
1.5. Matrix-Vector Products¶
$A \in \mathbb{R}^{m \times n}, x \in \mathbb{R}^{n} \Longleftrightarrow Ax \in \mathbb{R}^{m}$
Writing $A$ by rows, each entry of $Ax$ is an inner product between $x$ and a row of $A$
$$A = \begin{bmatrix} - &b_{1}^{T} & - \\ -& b_{2}^{T}&- \\ &\vdots& \\ -& b_{m}^{T}&- \end{bmatrix} ,\qquad Ax \in \mathbb{R}^{m} = \begin{bmatrix} b_{1}^{T}x \\ b_{2}^{T}x \\ \vdots \\ b_{m}^{T}x \end{bmatrix} $$
- Writing $A$ by columns, $Ax$ is a linear combination of the columns of $A$, with coefficients given by $x$
$$A = \begin{bmatrix} \mid&\mid&&\mid\\ a_{1} & a_{2} & \cdots & a_{n}\\ \mid&\mid&&\mid\\ \end{bmatrix} ,\qquad Ax \in \mathbb{R}^{m} = \sum\limits_{i=1}^{n}a_{i}x_{i}$$
2. Norms (strenth or distance in linear space)¶
A vector norm is any function $f : \mathbb{R}^{n} \rightarrow \mathbb{R}$ with
- $f(x) \geq 0 \;$ and $\;f(x) = 0 \quad \Longleftrightarrow \quad x = 0$
- $f(ax) = \lvert a \rvert f(x) \;$ for $\; a \in \mathbb{R}$
- $f(x + y) \leq f(x) + f(y)$
- $l_{2}$ norm
$$\left\lVert x \right\rVert _{2} = \sqrt{\sum\limits_{i=1}^{n}x_{i}^2}$$
- $l_{1}$ norm
$$\left\lVert x \right\rVert _{1} = \sum\limits_{i=1}^{n} \left\lvert x_{i} \right\rvert$$
- $\lVert x\rVert$ measures length of vector (from origin)
x = np.array([[4],
[3]])
np.linalg.norm(x, 2)
np.linalg.norm(x, 1)
2.1. Orthogonality¶
Two vectors $x, y \in \mathbb{R}^n$ are orthogonal if
$$x^Ty = 0$$They are orthonormal if, in addition,
$$\lVert x \rVert _{2} = \lVert y \rVert _{2} = 1 $$
x = np.matrix([[1],[2]])
y = np.matrix([[2],[-1]])
x.T*y
2.2. Angle between Vectors¶
For any $x, y \in \mathbb{R}^n, \lvert x^Ty \rvert \leq \lVert x \rVert \, \lVert y \rVert$
(Unsigned) angle between vectors in $\mathbb{R}^n$ defined as
$$ \begin{align*} \theta &= \angle(x,y) = \cos^{-1}\frac{x^Ty}{\lVert x \rVert \lVert y \rVert}\\ \\ \text{thus}\; x^Ty &= \lVert x \rVert \lVert y\rVert \cos\theta \end{align*} $$
- $\{ x \mid x^Ty \leq 0\} $ defines a halfspace with outward normal vector $y$, and boundary passing through 0
3. Matrix and Linear Transformation¶
- Vector
$$ \vec x = \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} $$
- Matrix and Transformation
$$
M=
\begin{bmatrix}
m_{11} & m_{12} & m_{13}\\
m_{21} & m_{22} & m_{23}\\
m_{31} & m_{32} & m_{33}\\
\end{bmatrix}
$$
$$\begin{array}\ \vec y& = &M \vec x\\ \begin{bmatrix}\space \\ \space \\ \space \end{bmatrix} & = &\begin{bmatrix} & & \\ & & \\ & &\end{bmatrix}\begin{bmatrix} \space \\ \space \\ \space \end{bmatrix} \end{array} $$
$$\begin{array}\ \qquad \quad \text{Given} & & \qquad \text{Interpret}\\ \text{linear transformation} & \longrightarrow & \text{matrix}\\ \text{matrix} & \longrightarrow & \text{linear transformation}\\ \end{array} $$
$$\begin{array}{c}\ \vec x\\ \text{input} \end{array} \begin{array}{c}\ \quad \text{linear transformation}\\ \implies \end{array} \quad \begin{array}{l} \vec y\\ \text{output} \end{array}$$
$$\text{transformation} =\text{rotate + stretch/compress}$$
$$T(x_1+x_2) = T(x_1)+T(x_2)$$
- Homogeneity
$$T(kx) = kT(x)$$
- Linear vs. Non-linear
$$\begin{array}{c}\\ \text{linear}& & \text{non-linear}\\ &\\ f(x) = 0 & & f(x) = x + c\\ f(x) = kx & & f(x) = x^2\\ f(x(t)) = \frac{dx(t)}{dt} & & f(x) = \sin x\\ f(x(t)) = \int_{a}^{b} x(t)dt & & \end{array}$$
$$ \vec y = R(\theta) \vec x$$
- Find matrix $R(\theta)$
$$\begin{array}\\ \Longrightarrow &\begin{bmatrix} \cos(\theta)& -\sin(\theta)\\ \sin(\theta)& \cos(\theta) \end{bmatrix}& =& R(\theta) \begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix}& =& R(\theta)\\ \\ \\ &\begin{array}\\ M\vec{x}_1 = \vec{y}_1\\ M\vec{x}_2 = \vec{y}_2\\ \end{array}& =& M \begin{bmatrix} \vec{x}_1 & \vec{x}_2 \end{bmatrix}& =& \begin{bmatrix} \vec{y}_1 & \vec{y}_2 \end{bmatrix} \end{array}$$
import numpy as np
theta = 90/180*np.pi
R = np.matrix([[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]])
x = np.matrix([[1],[0]])
y = R*x
print(y)
$$\begin{array}\\ \vec y = &k\vec x\\ & \uparrow\\ & \text{scalar (not matrix)}\\ \\ \vec y = &k I \vec x & \text{where } I = \text{ Identity martix}\\ \\ \vec y = &\begin{bmatrix}k&0\\0&k\end{bmatrix}\vec x \end{array}$$
Example
$T$: stretch $a$ along $\hat x$-direction & stretch $b$ along $\hat y$-direction
Compute the corresponding matrix $A$
$$\vec y = A \vec x$$
$$\begin{array}\\
\begin{bmatrix}ax_1\\ bx_2\end{bmatrix}& = A\begin{bmatrix}x_1\\ x_2\end{bmatrix} \Longrightarrow A = \,?\\\\
& = \begin{bmatrix}a & 0\\ 0 & b\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}
\end{array}$$
$$\begin{array}\\
A\begin{bmatrix}1\\0\end{bmatrix} & = \begin{bmatrix}a\\0\end{bmatrix} \\
A\begin{bmatrix}0\\1\end{bmatrix} & = \begin{bmatrix}0\\b\end{bmatrix} \\\\
A\begin{bmatrix}1 & 0\\ 0 &1\end{bmatrix} & = A = \begin{bmatrix}a & 0\\0 & b\end{bmatrix} \\
\end{array}$$
More importantly, by looking at $A = \begin{bmatrix}a & 0\\0 & b\end{bmatrix}$, can you think of transformation $T$?
3.4. Projection¶
- $P$: Projection onto $\hat x$ - axis
$$\begin{array}{c}\\ & P & \\ \begin{bmatrix}x_1\\x_2\end{bmatrix} & \implies & \begin{bmatrix}x_1\\ 0\end{bmatrix}\\ \vec x & & \vec y \end{array}$$
$$\vec y = P\vec x = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} x_1 \\ 0 \end{bmatrix}$$
$$
\begin{array}\\
P \begin{bmatrix} 1 \\ 0 \end{bmatrix} & = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\\
P \begin{bmatrix} 0 \\ 1 \end{bmatrix} & = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\\\\
P \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}
\end{array}
$$
import numpy as np
P = np.matrix([[1, 0],
[0, 0]])
x = np.matrix([[1],[1]])
y = P*x
print(y)
3.5. Linear Transformation¶
- If $\vec {v}_1$ and $\vec {v}_2$ are basis, and we know $T(\vec {v}_1) = \vec {\omega}_1$ and $T(\vec {v}_2) = \vec {\omega}_2$. Then, for any $\vec x$
$$\begin{array}{l} \vec x & = a_1\vec v_1 + a_2\vec v_2 & (a_1 \;\text{and } a_2 \;\text{unique})\\ \\ T(\vec x) & = T(a_1\vec v_1 + a_2\vec v_2) \\ & = a_1T(\vec v_1) + a_2T(\vec v_2)\\ & = a_1\vec {\omega}_1 + a_2\vec {\omega}_2\\ \end{array}$$
This is why a linear system makes our life much easier
Only thing that we need is to observe how basis are linearly-transformed
4. Eigenvalue and Eigenvector¶
$$ A \vec v = \lambda \vec v$$
$$ \begin{array}\\ \lambda & = &\begin{cases} \text{positive}\\ 0\\ \text{negative} \end{cases}\\ \lambda \vec v & : & \text{stretched vector}\\ &&\text{(same direction with } \vec v)\\ A \vec v & : &\text{linearly-transformed vector}\\ &&(\text{generally rotate + stretch}) \end{array}$$
- Intuitive interpretation of eigen-analysis
$$A \vec v \text{ parallel to } \vec v$$
- If $\vec {v}_1$ and $\vec {v}_2$ are basis and eigenvectors, and we know $T(\vec {v}_1) = \vec {\omega}_1 = \lambda_1 \vec{v}_1$ and $T(\vec {v}_2) = \vec {\omega}_2 = \lambda_2 \vec{v}_2$. Then, for any $\vec x$
$$\begin{array}{l} \vec x & = a_1\vec v_1 + a_2\vec v_2 & (a_1 \;\text{and } a_2 \;\text{unique})\\ \\ T(\vec x) & = T(a_1\vec v_1 + a_2\vec v_2) \\ & = a_1T(\vec v_1) + a_2T(\vec v_2)\\ & = a_1 \lambda_1\vec {v}_1 + a_2 \lambda_2 \vec {v}_2\\ & = \lambda_1 a_1 \vec {v}_1 + \lambda_2 a_2 \vec {v}_2\\ \end{array}$$
- Only thing that we need is to observe how each basis is independently scaled
4.1. How to Compute Eigenvalue & Eigenvector¶
$$A \vec{v} = \lambda \vec v = \lambda I \, \vec v$$
$$ \begin{array} \implies & A\vec v - \lambda I \vec v = (A - \lambda I)\vec v = 0\\ \\ \implies & A - \lambda I = 0 \text{ or }\\ &\vec v = 0 \text{ or } \\ & (A - \lambda I)^{-1} \text{ does not exist }\\ \\ \implies & \text{det}(A - \lambda I) = 0 \end{array} $$
- We can use its definition for eigen-analysis
Example
$A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ : projection onto $\hat x$- axis
Find eigenvalues and eigenvectors.
import numpy as np
A = np.array([[1, 0],
[0, 0]])
D, V = np.linalg.eig(A)
print('D :', D)
print('V :', V)
Example
Projection onto the plane. Find eigenvalues and eigenvectors.
For any $\vec x$ in the plane, $P\vec x = \vec x \Rightarrow \lambda = 1$
For any $\vec x$ perpendicular to the plane, $P\vec x = \vec 0 \Rightarrow \lambda = 0$
5. System of Linear Equations¶
- well-determined linear systems
- under-determined linear systems
- over-determined linear systems
5.1. Well-Determined Linear Systems¶
- System of linear equations
$$\begin{array}{c}\ 2x_1 + 3x_2 & = 7\\ x_1 + 4x_2 & = 6 \end{array} \; \implies \; \begin{array}{l}\begin{align*} x_1^{*} & = 2\\ x_2^{*} & = 1 \end{align*}\end{array}$$
- Geometric point of view
- Matrix form
$$\begin{array}{c}\ a_{11}x_1 + a_{12}x_2 = b_1\\ a_{21}x_1 + a_{22}x_2 = b_2 \end{array} \begin{array}{c}\ \quad \text{Matrix form}\\ \implies \end{array} \quad \begin{array}{l} \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = \begin{bmatrix} b_{1}\\ b_{2} \end{bmatrix} \end{array}$$
\begin{array}{l} \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \large AX = B \end{array}
\begin{array}{l} \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \therefore & X^{*} = A^{-1} B \quad\text{ if } A^{-1} \text{ exists} \end{array}
$$2x_1 + 3x_2 = 7 \quad \Longrightarrow \quad \text{Many solutions}$$
- Geometric point of view
- Matrix form
$$\begin{array}{c}\ a_{11}x_1 + a_{12}x_2 = b_1 \end{array} \begin{array}{c}\ \quad \text{Matrix form}\\ \implies \end{array} \quad \begin{array}{l} \begin{bmatrix} a_{11} & a_{12} \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = b_{1}\\ \end{array}$$
\begin{array}{l} \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \large AX = B \end{array}
\begin{array}{l} \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \therefore \; \text{ Many Solutions when $A$ is fat} \end{array}
$$\begin{array}{c}\ 2x_1 + 3x_2 & = 7\\ x_1 + 4x_2 & = 6\\ x_1 + x_2 & = 4 \end{array} \; \implies \; \begin{array}{l}\begin{align*} \text{No solutions} \end{align*}\end{array}$$
- Geometric point of view
- Matrix form
$$\begin{array}{c}\ a_{11}x_1 + a_{12}x_2 = b_1\\ a_{21}x_1 + a_{22}x_2 = b_2\\ a_{31}x_1 + a_{32}x_2 = b_3\\ \end{array} \begin{array}{c}\ \quad \text{Matrix form}\\ \implies \end{array} \quad \begin{array}{l} \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ a_{31} & a_{32}\\ \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix} \end{array}$$
\begin{array}{l} \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \large AX = B \end{array}
\begin{array}{l} \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \therefore \; \text{ No Solutions when $A$ is skinny} \end{array}
5.4. Summary of Linear Systems¶
$$\large AX = B$$
- Square: Well-determined Linear Systems
$$ \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = \begin{bmatrix} b_{1}\\ b_{2}\\ \end{bmatrix} $$
- Fat: Under-determined Linear Systems
$$ \begin{bmatrix} a_{11} & a_{12}\\ \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}=b_1 $$
- Skinny: Over-determined Linear Systems
$$ \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ a_{31} & a_{32}\\ \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3}\\ \end{bmatrix} $$
6. Optimization Point of View¶
6.1. Least-Norm Solution¶
- For under-determined linear system
$$ \begin{bmatrix} a_{11}&a_{12}\\ \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = b_1 \quad\text{ or }\quad AX = B$$
Find the solution of $AX = B$ that minimizes $\lVert X \rVert$ or $\lVert X \rVert^2$
i.e., optimization problem
$$\begin{align*} \min \; & \; \lVert X \rVert ^2\\ \text{s. t.} \; & AX = B \end{align*}$$
- Geometric point of view
- Select one solution among many solutions
$$X^{*} = A^T \left( AA^T \right)^{-1}B \quad \text{Least norm solution}$$
- Often control problem
6.2. Least-Square Solution¶
- For over-determined linear system
$$ \begin{align*} \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ a_{31} & a_{32} \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} &\neq \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix} \quad\text{ or }\quad AX \neq B \\ \\ x_1 \begin{bmatrix} a_{11} \\ a_{21} \\ a_{31} \end{bmatrix} + x_2 \begin{bmatrix} a_{12} \\ a_{22} \\ a_{32} \end{bmatrix} &\neq \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix} \end{align*} $$
Find $X$ that minimizes $\lVert E \rVert$ or $\lVert E \rVert^2$
i.e. optimization problem
$$ \begin{align*} \min\limits_{X}{\lVert E\rVert}^2 & = \min\limits_{X}{\lVert AX - B\rVert}^2\\ X^{*} & = \left( A^TA \right)^{-1}A^TB\\ B^{*} = AX^{*} & = A\left( A^TA \right)^{-1}A^TB \end{align*} $$
- Geometric point of view
- Often estimation problem
$$ \begin{align*} W & = \omega\hat{Y}= \omega \frac{Y}{\lVert Y \rVert}, \;\text{where } \omega = \lVert W \rVert\\ \omega & = \lVert X \rVert \cos \theta = \lVert X \rVert \frac{X \cdot Y}{\lVert X \rVert \lVert Y \rVert} = \frac{X \cdot Y}{\lVert Y \rVert}\\ W & = \omega \hat{Y} = \frac{X \cdot Y}{\lVert Y \rVert}\frac{Y}{\lVert Y \rVert} = \frac{X \cdot Y}{\lVert Y \rVert \lVert Y \rVert}Y = \frac{X^T Y}{Y^T Y}Y = \frac{\langle X, Y \rangle}{\langle Y, Y \rangle}Y\\ & = Y\frac{X^T Y}{Y^T Y} = Y\frac{Y^T X}{Y^T Y} = \frac{YY^T}{Y^T Y}X = PX \end{align*} $$
- Another way of computing $\omega$ and $W$
$$ \begin{align*} Y & \perp \left( X - W \right)\\ \implies & Y^T \left( X - W \right) = Y^T \left( X - \omega \frac{Y}{\lVert Y \rVert} \right) = 0\\ \implies & \omega = \frac{Y^T X}{Y^T Y}\lVert Y \rVert\\ & W = \omega \frac{Y}{\lVert Y \rVert} = \frac{Y^TX}{Y^TY}Y = \frac{\langle X, Y \rangle}{\langle Y, Y \rangle}Y \end{align*} $$
import numpy as np
X = np.matrix([[1],[1]])
Y = np.matrix([[2],[0]])
print(X)
print(Y)
print(Y.T*Y)
omega = (X.T*Y)/(Y.T*Y)
print(float(omega))
omega = float(omega)
W = omega*Y
print(W)
6.3.2. Orthogonal Projection onto a Subspace¶
Projection of $B$ onto a subspace $U$ of span of $A_1$ and $A_2$
Orthogonality
$$A \perp \left( AX^{*}-B\right)$$
$$ \begin{align*} A^T \left(AX^{*} - B \right) & = 0\\ A^TAX^{*} & = A^TB\\ X^{*} & = \left( A^TA \right)^{-1}A^TB\\ B^{*} = AX^{*} & = A\left( A^TA \right)^{-1}A^TB \end{align*} $$
import numpy as np
A = np.matrix([[1,0],[0,1],[0,0]])
B = np.matrix([[1],[1],[1]])
X = (A.T*A).I*A.T*B
print(X)
Bstar = A*X
print(Bstar)
6.3.3. Towards Minimization Problems¶
- Suppose $U$ is a subspace of $W$ and $\omega \in W$. Then
$$\lVert \omega - P_U\omega \rVert^2 \leq \lVert \omega - u \rVert$$
$\quad$ for every $u \in U$. Furthermore, if $u \in U$ and the inequality above is an equality, then $u = P_u\omega$
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