Orthogonality and Least Squares

$$A = \begin{bmatrix} 1 & -3 & 3 \\ 1 & 5 & 1 \\ 1 & 7 & 2 \end{bmatrix}, \quad c = \begin{bmatrix} 5 \\ -3 \\ 5 \end{bmatrix}$$ We solve the normal equations $A^TA\hat{x} = A^Tc$. --- $$A^T = \begin{bmatrix} 1 & 1 & 1 \\ -3 & 5 & 7 \\ 3 & 1 & 2 \end{bmatri...

Points: $(2,1), (5,2), (7,3), (8,3)$, with $n = 4$. $$\sum y = na0 + a1\sum x$$ $$\sum xy = a0\sum x + a1\sum x^2$$ $x$ $y$ $x^2$ $xy$ -------------- 2 1 4 2 5 2 25 10 7 3 49 21 8 3 64 24 $\sum x=22$ $\sum y=9$ $\sum x^2=142$ $\sum xy=57$ $$9 = 4a0 + 22a1 \...

Data points to fit with $y = a0 + a1 x$: $x$ $y$ ---------- 0 1 1 1 1 1 2 2 3 2 Number of data points: $n = 5$ (the point $(1,1)$ appears twice; both are used). $x$ $y$ $x^2$ $xy$ ----------------------- 0 1 0 0 1 1 1 1 1 1 1 1 2 2 4 4 3 2 9 6 $$\sum x = 7,...

$$A = \begin{bmatrix} -1 & 2 \\ 2 & -3 \\ -1 & 3 \end{bmatrix}, \qquad b = \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix}$$ Unknown: $\hat{x} = \begin{bmatrix} x1 \\ x2 \end{bmatrix}$ The least square solution satisfies the normal equations: $$A^T A \,\hat{x} = ...

$$A = \begin{bmatrix} 1 & -6 \\ 1 & -2 \\ 1 & 1 \\ 1 & 7 \end{bmatrix}, \quad b = \begin{bmatrix} -1 \\ 2 \\ 1 \\ 6 \end{bmatrix}$$ The least-square solution solves the normal equations: $$A^T A \hat{x} = A^T b$$ --- $$A^T = \begin{bmatrix} 1 & 1 & 1 & 1 \\...

$$A = \begin{bmatrix} 1 & 3 & 5 \\ 1 & 1 & 0 \\ 1 & 1 & 2 \\ 1 & 3 & 3 \end{bmatrix}, \qquad b = \begin{pmatrix} 3 \\ 5 \\ 7 \\ 3 \end{pmatrix}$$ The least-square solution $\hat{x}$ solves the normal equations: $$A^T A \hat{x} = A^T b$$ $$A^T = \begin{bmatr...

$$A = \begin{bmatrix} 1 & -3 & -3 \\ 1 & 5 & 1 \\ 1 & 7 & 2 \end{bmatrix}, \quad b = \begin{bmatrix} 5 \\ -3 \\ -5 \end{bmatrix}$$ The least square solution solves the normal equation $A^TA\hat{x} = A^Tb$. --- $$A^T = \begin{bmatrix} 1 & 1 & 1 \\ -3 & 5 & 7...

The line through u means the line passing through the origin in the direction of u, i.e., the set {tu : t ∈ ℝ}. The line segment from -v to v has its midpoint at the origin (since (-v + v)/2 = 0). The line through u is the perpendicular bisector of the segm...

- Vector $\mathbf{u} = (0, -2, 2, -3)$ in $\mathbb{R}^4$ - Task: Define unit vector; find unit vector in the direction of $\mathbf{u}$. All data present. A unit vector is a vector whose length (norm) is equal to $1$. For any nonzero vector $\mathbf{v} \in \...

$$u = (1,\ -2,\ 2,\ 3)$$ The unit vector in the direction of $u$ is: $$\hat{u} = \frac{u}{\u\}, \qquad \u\ = \sqrt{u1^2 + u2^2 + u3^2 + u4^2}$$ $$\u\ = \sqrt{(1)^2 + (-2)^2 + (2)^2 + (3)^2} = \sqrt{1 + 4 + 4 + 9} = \sqrt{18} = 3\sqrt{2}$$ $$\hat{u} = \frac{...

The line segment runs from -u to v. Its midpoint is: $$M = \frac{-\mathbf{u} + \mathbf{v}}{2}$$ The line through u (passing through the origin in the direction of u) is the perpendicular bisector of the segment from -u to v if and only if two conditions hol...

- Vectors: $\mathbf{u} = (1, -1)$ and $\mathbf{v} = (1, 1)$ - Required: State and prove the Pythagorean theorem for vectors, then verify. --- If $\mathbf{u}$ and $\mathbf{v}$ are two vectors in an inner product space, then $\mathbf{u}$ and $\mathbf{v}$ are ...

- Vector: $u = (1, -2, 2, 0)$ $$\u\ = \sqrt{u1^2 + u2^2 + u3^2 + u4^2}$$ $$\u\ = \sqrt{(1)^2 + (-2)^2 + (2)^2 + (0)^2} = \sqrt{1 + 4 + 4 + 0} = \sqrt{9} = 3$$ The unit vector in the same direction as $u$ is: $$v = \frac{u}{\u\} = \frac{1}{3}(1, -2, 2, 0)$$ ...

$$A = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$$ Columns: $$\mathbf{a1} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}, \qquad \mathbf{a2} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$ Goal: find orthogonal $Q$ and upper-triangular $R$ with $A = QR$. $$\mathbf{u1}...