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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 电气工程代写|数字信号过程代写digital signal process代考|The Scaling and Rotational Transformation

The scaling transformation stretches or compresses the length of each entry in the vector. Scaling is usually done from $R^{\mathrm{a}}$ to $R^{\mathrm{a}}$ space where the input and output vectors feature the same number of entries. In this case, we construct the transformation matrix $A$ by separately scaling the entries of the identity matrix.
\begin{aligned} \operatorname{In} &=\left[\begin{array}{l} x \ y \end{array}\right] \ O u t &=A \cdot \operatorname{In}=\left[\begin{array}{ll} \alpha & 0 \ 0 & \beta \end{array}\right] \cdot\left[\begin{array}{l} x \ y \end{array}\right]=\left[\begin{array}{l} \alpha x \ \beta y \end{array}\right] \end{aligned}
In the following example, we illustrate the scaling transformation by selecting a set of unit-length input vectors that form a circle around the origin. Once we transform these input vectors using the matrix $A$, the geometric properties of the transformation become especially clear. To understand the rotational transformation, recall from our discussion on complex numbers that we may rotate a point, $I n=x+j y$, in the complex plane by multiplying it by $e^{i \theta}=\cos (\theta)+j \cdot \sin (\theta)$, where $\theta$ is our angle of rotation.
\begin{aligned} O u t=e^{j \theta} \cdot I n &=(\cos \theta+j \sin \theta) \cdot(x+j y) \ &=(\cos \theta \cdot x-\sin \theta \cdot y)+j(\sin \theta \cdot x+\cos \theta \cdot y) \end{aligned}

The rotational transformation is easily captured in matrix notation. Note that the rotation transformation matrix is orthonormal (its column vectors are of unit length and perpendicular).
$$\text { Out }=\left[\begin{array}{l} x_{\text {out }} \ y_{\text {out }} \end{array}\right]=\left[\begin{array}{cc} \cos (\theta) & -\sin (\theta) \ \sin (\theta) & \cos (\theta) \end{array}\right] \cdot\left[\begin{array}{l} x \ y \end{array}\right]$$
Scaling and rotation are the two most important types of transformations that we will cover, and as it turns out, all other transformation matrices can be represented as a product of these two.

## 电气工程代写|数字信号过程代写digital signal process代考|Introducing the Matrix Inverse

At the beginning of this section we introduced a simple system of equations describing two lines in its traditional and matrix form.
$$\begin{gathered} a_{11} \cdot x+a_{12} y=b_1 \ a_{21} \cdot x+a_{22} y=b_2 \ {\left[\begin{array}{ll} a_{11} & a_{12} \ a_{21} & a_{22} \end{array}\right] \cdot\left[\begin{array}{l} x \ y \end{array}\right]=\left[\begin{array}{l} b_1 \ b_2 \end{array}\right]} \ A \cdot X=B \end{gathered}$$
Solving this system of equations (finding $X$ ) involves dividing by matrix $A$ or multiplying by its reciprocal, $A^{-1}$, or inverse, $\operatorname{inv} \nu(A)$.
\begin{aligned} A^{-1} \cdot A \cdot X &=A^{-1} \cdot B \ I \cdot X &=A^{-1} \cdot B \ X &=A^{-1} \cdot B=\operatorname{inv}(A) \cdot B \end{aligned}

If the inverse of $A$ exists, then one unique solution vector $X$ may be computed. If the inverse of $A$ does not exist, then either an infinite number of solution vectors $X$ exist or none at all. In terms of transformation, the inverse is very easy to understand. Assume a point $I n$ is transformed via matrix $A$ to the point Out. Can we come up with a matrix, $D$, that will transform the point Out back to the point In? If such a matrix $D$ exists, then $D$ is the inverse of $A$ and their product will result in the identity matrix $I$. Matrix invertibility may be determined by calculatiug the determinant or by proceeding with the inverse calculation directly. If the matrix is not invertible, the determinant will be zero and the inverse calculation will suffer an internal divide-by-zero error. We will meet the determinant calculation next and the Gaussian elimination, which we use to compute the inverse, in a later section. Let’s summarize the basic rules regarding the inverse of a square matrix $A$.

# 数字信号过程代考

## 电气工程代写|数字信号过程代写digital signal process代考|The Scaling and Rotational Transformation

$$\operatorname{In}=[x y] \text { Out } \quad=A \cdot \operatorname{In}=\left[\begin{array}{llll} \alpha & 0 & 0 & \beta \end{array}\right] \cdot\left[\begin{array}{ll} x y \end{array}\right]=\left[\begin{array}{ll} \alpha x & \beta y \end{array}\right]$$

$$\text { Out }=e^{j \theta} \cdot \text { In }=(\cos \theta+j \sin \theta) \cdot(x+j y) \quad=(\cos \theta \cdot x-\sin \theta \cdot y)+j(\sin \theta \cdot x+\cos \theta \cdot y)$$

## 电气工程代写|数字信号过程代写digital signal process代考|Introducing the Matrix Inverse

$$a_{11} \cdot x+a_{12} y=b_1 a_{21} \cdot x+a_{22} y=b_2\left[\begin{array}{lll} a_{11} & a_{12} a_{21} & a_{22} \end{array}\right] \cdot[x y]=\left[b_1 b_2\right] A \cdot X=B$$

$$A^{-1} \cdot A \cdot X=A^{-1} \cdot B I \cdot X \quad=A^{-1} \cdot B X=A^{-1} \cdot B=\operatorname{inv}(A) \cdot B$$

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