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Fundamental of Computer Graphics (third edition) Chapter 5 Exercises

2020-04-17 18:08:42 作者：互联网

Exercises

Write an implicit equation for the 2D line through points($x_0,y_0$) and ($x_1,y_1$) using a 2D determinant.

有一点$P(x,y)$,在$P_0,P_1$构成的直线上,则三点构成的平行四边形面积一定为0.即$P_1P_0 \times PP_0 = 0$,写成行列式为

\[\begin{vmatrix} x-x_0 & x_1-x_0 \\ y-y_0 & y_1-y_0 \end{vmatrix} = 0\]

Show that if the columns of a matrix are orthonormal, then so are the rows.

根据正交矩阵的性质$R^TR = I = RR^T$

如果一个矩阵是正交的,那么他的转置矩阵也是正交的

Prove the properties of matrix determinants state in Equations (5.5)-(5.7).

(5.5): $|\mathbf{AB}| = |\mathbf{A}||\mathbf{B}|$

根据$\begin{vmatrix}A && O \\ C && B\end{vmatrix} = |\mathbf{A}||\mathbf{B}|,\begin{vmatrix}C && A \\ B && O\end{vmatrix} = (-1)^{n^2}|\mathbf{A}||\mathbf{B}|$

构造矩阵
$\begin{pmatrix}A && O\\ -E && B \end{pmatrix}\begin{pmatrix}E && B\\ O && E \end{pmatrix} = \begin{pmatrix}A && AB\\ -E && O \end{pmatrix}$

对右边取行列式

\[(-1)^{n^2}|-E||AB|=(-1)^{n^2+n}|AB| = |AB| \]

左边两个矩阵的行列式乘积为$|A||B|$,于是(5.5)得证

(5.6) $|A^{-1}| = \frac{1}{|A|}$

$AA^{-1} = E$ 根据(5.5) $|A||A^{-1}|=1$
\because $A^{-1}存在,|A|\ne 0 \therefore |A^{-1}| = \frac{1}{|A|} (5.6)得证 $

(5.7) $|A^T| = |A|$

考虑一下数学归纳,对于$n=1$,显然成立
对于$n=k$时,对$A$对$i$行进行展开
$|A| = \sum_{j=1}^{k} a_{ij}A_{ij}$
对$A^T$对$i$列进行展开
$|A^T| = \sum_{j=1}^{k} b_{ji}A^T_{ji}$

其中$a_{ij} = b_{ji}, A^T_{ji} = (A_{ij})^T, A_{ij}$是n=k-1的情况,我们假设他成立,则n=k也成立

Show that the eigenvalues of a diagonal matrix are its diagonal elements.

$0 = |A-\lambda E|$ 解得 $\lambda$
对于对角阵,显然多项式的零点就是A对角线上的元素

Show that for a square matrix $A$, $AA^T$ is a symmetric matrix.

设$AA^T=B, [a_i]$是$A$的行向量组,则也是$A^T$的列向量组
$\because B_{ij} = a_i\cdot a_j = B_{ji}$
$\therefore B = B^T$

Show that for three 3D vector $a,b,c$, the following identity holds: $|abc| = (a \times b) \cdot c$.

$M = |abc| = \begin{vmatrix}a_x&&b_x&& c_x\\a_y&&b_y&&c_y\\a_z&&b_z&&c_z\end{vmatrix} = c_x*M_{13} + c_y*M_{23} + c_z * M_{33} = M_{i3} \cdot c$

$\because M_{13} = \begin{vmatrix}a_y && b_y\\a_z && b_z\end{vmatrix}, M_{23} = -\begin{vmatrix}a_x && b_x\\a_y && b_y\end{vmatrix}, M_{33} = \begin{vmatrix}a_x && b_x\\a_z && b_z\end{vmatrix}$
$\therefore M_{i3} = (a \times b), |abc| = (a \times b) \cdot c$

Explain why the volume of the tetrahedron with side vectors $a,b,c$(see Figure 5.2) is given by $|abc|/6$.

四面体的体积$V = \frac{1}{3}S*h$

我们让$b,c$构成的平面做底面 $S = \frac{1}{2}b \times c$
设$a$与底面单位法向量$n$的夹角为$\theta$ 则$h = a*cos\theta = a \cdot n$
于是 $V = \frac{1}{6} a \cdot (b \times c) = \frac{1}{6}|abc|$

Demonstrate the four interpretations of matrix-matrix multiplication by taking the following matrix-matrix multiplication code, rearranging the nested loops, and interpreting the resulting code in terms of matrix and vector operations

function mat-mult(in a[m][p], in b[p][n], out c[m][n]){
    // the array c is initialized to zero
    for i = 1 to m
        for j = 1 to n
            for k = 1 to p
                c[i][j] += a[i][k] * b[k][j]
}

三重循环是相互无关的,调节循环的位置可以得到对矩阵乘法不同的解释

$c_{ij} = \vec{a_i} \cdot \vec{b_j}$
$c^{col}_{j} = A \cdot \vec{b_j} $
$c^{row}_{i} = \vec{a_i} \cdot B$
$C = A \cdot B$

Prove that if A,Q, and D satisfy Equation(5.14), v is the $i$th row of Q, and $\lambda$ is the $i$th entry on the diagonal of D, then v is an eigenvector of A with eigenvalue $\lambda$.

感性理解一下 $\mathbf{A} = \mathbf{QDQ}^T \rightarrow \mathbf{A}v = \mathbf{QDQ}^Tv$
$\mathbf{Q}$是正交矩阵$Q^Tv$只有在第$i$行为1其他全为0,再右乘对角阵,得到$D'_{i} = \lambda$ 其余全为0 的向量
所以 $\mathbf{A}v = \mathbf{QDQ}^Tv = QD'$,$QD'$就相当于在$Q$第$i$列的向量乘上$\lambda$即$\lambda v$

Prove that if A,Q, and D satisfy Equation(5.14), the eigenvalues of A are all distince, and v is an eigenvector of A with eigenvalue $\lambda$, then for some $i$, v is the row of Q and $\lambda$ is the $i$th entry on the diagonal of D

设$\lambda_1,\lambda_2$是$A$的不同特征值,$a_1,a_2$是他们对应的特征向量
则有$Aa_1 = \lambda_1a_1,Aa_2 = \lambda_2a_2 $ 对第一个式子左乘$a^T_2$得$a_2TAa_1 = \lambda_1a_2^Ta_1$
继续化简 $a_2^TAa_1 = (A^Ta_2)^Ta_1=(Aa_2)^Ta_1=(\lambda_2a_2)^Ta_1=\lambda_2a^T_2a_1$
于是 $\lambda_2a^T_2a_1 = \lambda_1a_2^Ta_1 \rightarrow (\lambda_1 - \lambda_2)a^T_2a_1 = 0$
$\lambda_1 \ne \lambda_2 \therefore a_2 a_1$ 正交

Given the ($x,y$) corordinates of the three vertices of a 2D triangle, explain why the area is given by

\[\frac{1}{2}\begin{vmatrix}x_0 && x_1 && x_2 \\ y_0 && y_1 && y_2 \\ 1 && 1 && 1\end{vmatrix} \]

对原式进行化简,$c_1-c_3,c_2-c_3$

\[\frac{1}{2}\begin{vmatrix}x_0-x_2 && x_1-x_2 && x_2 \\ y_0-y_2 && y_1-y_2 && y_2 \\ 0 && 0 && 1\end{vmatrix} \]

对$r3$展开

\[\frac{1}{2}\begin{vmatrix}x_0-x_2 && x_1-x_2 \\ y_0-y_2 && y_1-y_2 \end{vmatrix} \]

即为$x_0x_2,x_1x_2$两个向量点乘乘二分之一的形式,即为组成的三角形的面积

标签：Chapter,begin,mathbf,third,edition,vmatrix,&&,end,lambda
来源： https://www.cnblogs.com/xxrlz/p/12721495.html