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 imes PP_0 = 0),写成行列式为
- 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(egin{vmatrix}A && O \ C && Bend{vmatrix} = |mathbf{A}||mathbf{B}|,egin{vmatrix}C && A \ B && Oend{vmatrix} = (-1)^{n^2}|mathbf{A}||mathbf{B}|)
构造矩阵
(egin{pmatrix}A && O\ -E && B end{pmatrix}egin{pmatrix}E && B\ O && E end{pmatrix} = egin{pmatrix}A && AB\ -E && O end{pmatrix})
对右边取行列式
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)
ecause $A^{-1}存在,|A|
e 0 herefore |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
(ecause B_{ij} = a_icdot a_j = B_{ji})
( herefore B = B^T)
- Show that for three 3D vector (a,b,c), the following identity holds: (|abc| = (a imes b) cdot c).
(M = |abc| = egin{vmatrix}a_x&&b_x&& c_x\a_y&&b_y&&c_y\a_z&&b_z&&c_zend{vmatrix} = c_x*M_{13} + c_y*M_{23} + c_z * M_{33} = M_{i3} cdot c)
(ecause M_{13} = egin{vmatrix}a_y && b_y\a_z && b_zend{vmatrix}, M_{23} = -egin{vmatrix}a_x && b_x\a_y && b_yend{vmatrix}, M_{33} = egin{vmatrix}a_x && b_x\a_z && b_zend{vmatrix})
( herefore M_{i3} = (a imes b), |abc| = (a imes 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 imes c)
设(a)与底面单位法向量(n)的夹角为( heta) 则(h = a*cos heta = a cdot n)
于是 (V = frac{1}{6} a cdot (b imes 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
ightarrow 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$Aa_1 = lambda_1a_1,Aa_2 = lambda_2a_2 (
对第一个式子左乘)aT_2$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
ightarrow (lambda_1 - lambda_2)a^T_2a_1 = 0)
(lambda_1
e lambda_2 herefore 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
对原式进行化简,(c_1-c_3,c_2-c_3)
对(r3)展开
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