向量空间

内积 `(: alpha, beta :)`

内积定义为具有正定性的对称双线性函数:

1. `(: alpha, alpha :) ge 0`, 等号成立 `iff alpha = 0`
2. `(: alpha, beta :) = (: beta, alpha :)`
3. `k (: alpha, beta :) = (: k alpha, beta :)`
4. `(: alpha+beta, gamma :) = (: alpha, gamma :) + (: beta, gamma :)`

设 `X, Y in RR^n`, 则 `(:X, Y:) = sum_(i=1)^n x_i y_i in RR`.

设 `X in RR^3`, 则 `X = (:i, X:)i + (:j, X:)j + (:k, X:)k`.

称 `alpha, beta` 正交 (或垂直), 如果 `(:alpha, beta:) = 0`.

模（长度） `|alpha| = sqrt((: alpha, alpha :))`

1. `|alpha| ge 0`, 等号成立 `iff alpha = 0`
2. `|k alpha| = |k| |alpha|`
3. `|alpha + beta| le |alpha| + |beta|`

设 `X in RR^n`, 则 `|X| = sqrt(sum_(i=1)^n x_i^2)`.

外积 `alpha ^^ beta`

外积是一向量, 模等于两向量所夹的平行四边形的面积, 方向满足右手定则.

1. `alpha ^^ beta = -beta ^^ alpha`
2. `(alpha+beta) ^^ gamma = alpha ^^ gamma + beta ^^ gamma`
3. `gamma ^^ (alpha+beta) = gamma ^^ alpha + gamma ^^ beta`
4. `k(alpha ^^ beta) = (k alpha) ^^ beta = alpha ^^(k beta)`

设 `X, Y in RR^3`, 则 `X ^^ Y = | i, j, k; x_1, x_2, x_3; y_1, y_2, y_3; | in RR^3`.

`alpha, beta` 线性相关 (共线) 当且仅当 `alpha ^^ beta = bb 0`.

混合积 `(alpha, beta, gamma) = (:alpha ^^ beta, gamma:) in RR^3`

`(alpha, beta, gamma) = (beta, gamma, alpha) = (gamma, alpha, beta)`

设 `X, Y, Z in RR^3`, 则 `(X, Y, Z) = (:| i, j, k; x_1, x_2, x_3; y_1, y_2, y_3; |, Z:) = | x_1, x_2, x_3; y_1, y_2, y_3; z_1, z_2, z_3; |.`

`alpha, beta, gamma` 线性相关 (共面) 当且仅当 `(alpha, beta, gamma) = 0`.

恒等式 / 不等式

1. (由定义或由行列式性质) `(:alpha ^^ beta, alpha:) = (:alpha ^^ beta, beta:) = 0`;
2. (平行四边形等式) `|alpha+beta|^2 + |alpha-beta|^2 = 2(|alpha|^2+|beta|^2)`;
3. (二重外积) `alpha ^^ (beta ^^ gamma) = | beta, gamma; (:alpha,beta:),(:alpha,gamma:); |`;
4. (推论, 注意联系几何意义) 若 `(:alpha, beta:) = (:alpha, gamma:) = 0`, 则 `alpha ^^ (beta ^^ gamma) = bb 0`.
5. (Lagrange) `(:alpha ^^ beta, gamma ^^ delta:) = | (:alpha, gamma:), (:alpha, delta:); (:beta, gamma:), (:beta, delta:); |`;
6. (Cauchy) `|(:alpha, beta:)| le |alpha||beta|`, 等号成立当且仅当 `alpha, beta` 线性相关;

Lagrange 恒等式的证明: 左 = `(:beta ^^ (gamma ^^ delta), alpha:) = (:|gamma, delta; (:beta, gamma:), (:beta, delta:)|, alpha:)` = 右. 由 Lagrange 恒等式推出外积公式: ` X ^^ Y = sum (:X ^^ Y, epsi_i:) epsi_i = sum(:X ^^ Y, epsi_j ^^ epsi_k:) epsi_i` `= sum | (:X, epsi_j:), (:X, epsi_k:); (:Y, epsi_j:), (:Y, epsi_k:)| epsi_i = | i, j, k; (:X, i:), (:X, j:), (:X, k:); (:Y, i:), (:Y, j:), (:Y, k:)|`.

向量分析

向量/矩阵的导数

定义 `alpha(t) = (a_1(t), a_2(t), a_3(t))` 的导数: `d/dt alpha(t) = ((da_1)/dt, (da_2)/dt, (da_3)/dt)`.

设 `lambda = lambda(t)` 是数值函数, 有:

1. `d/dt (alpha+beta) = (d alpha)/dt + (d beta)/dt`
2. `d/dt (lambda alpha) = (d lambda)/dt alpha + lambda (d alpha)/dt`
3. `d/dt (:alpha, beta:) = (: (d alpha)/dt, beta :) + (:alpha, (d beta)/dt:)`
4. `d/dt (alpha ^^ beta) = (d alpha)/dt ^^ beta + alpha ^^ (d beta)/dt`
5. `d/dt (alpha, beta, gamma) = ((d alpha)/dt, beta, gamma) + (alpha, (d beta)/dt, gamma) + (alpha, beta, (d gamma)/dt)`
6. A(t), B(t) 为矩阵. `d/dt (A(t)B(t)) = (dA)/dt B + A (dB)/dt`

1, 2 验证分量即可. 3 由 `(:alpha, beta:) = sum a_i b_i` 可证. 4 由行列式求导的法则可得. 5 由 3, 4 可得.

Nabla 算子 `grad = (del/(del x), del/(del y), del/(del z))`

设 `f(x, y, z)` 为数量函数, `F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))` 为向量函数 (或称向量场).

梯度 `"grad" f = grad f`, 散度 `"div" F = (:grad, F:)`, 旋度 `"rot" F = grad ^^ F`.

1. `(:grad, F+G:) = (:grad, F:) + (:grad, G:)`
2. `(:grad, fF:) = f(:grad, F:) + (:grad f, F:)`
3. `grad ^^ (F+G) = grad ^^ F + grad ^^ G`
4. `grad ^^ (fF) = f(grad ^^ F) + (grad f) ^^ F`
5. (梯度场是无旋场) `grad ^^ (grad f) = bb"0"`
6. (旋度场是无源场) `(: grad, grad ^^ F:) = 0`

2, 4 相当于将 `grad` 分配到 f 与 F. 5 假定混合偏导相等, 故 `| i, j, k; del/(del x), del/(del y), del/(del z); (del f)/(del x), (del f)/(del y), (del f)/(del z); | = bb"0"`. 6 同样假定混合偏导相等, 有 `| del/(del x), del/(del y), del/(del z); del/(del x), del/(del y), del/(del z); P, Q, R; | = 0`.

合同变换 (保距变换)

具有形式 `cc"T"(X) = cc"T" X + X_0`, 其中 `cc"T"` 为正交实矩阵.

设 `T` 为 `n` 阶正交阵, `t_i` 是其第 `i` 列 (行), 有 `(:t_i, t_j:) = delta_(ij)`.

在上述定理中特别取 `n = 3` 时, `t_1 ^^ t_2 = "sgn"|T| t_3`.

不妨设 `|T| = 1`, 因为 `(:t_1 ^^ t_2, t_3:) = |T| = 1`, 但 `|t_1 ^^ t_2|^2 = (:t_1 ^^ t_2, t_1 ^^ t_2:) = | (:t_1, t_1:), (:t_1, t_2:); (:t_2, t_1:), (:t_2, t_2:); | = 1`,
`|t_3|^2 = (:t_3, t_3:) = 1`. 由 Cauchy 不等式知 `t_1 ^^ t_2 = t_3`.

设 `T` 为正交阵, 则 `T (alpha ^^ beta) = T alpha ^^ T beta`.

记 `t_i` 为 T 的第 i 行, 左 = `sum (:t_i, alpha ^^ beta:) = sum (:t_j ^^ t_k, alpha ^^ beta:) epsi_i` `= sum | (:t_j, alpha:), (:t_j, beta:); (:t_k, alpha:), (:t_k, beta:); | epsi_i = | i, j, k; (:alpha, t_1:), (:alpha, t_2:), (:alpha, t_3:); (:beta, t_1:), (:beta, t_2:), (:beta, t_3:); | = T alpha ^^ T beta`.