正交化得正交向量組 β1,β2,⋯ ,βn\boldsymbol{\beta}_1,\boldsymbol{\beta}_2,\cdots,\boldsymbol{\beta}_nβ1,β2,⋯,βn: β1=α1,β2=α2−(α2,β1)(β1,β1)β1,⋯βn=αn−(αn,βn−1)(βn−1,βn−1)βn−1−(αn,βn−2)(βn−2,βn−2)βn−2−⋯−(αn,β1)(β1,β1)β1 \begin{aligned} & \boldsymbol{\beta}_1 = \boldsymbol{\alpha}_1,\\ & \boldsymbol{\beta}_2 = \boldsymbol{\alpha}_2 - \frac{(\boldsymbol{\alpha}_2,\boldsymbol{\beta}_1)}{(\boldsymbol{\beta}_1,\boldsymbol{\beta}_1)}\boldsymbol{\beta}_1,\\ & \cdots\\ & \boldsymbol{\beta}_n = \boldsymbol{\alpha}_n - \frac{(\boldsymbol{\alpha}_n,\boldsymbol{\beta}_{n-1})}{(\boldsymbol{\beta}_{n-1},\boldsymbol{\beta}_{n-1})}\boldsymbol{\beta}_{n-1} - \frac{(\boldsymbol{\alpha}_n,\boldsymbol{\beta}_{n-2})}{(\boldsymbol{\beta}_{n-2},\boldsymbol{\beta}_{n-2})}\boldsymbol{\beta}_{n-2} - \cdots - \frac{(\boldsymbol{\alpha}_n,\boldsymbol{\beta}_{1})}{(\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{1})}\boldsymbol{\beta}_{1} \end{aligned} β1=α1,β2=α2−(β1,β1)(α2,β1)β1,⋯βn=αn−(βn−1,βn−1)(αn,βn−1)βn−1−(βn−2,βn−2)(αn,βn−2)βn−2−⋯−(β1,β1)(αn,β1)β1
單位化得標準正交向量組 η1,η2,⋯ ,ηn\boldsymbol{\eta}_1,\boldsymbol{\eta}_2,\cdots,\boldsymbol{\eta}_nη1,η2,⋯,ηn: η1=β1∥β1∥,η2=β2∥β2∥,⋯ηn=βn∥βn∥ \begin{aligned} &\boldsymbol{\eta}_1 = \frac{\boldsymbol{\beta}_1}{\begin{Vmatrix} \boldsymbol{\beta}_1 \end{Vmatrix}},\\ &\boldsymbol{\eta}_2 = \frac{\boldsymbol{\beta}_2}{\begin{Vmatrix} \boldsymbol{\beta}_2 \end{Vmatrix}},\\ &\cdots\\ &\boldsymbol{\eta}_n = \frac{\boldsymbol{\beta}_n}{\begin{Vmatrix} \boldsymbol{\beta}_n \end{Vmatrix}} \end{aligned} η1=∥∥β1∥∥β1,η2=∥∥β2∥∥β2,⋯ηn=∥∥βn∥∥βn
1 齊次線性方程組 mmm 個方程 nnn 個未知量的齊次線形方程組: {a11x1+a12x2+⋯+a1nxn=0a21x1+a22x2+⋯+a2nxn=0⋯⋯am1x1+am2x2+⋯+amnxn=0 \begin{cas
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如果矩陣滿足,則矩陣P稱爲對稱矩陣,對稱矩陣有很多優秀的屬性,可以說是最重要的矩陣。 1.對稱矩陣的對角化 如果一個矩陣有n個線性無關的特徵向量,則矩陣是可對角化的,矩陣可表示成,相應的。因爲,很有可能A的逆等於A的轉置。同樣的,就可能有
How to calculate the rank of a matrix? rank: number of pivots in a matrix. pivot position: the location that correspon
1 等價矩陣 設 A,B\boldsymbol{A},\boldsymbol{B}A,B 均是 m×nm \times nm×n 矩陣,若存在可逆矩陣 Pm×m,Qn×n\boldsymbol{P}_{m \times m},