## 行列式的幾何意義

$\det (AB)=(\det A)(\det B)$

$A=\left[\!\!\begin{array}{cr} 1&-2\\ 0&2\\ 1&-1 \end{array}\!\!\right]$

$B^TB=\begin{bmatrix} \mathbf{a}^T\\ \mathbf{b}^T\\ \mathbf{c}^T \end{bmatrix}\begin{bmatrix} \mathbf{a}&\mathbf{b}&\mathbf{c} \end{bmatrix}=\begin{bmatrix} \mathbf{a}^T\mathbf{a}&\mathbf{a}^T\mathbf{b}&\mathbf{a}^T\mathbf{c}\\ \mathbf{b}^T\mathbf{a}&\mathbf{b}^T\mathbf{b}&\mathbf{b}^T\mathbf{c}\\ \mathbf{c}^T\mathbf{a}&\mathbf{c}^T\mathbf{b}&\mathbf{c}^T\mathbf{c} \end{bmatrix}=\begin{bmatrix} \mathbf{a}^T\mathbf{a}&\mathbf{a}^T\mathbf{b}&0\\ \mathbf{b}^T\mathbf{a}&\mathbf{b}^T\mathbf{b}&0\\ 0&0&1 \end{bmatrix}=\begin{bmatrix} A^TA&0\\ 0&1 \end{bmatrix}$

\displaystyle \begin{aligned} \det B&=\det\begin{bmatrix} \mathbf{a}&\mathbf{b}&\mathbf{c} \end{bmatrix}=(\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}\\ &=(\mathbf{a}\times\mathbf{b})\cdot\frac{\mathbf{a}\times\mathbf{b}}{\Vert\mathbf{a}\times\mathbf{b}\Vert}=\Vert\mathbf{a}\times\mathbf{b}\Vert. \end{aligned}

\begin{aligned} V_2&=v_1v_2=\Vert\mathbf{a}_1\Vert\cdot\Vert(I-P_1)\mathbf{a}_2\Vert\end{aligned}

\begin{aligned} V_3&=v_1v_2v_3=\Vert\mathbf{a}_1\Vert\cdot\Vert(I-P_1)\mathbf{a}_2\Vert\cdot\Vert(I-P_{1,2})\mathbf{a}_3\Vert\end{aligned}

\begin{aligned} V_n&=v_1v_2v_3\cdots v_n\\ &=\Vert\mathbf{a}_1\Vert\cdot\Vert(I-P_1)\mathbf{a}_2\Vert\cdot\Vert(I-P_{1,2})\mathbf{a}_3\Vert\cdots\Vert(I-P_{1,2,\ldots,n-1})\mathbf{a}_n\Vert,\end{aligned}

\begin{aligned} \mathbf{a}_1&=r_{11}\mathbf{q}_{1}\\ \mathbf{a}_2&=r_{12}\mathbf{q}_{1}+r_{22}\mathbf{q}_2\\ &\vdots\\ \mathbf{a}_n&=r_{1n}\mathbf{q}_1+r_{2n}\mathbf{q}_2+\cdots+r_{nn}\mathbf{q}_n.\end{aligned}

\begin{aligned} \mathbf{q}_i^T\mathbf{a}_j&=\mathbf{q}_i^T(r_{1j}\mathbf{q}_1+\cdots+r_{jj}\mathbf{q}_j)=r_{ij}\mathbf{q}_i^T\mathbf{q}_i=r_{ij}.\end{aligned}

$P_{1,\ldots,k-1}=Q_{k-1}Q_{k-1}^{T}$

\begin{aligned} (I-P_{1,\ldots,k-1})\mathbf{a}_k&=\mathbf{a}_k-Q_{k-1}Q_{k-1}^{T}\mathbf{a}_k\\ &=\mathbf{a}_k-\begin{bmatrix} \mathbf{q}_1&\cdots&\mathbf{q}_{k-1} \end{bmatrix}\begin{bmatrix} \mathbf{q}_1^T\mathbf{a}_k\\ \vdots\\ \mathbf{q}_{k-1}^{T}\mathbf{a}_k \end{bmatrix}\\ &=\mathbf{a}_k-\begin{bmatrix} \mathbf{q}_1&\cdots&\mathbf{q}_{k-1} \end{bmatrix}\begin{bmatrix} r_{1k}\\ \vdots\\ r_{k-1,k} \end{bmatrix}\\ &=\mathbf{a}_k-r_{1k}\mathbf{q}_1-\cdots-r_{k-1,k}\mathbf{q}_{k-1}\\ &=r_{kk}\mathbf{q}_k,\end{aligned}

\begin{aligned} v_k&=\Vert(I-P_{1,\ldots,k-1})\mathbf{a}_k\Vert=\Vert r_{kk}\mathbf{q}_k\Vert=\vert r_{kk}\vert\end{aligned}

\begin{aligned} V_n&=v_1v_2\cdots v_n=\vert r_{11}r_{22}\cdots r_{nn}\vert=\vert\det R\vert\end{aligned}

\begin{aligned} \det(A^{T}A)&=\det(R^TQ^TQR)=\det(R^TR)=(\det R^T)(\det R)\\ &=(\det R)(\det R)=(\det R)^2\\ &=(r_{11}\cdots r_{nn})^2=V_n^2,\end{aligned}

$\sqrt{\det(A^{T}A)}=\vert\det R\vert=V_n$。若 $m=n$$A$$n$ 階方陣，則

\begin{aligned} \det(A^{T}A)&=(\det A^{T})(\det A)=(\det A)^2\end{aligned}

### 3 Responses to 行列式的幾何意義

1. wOOL 說道：

‘A的n个行向量’于R^m空间所张开的平行多面体体积 -> 这里应该是’A的n个列向量’吧

2. ccjou 說道：

這是用語不同造成的誤會。在中國大陸，橫向稱為行，縱向稱為列；在台灣，橫向稱為列，縱向稱為行。

3. suehang 說道：

上述混淆问题的只有靠数学符号自身来解决了.