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docs: 二次型重新修改 #95

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2 changes: 1 addition & 1 deletion 讲义/专题/10 行列式.tex
Original file line number Diff line number Diff line change
Expand Up @@ -1915,7 +1915,7 @@ \section{行列式的秩}
\subsection{行列式的秩}

首先我们需要给出矩阵的子式、主子式的定义,然后给出相关的顺序主子式的定义.
\begin{definition}{}{}
\begin{definition}{}{子式、主子式、顺序主子式}
矩阵$A=(a_{ij})_{n \times n}$的任意$k$行($i_1<i_2<\cdots<i_k$行)和任意$k$列($j_1<j_2<\cdots<j_k$列)的交点上的$k^2$个元素排成的行列式
\[\begin{vmatrix}
a_{i_1j_1} & a_{i_1j_2} & \cdots & a_{i_1j_k} \\
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9 changes: 9 additions & 0 deletions 讲义/专题/19 内积空间.tex
Original file line number Diff line number Diff line change
Expand Up @@ -93,6 +93,15 @@ \subsection{正交的定义 \quad 基于正交的性质}
设 $u, v$ 是 $V$ 中的正交向量,则 $\lVert u + v \rVert^2 = \lVert u \rVert^2 + \lVert v \rVert^2 $.
\end{theorem}

\begin{proof}
因为 $u, v$ 互相正交,所以 $\langle u, v \rangle = \langle v, u \rangle = 0$.
\begin{align*}
\lVert u + v \rVert^2 & = \langle u + v, u + v \rangle \\
& = \langle u, u \rangle + \langle u, v \rangle + \langle v, u \rangle + \langle v, v \rangle \\
& = \lVert u \rVert^2 + \lVert v \rVert^2
\end{align*}
\end{proof}

注意勾股定理的逆定理仅在实内积空间上成立.

借助于正交的性质,我们能够简化很多与内积相关的计算,进而会很自然的思考这样一个问题:一个向量能否分解两个互相正交的向量?
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