upgrade structures and migrate to nextra v4

content/Math429/Math429_L3.md

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+# Lecture 3
+
+## Chapter I Vector Spaces
+
+### Subspaces 1C
+
+Given a vector space $V$, a subset $W\subset V$ is called a subspace if 
+
+$$
+\begin{cases}
+W\neq \phi\\
+W\textup{ is closed under addition and scalar multiplication}
+\end{cases}
+$$
+
+#### Definition 1.41
+
+ Direct Sum
+
+Suppose $V_1,...,V_m$ are subspace of $V$. Their sum $V_1+...+V_m$ is called a direct sum if **each element** $\vec{v_1}+...+\vec{v_m}\in V_1+...+V_m$ in a unique way.
+
+If $v_1+....+v_m$ is a direct sum, we write it as 
+
+$$
+v_1\oplus v_2\oplus ...\oplus v_m
+$$
+
+Example:
+
+$V=\mathbb{R}^3$
+
+$$
+V_1=\{(x,y,0):x,y\in \mathbb{R}\}\\
+V_2=\{(0,a,b):a,b\in \mathbb{R}\}
+$$
+
+Is $V_1+V_2$ a direct sum?
+
+**No, because there are other ways to build (0,0,0) in such space, which is not unique**
+
+For vector $(0,0,0)=(x,y,0)+(0,a,b)$, as long as $y=-a$, there are other ways to build up the vector.
+
+### Theorem 1.45
+
+Suppose $V_1,...,V_m$ are subspaces of $V$, then $V_1+...+V_m$ is a direct sum if and only if the only way to write $\vec{0}=\vec{v_1}+...+\vec{v_m}$ with $\vec{v_1}\in V_1,...,\vec{v_m}\in V_m$. is $\vec{v_1}=...=\vec{v_m}=\vec{0}$
+
+Proof:
+
+$\Rightarrow$
+
+If $\vec{v_1}=...=\vec{v_m}$ is a direct sum, then the only way to write $\vec{0}=\vec{v_1}+...+\vec{v_m}$ where $\vec{v_i}\in V_i$ is $\vec{0}=\vec{0}+...+\vec{0}$ **follows from the definition of direct sum**
+
+$\Leftarrow$
+
+Need to show if the property holds for $\vec{0}$, then it holds for any $\vec{v}\in V_1+...+V_m$ $\iff$ If the property fails for any $\vec{v}\in V_1+...+V_m$, then it fails for $\vec{0}$
+
+If a vector $\vec{v}\in V_1+...+V_m$ satisfies $\vec{v}=\vec{v_1}+\vec{v_2}+...+\vec{v_m}=\vec{u_1}+\vec{u_2}+...+\vec{u_m}$, $\vec{v_i},\vec{u_i}\in V_i$ and there exists $i\in\{1,...,m\}\vec{v_i}\neq \vec{u_i}$,
+
+then $(\vec{v_1}+\vec{v_2}+...+\vec{v_m})-(\vec{u_1}+\vec{u_2}+...+\vec{u_m})=\vec{0}$