1.8 KiB
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}