2.8 KiB
Lecture 7
Chapter II Finite Dimensional Subspaces
Dimension 2C
Intuition: \mathbb{R}^2 is two dimensional. \mathbb{R}^n is n dimensional.
Definition 2.35
The dimension of a finite dimensional vector space denoted dim(V) is the length of any basis of V.
Potential issue:
- Why does it not matter which basis I take...
Theorem 2.34
Any two basis of a finite dimensional vector spaces have the same length.
Proof:
Let V be a finite dimensional vector space, and let B_1,B_2 (list of vectors) be two basis of V. B_1 is linearly independent and B_2 spans V, so by (Theorem 2.22) the length of B_1 is less than or equal to B_2, By symmetry the length of B_2 is less than or equal to the length of B_1 so length of B_1 = length of B_2.
Examples:
dim\{\mathbb{F}^2\}=2 because (0,1),(1,0) forms a basis
dim\{\mathscr{P}_m\}=m+1 because z^0,...,z^m forms a basis
dim_{\mathbb{C}}\{\mathbb{C}\}=1 as a \mathbb{C} vector space, because 1 forms a basis
dim_{\mathbb{R}}\{\mathbb{C}\}=2 as a \mathbb{R} vector space, because 1,i forms a basis
Proposition 2.37
If a vector space is finite dimensional, then every linearly independent list of length dim\{V\} is a basis.
Proposition 2.42
If a vector space is finite dimensional, then every spanning list of length dim\{V\} is a basis for V.
Sketch of Proof:
If it's not a basis, extend reduce to a basis, but then that contradicts with Theorem 2.34
Proposition 2.39
If U is a subspace of a finite dimensional vector space V and dim\{V\}=dim\{U\} then U=V
Proof:
Suppose u_1,...,u_n is basis for U, then it is linearly independent in V. but dim\{V\}=dim\{U\}, by Proposition 2.37, u_1,...,u_n is a basis of V.
So U=V.
Theorem 2.43
Let V_1 and V_2 be subspaces of a finite dimensional vector space V, then dim\{V_1+V_2\}=dim\{V_1\}+dim\{V_2\}-dim\{V_1\bigcap V_2\}
Proof:
Let u_1,...,u_m be a basis for V_1\bigcap V_2,
then extend by v_1,...,v_k,u_1,...,u_m to a basis of V_1,
then extend to u_1,...,u_m,w_1,..,w_l a basis of V_2.
Then I claim v_1,...,v_k,u_1,...,u_m,w_1,...,w_l is a basis of V_1+V_2
Note: given the above statement, we have dim\{V_1+V_2\}=k+m+l=(k+m)+(m+l)-m=dim\{V_1\}+dim\{V_2\}-dim\{V_1\bigcap V_2\}
So showing v_1,...,v_k,u_1,...,u_m,w_1,...,w_l is a basis suffices.
Since V_1,V_2\subseteq Span\{v_1,...,v_k,u_1,...,u_m,w_1,...,w_l\}, V_1+V_2\subseteq Span\{v_1,...,v_k,u_1,...,u_m,w_1,...,w_l\}.
Since V_1+V_2 is the smallest subspace contains both V_1,V_2, v_i,u_k,w_j\in V_1+V_2, V_1+V_2= Span\{v_1,...,v_k,u_1,...,u_m,w_1,...,w_l\}
So the list above spans V_1+V_2.
Suppose a_1 v_1+...+a_k v_k=-b_1 u_1-...-b_m u_m-c_1 w_1-...- c_e w_l\in V_1\bigcap V_2...