Dot Product and Span Summary

Dot Product

\text{span}\{\mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k}\}=\{c_1\mathbf{u_1}+c_2\mathbf{u_2}+\dots+c_k\mathbf{u_k}\mid c_1,c_2,\dots,c_k\in\mathbb{R}\}=\text{set of all linear combinations of } \{\mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k}\}.

V\subseteq\mathbb{R}^n is a subspace of \mathbb{R}^n if
1) V=\text{span}\{\mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k}\} for some vectors \mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k}.
2) V satisfies the closure properties:

(i) for all \mathbf{u},\mathbf{v}\in V, we must have \mathbf{u}+\mathbf{v}\in V.

(ii) for all \mathbf{u}\in V and c\in\mathbb{R}, we must have c\mathbf{u}\in V.

3) V is the solution set of a homogeneous system.

(Sufficient to check either one of Condition 1, 2, 3.)

For V to be a subspace, zero vector \mathbf{0} must be in V. (Since for \mathbf{u}\in V, 0\in\mathbb{R}, we have 0\mathbf{u}\in V.)

Linear Independence and Dependence
\mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k} are linearly independent if the system \displaystyle c_1\mathbf{u_1}+c_2\mathbf{u_2}+\dots+c_k\mathbf{u_k}=0 has only the trivial solution, i.e. c_1=c_2=\dots=c_k=0.

If the system has non-trivial solutions, i.e. at least one c_i not zero, then \mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k} are linearly dependent.

Author: mathtuition88

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