Basis

A Set of Basis Vectors. A set of Linearly Indepedent Vectors whose Linear Combinations can represent any vector in that space.

Standard Basis Vectors in ${\mathbb{R}}^3$ are i, j, k, but you can use other vectors to span the same amount of space if you want.

• Let $\mathcal{B}\in H$
• $\mathcal{B}=\{\overrightarrow{b_1},...,\overrightarrow{b_n}\}$
• $\mathcal{B}$ is some basis for the subspace $H$

$$\begin{array}{c}\vec{x}\in H⟹\\ \text{coords of x relative to B are c1,...,cn}\vec{x}=c_1\overrightarrow{b_1}+...+c_n\overrightarrow{b_n}\\ \\ \wedge \\ \text{coord vector of }\vec{x}\text{ relative to }\mathcal{B}[\vec{x}{]}_{\mathcal{B}}=\left[\begin{array}{c}{c}_1\\ ...\\ c_n\end{array}\right]\end{array}$$

Dimension

The dimension of a Subspace $H$ is the Cardinality of its Basis $\mathcal{B}$

$$\dim H=|\mathcal{B}|$$

$\dim \{\vec{0}\}$

$$\dim \vec{0}=0$$

This is because a basis requires a Set of Linearly Indepedent Vectors, and $\vec{0}$ does not meet this requirement. Therefore, its only basis is $\varnothing$. The cardinality of the empty set is 0, therefore so is $\dim \vec{0}$

Examples

• dim ${\mathbb{R}}^n$
  • $n$
• dim(Nul A) is the number of
  • number of free vars of A
• dim(Col A) is the number of
  • number of pivots of A
• H = $\{(x_1....,x_n):x_1+...+x_n=0\}$
  • n variables, solve for $x_1$ ito everything else. → one pivot everything else free vars. Therefore $n-1$ free vars

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Basis Theorem

Any two bases for a Subspace have the same Dimension

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