AcademySubspaces And Dimension
Academy
Bases In Rn
Level 1 - Linear Algebra topic page in Subspaces And Dimension.
Principle
A basis is a minimal set of building blocks for a subspace. It spans the subspace and is linearly independent. Spanning ensures every vector can be built; independence ensures the building instructions are unique.
For \(\mathbb R^n\), a basis contains exactly \(n\) vectors.
Notation
\(B\)
a proposed basis
\(\mathbf v_1,\ldots,\mathbf v_k\)
vectors in the proposed basis
\(W\)
the subspace being described
\(\operatorname{span}(B)\)
the span of all vectors in B
\(\mathcal E\)
the standard basis, when used
The Core Test
A list \(B=(\mathbf v_1, \ldots,\mathbf v_k)\) is a basis for a subspace \(W\) when both conditions hold:
- \(\operatorname{span}(B)=W\).
- The vectors in \(B\) are linearly independent.
For \(\mathbb R^n\), if you have exactly \(n\) vectors, then either of these tests is enough:
Basis test in Rn
\[n\text{ vectors in }\mathbb R^n:\quad \text{independent}\Longleftrightarrow \text{span }\mathbb R^n\Longleftrightarrow \text{basis}\]
The standard basis of \(\mathbb R^n\) is
Standard basis
\[\mathcal E=(\mathbf e_1,\ldots,\mathbf e_n)\]
Worked Cases
Question
Is \(((1,2),(3,4))\) a basis of \(\mathbb R^2\)?
Answer
Place the vectors as columns: \(A=\begin{pmatrix}1&3\\2&4\end{pmatrix}\). Compute the determinant: \(\det A=1\cdot4-3\cdot2=4-6=-2\). Since the determinant is non-zero, the two columns are independent and span \(\mathbb R^2\). Therefore they form a basis of \(\mathbb R^2\).
Examples
Question
Write the standard basis of \(\mathbb R^4\).
Answer
The standard basis is \((\mathbf e_1,\mathbf e_2,\mathbf e_3,\mathbf e_4)\), where \(\mathbf e_1=(1,0,0,0)\), \(\mathbf e_2=(0,1,0,0)\), \(\mathbf e_3=(0,0,1,0)\), and \(\mathbf e_4=(0,0,0,1)\).