4.2-basis_and_dimension.ipynb

Kernel: SageMath 8.0

Basis and Dimension Round 2

Definition: A set $B=\{u_1,\ldots,u_m\}$ is a basis for a subspace $S$ if

• $B$ spans $S$,

• $B$ is linearly independent

Theorem: Let $B=\{u_1,\ldots,u_m\}$ be a basis for a subspace $S$. Then every $s\in S$ can be written as a linear combination of $u_1,\ldots,u_m$ in a unique way.

Example: We'll use the same subspace as last time. Let $S\subseteq \mathbb{R}^4$ be the subspace spanned by $u_1=(-1,2,3,1), u_2=(-6,7,5,2), u_3=(4,-3,1,0)$. From last class, we determined that a basis for $S$ is given by $\{v_1,v_2\}$ where $v_1=(-1,2,3,1),v_2=(0,5,13,4)$. It is clear that $u_3\in S$. How do we express $u_3$ as a linear combination of $v_1,v_2$? This amounts to solving $[v_1,\; v_2 | u_3]$.

v1 = vector([-1,2,3,1])
v2 = vector([0,5,13,4])
u3 = vector([4,-3,1,0])
A = matrix([v1,v2]).transpose()
A, u3

(
[-1  0]               
[ 2  5]               
[ 3 13]               
[ 1  4], (4, -3, 1, 0)
)

A \ u3

(-4, 1)

Most sets of $n$ vectors in $\mathbb{R}^n$ are a basis.

Example: Take $S$ as before with basis $B=\{v_1,v_2\}$. How can we extend $B$ to be a basis for $\mathbb{R}^n$?

• Eyeball it

• Add 2 random vectors

In this case, we see that $\{v_1,v_2,e_1,e_2\}$ will form a basis for $\mathbb{R}^4$. But so will $\{v_1,v_2,(-213,\pi,4,2),(4,\pi^2,3,4)\}$.