Linear Independence, Basis and Dimension

$\hat{i}=(1,0),\hat{j}=(0,1)\in {\mathbb{R}}^2$

Any vector in this space can be uniquely expressed as a linear combination of these basis vectors: $(a,b)=a(1,0)+b(0,1)$.

Linear Independence

Q: Can any vector be removed from a set without affecting its span? (i.e. a “redundant” vector)

If a vector can be removed from a set without affecting span, the vector is expressible as a linear combination.

Let $S=\{v_1,v_2,v_3,\dots ,v_n\}\in V$. If removing any vector results in a change in span, the set is linearly independent. The set $S$ is said to be linearly dependent if there exist scalars $c_1,c_2,c_3,\dots ,c_n$ not all zero, s.t.:

$c_1{v}_1+c_2{v}_2+\cdots +c_n{v}_n=0$

Eg:

$k_1(1,2,3)+k_2(4,5,6)+k_3(7,8,9)=(0,0,0)$

$\{\begin{array}{l}{k}_1+4k_2+7k_3=0\\ 2k_1+5k_2+8k_3=0\\ 3k_1+6k_2+9k_3=0\end{array}⟹(k_1,k_2,k_3)=(t,-2t,t)\text{for any }t\in \mathbb{R}$

Row 1 + Row 3 = 2 × Row 2

$\therefore (k_1,k_2,k_3)=(t,-2t,t)$

Geometrically, think of line through origin in ${\mathbb{R}}^3$.

Basis

A basis is a minimal set of vectors that span a vector space.

$S=\{v_1,v_2,\dots ,v_n\}\in V$ is a basis of $V$ if $S$ spans $V$ and is linearly independent.

$(7,8,9)$ as a linear combination of $(1,2,3)$ and $(4,5,6)$:

$a(1,2,3)+b(4,5,6)=(7,8,9)$

$\{\begin{array}{l}a+4b=7\\ 2a+5b=8\\ 3a+6b=9\end{array}\begin{array}{rl}a& =7-4b\\ 2(7-4b)+5b& =8\\ 14-3b& =8\\ b=2,a& =-1\end{array}$

$3(-1)+6(2)=-3+12=9✓$

$(7,8,9)=-1(1,2,3)+2(4,5,6)=(-1+8,-2+10,-3+12)=(7,8,9)✓$

• If $S$ is a basis of a vector space $V$, then every vector $w\in V$ has a unique representation as a linear combination of $S$.

Let $w\in V$ and $S=\{v_1,v_2,v_3,\dots ,v_n\}$. Since $S$ spans $V$, $w$ can be expressed as

$w=k_1{v}_1+k_2{v}_2+\cdots +k_n{v}_n$

Assume another representation exists:

$w=m_1{v}_1+m_2{v}_2+\cdots +m_n{v}_n$

Subtracting gives

$0=(k_1-m_1)v_1+\cdots +(k_n-m_n)v_n$

Since $S$ is linearly independent:

$k_1-m_1=k_2-m_2=\cdots =k_n-m_n=0$

$\therefore k_i=m_i\text{ for all }i$

Show that $S=\{(1,0,0),(1,1,0),(1,1,1)\}$ is a basis of ${\mathbb{R}}^3$.

For a general vector $v=(a,b,c)\in {\mathbb{R}}^3$:

$(a,b,c)=k_1(1,0,0)+k_2(1,1,0)+k_3(1,1,1)$

where $k_1,k_2,k_3\in \mathbb{R}$. This leads to

$\{\begin{array}{l}{k}_1+k_2+k_3=a\\ k_2+k_3=b\\ k_3=c\end{array}⟹\begin{array}{rl}{k}_1& =a-b\\ k_2& =b-c\\ k_3& =c\end{array}$

$\therefore k_1(1,0,0)+k_2(1,1,0)+k_3(1,1,1)=(0,0,0)$

is $k_1=k_2=k_3=0$, $\therefore$ proving linear independence. Hence, $S$ is a basis. $■$