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Linear Independence

Definition

Suppose you have a set of vectors {v1,v2,,vn}\{\vec v_1,\vec v_2,\dots,\vec v_n\}, and let c1,c2,,cnRc_1, c_2, \dots, c_n \in \reals.
Consider the equation: c1v1+c2v2++cnvn=0c_1\vec v_1 + c_2\vec v_2 + \dots + c_n\vec v_n = \vec 0.
If the only solution to this equation is c1=c2==cn=0c_1 = c_2 = \ldots = c_n = 0, then the set {v1,v2,,vn}\{\vec v_1,\vec v_2,\dots,\vec v_n\} is said to be linearly independent.
If there are any non-zero values cic_i that satisfy the equation, the set is linearly dependent.
Wize Tip
If any vector can be written as a linear combination of other vectors, then the set is linearly dependent.

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Testing Linear Independence

Asking if the set of vectors {v1,v2,,vn}\{\vec v_1,\vec v_2,\dots,\vec v_n\} is linearly independent is equivalent to solving a homogeneous linear system.
Taking the unknowns to be the coefficients c1,c2,,cnc_1,c_2,\dots,c_n, we can write the augmented matrix of the system:

[      v1v2vn0      ]\left[\begin{array}{rrrr|r} |\ \ &|\ \ & & |\ \ & | \\ \vec v_1&\vec v_2&\cdots&\vec v_n& \vec 0\\ |\ \ &|\ \ & & |\ \ & | \\ \end{array}\right]
The type of solution determines linear independence:
  • Only the trivial solution (c1=c2==cn=0c_1=c_2=\cdots=c_n=0)     \implies linearly independent
  • Non-trivial solutions     \implies linearly dependent
Shortcut: Determinant
The determinant can be used to determine the number of solutions to a homogeneous linear system.

Let A=[v1v2vn]A=\left[\begin{array}{rrrr} \\ \vec v_1&\vec v_2&\cdots&\vec v_n\\ \\ \end{array}\right].
  • det(A)0     \boxed{\text{det}(A)\textcolor{red}{\neq}0 } \ \implies AA is invertible     \impliestrivial solution only     \implies linearly independent
  • det(A)=0     \boxed{\text{det}(A)=0} \ \implies AA is not invertible     \implies infinitely many solutions     \implies linearly dependent
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Examples

Linearly Independent
In R2\mathbb{R}^2, the set {1,1,1,2}\{\lang 1,1\rang , \lang 1,2 \rang\} is linearly independent.
c1[11]+c2[12]=[00]    1c1+1c2=01c1+2c2=0\begin{array}{rcl} c_1 \begin{bmatrix} 1\\ 1\\ \end{bmatrix} +c_2 \begin{bmatrix} 1\\ 2\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix} &\implies& 1c_1&+&1c_2&=&0\\ &&1c_1&+&2c_2&=&0\end{array}
The only solution to this linear system is c1=c2=0c_1=c_2=0.
Check using the determinant:
1112=(1)(2)(1)(1)=10\left| \begin{array}{rrr} 1 & 1\\ 1 & 2\\ \end{array} \right| =(1)(2)-(1)(1)=1\textcolor{red}{\ne}0
Linearly Dependent
In R2\mathbb{R}^2, the set {1,2,3,6}\{\lang 1,2 \rang, \lang 3,6 \rang\} is linearly dependent.
c1[12]+c2[36]=[00]    1c1+3c2=02c1+6c2=0\begin{array}{rcl} c_1 \begin{bmatrix} 1\\ 2\\ \end{bmatrix} + c_2 \begin{bmatrix} 3\\ 6\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix} &\implies& 1c_1&+&3c_2&=&0\\ &&2c_1&+&6c_2&=&0 \end{array}
This linear system has non-zero solutions, such as c1=3, c2=1c_1=-3,\ c_2=1.
Moreover, we can see that one vector 3,6\lang 3,6 \rang is a linear combination of the other:3,6=31,2\lang 3,6 \rang = 3\cdot\lang 1,2 \rang.
Check using the determinant:
1326=(1)(6)(3)(2)=0\left| \begin{array}{rrr} 1 & 3\\ 2 & 6\\ \end{array} \right| =(1)(6)-(3)(2)=0
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Example: Linear Independence

Let u=[402]\vec u = \begin{bmatrix} 4\\ 0\\ -2\\ \end{bmatrix} , v=[137]\vec v= \begin{bmatrix} 1\\ 3\\ 7\\ \end{bmatrix}, and w=[335]\vec w = \begin{bmatrix} 3\\ 3\\ 5\\ \end{bmatrix}.
Determine if the set {u,v,w}\{\vec u, \vec v, \vec w\} is linearly independent.
Using the definition of linear independence, we are looking to find what values of c1,c2c_1, c_2, and c3c_3 are solutions to the equation:
c1u+c2v+c3w=0c_1\vec u + c_2\vec v + c_3\vec w = \vec 0
    c1[402]+c2[137]+c3[335]=[000]\implies c_1 \begin{bmatrix} 4\\ 0\\ -2\\ \end{bmatrix} + c_2 \begin{bmatrix} 1\\ 3\\ 7\\ \end{bmatrix} + c_3 \begin{bmatrix} 3\\ 3\\ 5\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}
    {4c1+c2+3c3=03c2+3c3=02c1+7c2+5c3=0\implies \left\{\begin{array}{rcl} 4c_1+c_2+3c_3&=&0\\[0.5em] 3c_2+3c_3&=&0\\[0.5em] -2c_1+7c_2+5c_3&=&0 \end{array}\right.
    [413003302750]\implies \left[\begin{array}{rrr|r}4&1&3&0\\0&3&3&0\\-2&7&5&0\end{array}\right]
The columns of this augmented matrix are the vectors u,v,\vec u, \vec v, and w\vec w.
We can use Gauss-Jordan elimination to find the RREF:[100001000010]\left[\begin{array}{rrr|r}1&0&0&0\\0&1&0&0\\0&0&1&0\end{array}\right]
We can then read of the solution directly: c1=c2=c3=0c_1=c_2=c_3=0.
Instead, let's simply find the determinant of the coefficient matrix:
413033275=0() + 3(4(5)3(2))  3(4(7)1(2))=3(26)3(30)=12  0\begin{array}{rcl} \left|\begin{array}{rrr}4&1&3\\\colorOne0&\colorOne3&\colorOne3\\-2&7&5\end{array}\right| &=& -\colorOne0(\dots) \ +\ \colorOne3\big(4(5)-3(-2)\big) \ -\ \colorOne3\big(4(7)-1(-2)\big)\\[1em] &=& 3(26)-3(30)\\[1em] &=&-12 \ \textcolor{red}{\neq} \ 0 \end{array}
Since the determinant is not 0, the system has only the trivial solution c1=c2=c3=0c_1=c_2=c_3=0.
Therefore, {u,v,w}\{\vec u, \vec v,\vec w\} is a linearly independent set.
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Example: Linear Independence

Let u=(1,2,1)\vec u = (1,2,1), v=(1,3,2)\vec v=(1,3,2), and w=(7,17,a2+1)\vec w = (7,17,a^2+1).
Determine the value(s) of aa that will make the set {u,v,w}\{\vec u, \vec v, \vec w\} linearly independent.
Let's find the determinant of the matrix whose columns are u,v,\vec u, \vec v, and w\vec w: [117231712a2+1]\left[ \begin{array}{rrc} 1 & 1 & 7\\ 2 & 3 & 17\\ 1 & 2 & a^2+1\\ \end{array} \right]
117231712a2+1=1(3(a2+1)17(2))1(2(a2+1)17(1))+7(2(2)3(1))=3a2+3342a22+17+7=a29\begin{array}{rcl} \left|\begin{array}{ccc} \colorOne1&\colorOne1&\colorOne7\\ 2&3&17\\ 1&2&a^2+1 \end{array}\right| &=& \colorOne1\Big(3(a^2+1)-17(2)\Big) -\colorOne1\Big(2(a^2+1)-17(1)\Big) +\colorOne7\Big(2(2)-3(1)\Big)\\[1em] &=&3a^2+3-34-2a^2-2+17+7\\[1em] &=&a^2-9 \end{array}
Since we want the vectors to be linearly independent, we must make sure the determinant is not 0:
a290    (a3)(a+3)0    a3anda3a^2-9\textcolor{red}{\neq}0 \implies (a-3)(a+3)\textcolor{red}{\neq}0 \implies a\textcolor{red}{\neq}3 \quad\text{and} \quad a\textcolor{red}{\neq}-3.
Therefore, as long as aa is not equal to 3 or -3, the vectors will be linearly independent.

Practice: Linear Independence

Let {v1,v2,v3}\{\vec v_1,\vec v_2,\vec v_3\} be a linearly independent set of vectors in a vector space VV.
Define the following vectors in VV, where kRk \in \reals:
w1=2v2+kv3w2=v1+kv2w3=kv14v3\vec w_1=2\vec v_2+k\vec v_3\qquad \vec w_2=\vec v_1+k\vec v_2 \qquad \vec w_3=k\vec v_1-4\vec v_3
Determine what conditions kk must satisfy in order for the set {w1,w2,w3}\{\vec w_1,\vec w_2,\vec w_3\} to be linearly independent.
Extra Practice