How to Determine if Vectors are Linearly Dependent or Independent?
To ascertain whether vectors are linearly dependent or independent, I check if there exists a linear combination of vectors equal to zero.
$$ a_1 v_1 + ... + a_n v_n = 0 $$
If such a combination exists, the vectors are linearly dependent.
And if it doesn't? If the problem has no solution, then the vectors are linearly independent. If the only possible combination to achieve a1 v1 +...+ an vn = 0 is the trivial one (all scalars equal to zero), then the vectors are linearly independent.
Practical Exercise
Example 1
In the vector space R4 within the field K=R, I have a vector subspace W defined by a set of generators consisting of three vectors.
$$W = L_R (w_1, w_2, w_3) \:\:\:where \begin{cases} w_1 = (1,1,0,0) \\ w_2 = (1,2,0,1) \\ w_3= (0,1,0,1) \end{cases} $$
I check the zero linear combination of a generic vector v in W
$$ v = a_1 w_1 + a_2 w_2 + a_3 w_3 = 0 $$
$$ a_1 \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + a_2 \begin{pmatrix} 1 \\ 2 \\ 0 \\ 1 \end{pmatrix} + a_3 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \end{pmatrix} = 0 $$
$$ \begin{pmatrix} a_1 \\ a_1 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} a_2 \\ 2a_2 \\ 0 \\ a_2 \end{pmatrix} + \begin{pmatrix} 0 \\ a_3 \\ 0 \\ a_3 \end{pmatrix} = 0 $$
$$ \begin{pmatrix} a_1 + a_2 \\ a_1 + 2a_2 + a_3 \\ 0 \\ a_2+ a_3 \end{pmatrix} = 0 $$
I ignore the equation 0=0 in the system as it's a mere identity and doesn't influence the solution.
Rewriting the vector equation as a system of equations.
$$ \begin{cases} a_1+a_2=0 \\ a_1+2a_2+a_3=0 \\ a_2+a_3=0 \end{cases} $$
Then, I determine if the system has a solution using Cramer's rule.
$$ det \begin{pmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{pmatrix} = 0 $$
The determinant of the coefficient matrix equals zero.
Hence, the system has no solutions.
Since it's impossible to find a solution to the system, I conclude that the vectors are linearly dependent.
Note. The same conclusion could have been reached much more swiftly with an analysis of the rank of the system's coefficient matrix. $$ \begin{pmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix} $$ The highest order complementary minor with a nonzero determinant is of order two, while the vector space equals n=4. $$ det \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \ne 0 $$ $$ det \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \ne 0 $$ $$ det \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \ne 0 $$ Therefore, the vectors are linearly independent in pairs (e.g., w1 and w2) but are linearly dependent all together since the rank is lower than the dimension of the vector space.
