How is linear dependence related to rank

[!Rank and Lin. Dep] Rank and Linear Independence A set is linearly independent iff the rank of the matrix with the vectors as its columns is equal to the number of vectors.

• If we try to solve the linear dependence equation, we get a homogenous system:

$$c_1{\vec{v}}_1+...+c_2{\vec{v}}_n=0$$

• We can create an augmented matrix to solve this system where the columns of the matrix are basically the vectors ${\vec{v}}_1,{\vec{v}}_2,...,{\vec{v}}_n$. The vector on the right side will be full of zeroes.

• Taking our matrix to RREF/REF and converting back to equations, we can see that each of $c_1,c_2,...,c_n$ is equal to zero (since the vector on the right is full of zeroes).

• The only way our coefficients are not zero is if one of them can be expressed as a linear combination of the others (i.e. one of the cols doesn’t have a pivot ⇒ rank is not equal to number of columns).