AcademySubspaces And Dimension
Academy
Linear Independence
Level 1 - Linear Algebra topic page in Subspaces And Dimension.
Principle
Vectors are linearly independent when none of them is redundant. Algebraically, the only way to make the zero vector as a linear combination is to use all zero coefficients.
Linear dependence means at least one vector can be built from the others.
Notation
\(\mathbf v_1,\ldots,\mathbf v_k\)
vectors being tested
\(c_1,\ldots,c_k\)
coefficients in a linear combination
\(\mathbf 0\)
the zero vector
\(A\)
matrix whose columns are the vectors being tested
\(A\mathbf c=\mathbf 0\)
homogeneous system for testing independence
The Core Test
The vectors \(\mathbf v_1, \ldots,\mathbf v_k\) are linearly independent when
Independence test
\[c_1\mathbf v_1+\cdots+c_k\mathbf v_k=\mathbf 0\Longrightarrow c_1=\cdots=c_k=0\]
They are linearly dependent when there is a non-zero coefficient vector \((c_1,\ldots,c_k)\) such that
Dependence test
\[c_1\mathbf v_1+\cdots+c_k\mathbf v_k=\mathbf 0\]
with at least one coefficient non-zero.
Put the vectors as columns of a matrix \(A\). Then independence is equivalent to the homogeneous system \(A\mathbf c=\mathbf 0\) having only the zero solution.
Worked Cases
Question
Are \(\mathbf v_1=(1,2)\) and \(\mathbf v_2=(3,6)\) linearly independent?
Answer
Notice that \(\mathbf v_2=3\mathbf v_1\). Therefore \(-3\mathbf v_1+\mathbf v_2=\mathbf0\). The coefficients \(-3\) and \(1\) are not both zero, so there is a non-trivial linear combination giving the zero vector. The vectors are linearly dependent.
Examples
Question
Can four vectors in \(\mathbb R^3\) be linearly independent?
Answer
No. At most three vectors can be linearly independent in \(\mathbb R^3\). Four vectors in \(\mathbb R^3\) must be linearly dependent because there are more vectors than available coordinate dimensions.