1.4.  Linear Dependence and Linear Independence

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Definition 1.4.1. Let \(V\) be a vector space over \(F\), and let \(S \subseteq V\). Then, \(S\) is linearly dependent if and only if for \(n\in \mathbb N^{+}\) there exist distinct vectors \(v_1,\dots,v_n\in S\) and \(a_1,\dots,a_n\in F\), not all zero, such that

\[a_1v_1 + a_2v_2 + \cdots + a_nv_n = 0_V.\]

In this case, \(a_1v_1 + \cdots + a_nv_n\) is called a nontrivial representation of \(0_V\) as a linear combination of vectors in \(S\). If \(S\) is not linearly dependent, then \(S\) is linearly independent.

Proposition 1.4.2. Let \(V\) be a vector space over \(F\), and let \(S \subseteq V\).

1. \(S\) is linearly dependent if \(0_V\in S\).
2. \(\varnothing\) is linearly independent.
3. \(S\) is linearly independent if it consists of a single nonzero vector.
4. \(S\) is linearly independent if and only if every representation of \(0_V\) as a linear combination of its vectors is a trivial representation.

Proof. (1) Since \(S\) has a nontrivial representation of \(0_V\), namely \(1\cdot 0_V\), \(S\) is linearly dependent. (2) Since \(\varnothing\) has no vectors, it does not have nontrivial representation of \(0_V\). (3) Let \(S=\lbrace v\rbrace\). If \(S\) is linearly dependent, then \(av=0_V\) for some nonzero \(a\in F\). It follows that \(v = a^{-1}(av) = a^{-1}0_V=0_V\), which contradicts the fact that \(v\) is nonzero. (4) If \(S\) is linearly independent, then there are no trivial representations of \(0_V\). If \(S\) has only trivial representations of \(0_V\), then it is linearly independent.\(\square\)

Example 1.4.3. We consider the vector space \(\mathbb R^4\) over \(\mathbb R\) and the subset \(S\) of \(\mathbb R^4\) defined by

\[S = \lbrace (1,0,0,-1), (0,1,0,-1), (0,0,1,-1), (0,0,0,1)\rbrace.\]

We show that \(S\) is linearly independent. Indeed, suppose that

\[a(1,0,0,-1)+b (0,1,0,-1)+c (0,0,1,-1)+d (0,0,0,1) = (0,0,0,0)\]

where \(a,b,c,d\in \mathbb R.\) Then \(a=b=c=d=0\). Hence, by (4) of Proposition 1.4.2, \(S\) is linearly independent.

Example 1.4.4. Let \(n\in \mathbb N\). For \(k\le n\), define

\[p_k(x) = x^n + x^{n-1} + \cdots + x^k.\]

Then the set

\[S = \lbrace p_0(x),p_1(x),\dots,p_n(x)\rbrace\]

is linearly independent in \(F_n[x]\). Indeed, suppose that

\[a_0p_0(x) + a_1p_1(x) + \cdots + a_np_n(x) = \mathbf 0\]

for some \(a_0,\dots,a_n\in F\). Then

\[(a_0 + \cdots + a_n)x^n + (a_0 + \cdots + a_{n-1})x^{n-1} + \cdots + (a_0 + a_1)x + a_0 = \mathbf 0.\]

It follows that \(a_0 = \cdots =a_n = 0\). Hence, by (4) of Proposition 1.4.2, \(S\) is linearly independent.

References

1. Friedberg, S. H., Insel, A. J., & Spence, L. E. (2025). Linear algebra (5th ed.). Pearson Education South Asia Pte Ltd.

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