Linear independence

In order to allow the number of linearly independent vectors in a vector space to be countably infinite, it is useful to define linear dependence as follows. More generally, let V be a vector space over a field K, and let {v_i | i ∈ I} be a family of elements of V indexed by the set I. The family is linearly dependent over K if there exists a non-empty finite subset J ⊆ I and a family {a_j | j ∈ J} of elements of K, all non-zero, such that

${\displaystyle \sum _{j\in J}a_{j}v_{j}=0.}$

A set X of elements of V is linearly independent if the corresponding family {x}_x∈X is linearly independent. Equivalently, a family is dependent if a member is in the closure of the linear span of the rest of the family, i.e., a member is a linear combination of the rest of the family. The trivial case of the empty family must be regarded as linearly independent for theorems to apply.

A set of vectors which is linearly independent and spans some vector space, forms a basis for that vector space. For example, the vector space of all polynomials in x over the reals has the (infinite) subset {1, x, x², ...} as a basis.

Three vectors: Consider the set of vectors v₁ = (1, 1), v₂ = (−3, 2) and v₃ = (2, 4), then the condition for linear dependence seeks a set of non-zero scalars, such that

${\displaystyle a_{1}{\begin{Bmatrix}1\\1\end{Bmatrix}}+a_{2}{\begin{Bmatrix}-3\\2\end{Bmatrix}}+a_{3}{\begin{Bmatrix}2\\4\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}},}$

or

${\displaystyle {\begin{bmatrix}1&-3&2\\1&2&4\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\\a_{3}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}.}$

Row reduce this matrix equation by subtracting the first row from the second to obtain,

${\displaystyle {\begin{bmatrix}1&-3&2\\0&5&2\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\\a_{3}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}.}$

Continue the row reduction by (i) dividing the second row by 5, and then (ii) multiplying by 3 and adding to the first row, that is

${\displaystyle {\begin{bmatrix}1&0&16/5\\0&1&2/5\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\\a_{3}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}.}$

We can now rearrange this equation to obtain

${\displaystyle {\begin{bmatrix}1&0\\0&1\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\end{Bmatrix}}={\begin{Bmatrix}a_{1}\\a_{2}\end{Bmatrix}}=-a_{3}{\begin{Bmatrix}16/5\\2/5\end{Bmatrix}}.}$

which shows that non-zero a_i exist such that v₃ = (2, 4) can be defined in terms of v₁ = (1, 1), v₂ = (−3, 2). Thus, the three vectors are linearly dependent.

Two vectors: Now consider the linear dependence of the two vectors v₁ = (1, 1), v₂ = (−3, 2), and check,

${\displaystyle a_{1}{\begin{Bmatrix}1\\1\end{Bmatrix}}+a_{2}{\begin{Bmatrix}-3\\2\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}},}$

or

${\displaystyle {\begin{bmatrix}1&-3\\1&2\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}.}$

The same row reduction presented above yields,

${\displaystyle {\begin{bmatrix}1&0\\0&1\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}.}$

This shows that a_i = 0, which means that the vectors v₁ = (1, 1) and v₂ = (−3, 2) are linearly independent.

In order to determine if the three vectors in R⁴,

${\displaystyle \mathbf {v} _{1}={\begin{Bmatrix}1\\4\\2\\-3\end{Bmatrix}},\mathbf {v} _{2}={\begin{Bmatrix}7\\10\\-4\\-1\end{Bmatrix}},\mathbf {v} _{3}={\begin{Bmatrix}-2\\1\\5\\-4\end{Bmatrix}}.}$

are linearly dependent, form the matrix equation,

${\displaystyle {\begin{bmatrix}1&7&-2\\4&10&1\\2&-4&5\\-3&-1&-4\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\\a_{3}\end{Bmatrix}}={\begin{Bmatrix}0\\0\\0\\0\end{Bmatrix}}.}$

Row reduce this equation to obtain,

${\displaystyle {\begin{bmatrix}1&7&-2\\0&-18&9\\0&0&0\\0&0&0\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\\a_{3}\end{Bmatrix}}={\begin{Bmatrix}0\\0\\0\\0\end{Bmatrix}}.}$

Rearrange to solve for v₃ and obtain,

${\displaystyle {\begin{bmatrix}1&7\\0&-18&\end{bmatrix}}{\begin{Bmatrix}a_{1}\\a_{2}\end{Bmatrix}}=-a_{3}{\begin{Bmatrix}-2\\9\end{Bmatrix}}.}$

This equation is easily solved to define non-zero a_i,

${\displaystyle a_{1}=-3a_{3}/2,a_{2}=a_{3}/2,}$

where a₃ can be chosen arbitrarily. Thus, the vectors v₁, v₂ and v₃ are linearly dependent.

An alternative method relies on the fact that n vectors in ${\displaystyle \mathbb {R} ^{n}}$ are linearly independent if and only if the determinant of the matrix formed by taking the vectors as its columns is non-zero.

In this case, the matrix formed by the vectors is

${\displaystyle A={\begin{bmatrix}1&-3\\1&2\end{bmatrix}}.}$

We may write a linear combination of the columns as

${\displaystyle A\Lambda ={\begin{bmatrix}1&-3\\1&2\end{bmatrix}}{\begin{bmatrix}\lambda _{1}\\\lambda _{2}\end{bmatrix}}.}$

We are interested in whether AΛ = 0 for some nonzero vector Λ. This depends on the determinant of A, which is

${\displaystyle \det A=1\cdot 2-1\cdot (-3)=5\neq 0.}$

Since the determinant is non-zero, the vectors (1, 1) and (−3, 2) are linearly independent.

Otherwise, suppose we have m vectors of n coordinates, with m < n. Then A is an n×m matrix and Λ is a column vector with m entries, and we are again interested in AΛ = 0. As we saw previously, this is equivalent to a list of n equations. Consider the first m rows of A, the first m equations; any solution of the full list of equations must also be true of the reduced list. In fact, if 〈i₁,...,i_m〉 is any list of m rows, then the equation must be true for those rows.

${\displaystyle A_{{\langle i_{1},\dots ,i_{m}}\rangle }\Lambda =\mathbf {0} .}$

Furthermore, the reverse is true. That is, we can test whether the m vectors are linearly dependent by testing whether

${\displaystyle \det A_{{\langle i_{1},\dots ,i_{m}}\rangle }=0}$

for all possible lists of m rows. (In case m = n, this requires only one determinant, as above. If m > n, then it is a theorem that the vectors must be linearly dependent.) This fact is valuable for theory; in practical calculations more efficient methods are available.

If there are more vectors than dimensions, the vectors are linearly dependent. This is illustrated in the example above of three vectors in R².

Suppose that a₁, a₂, ..., a_n are elements of R such that

${\displaystyle a_{1}\mathbf {e} _{1}+a_{2}\mathbf {e} _{2}+\cdots +a_{n}\mathbf {e} _{n}=\mathbf {0} .}$

Since

${\displaystyle a_{1}\mathbf {e} _{1}+a_{2}\mathbf {e} _{2}+\cdots +a_{n}\mathbf {e} _{n}=(a_{1},a_{2},\ldots ,a_{n}),}$

then a_i = 0 for all i in {1, ..., n}.

Suppose ${\displaystyle a}$ and ${\displaystyle b}$ are two real numbers such that

${\displaystyle ae^{t}+be^{2t}=0}$

Take the first derivative of the above equation such that

${\displaystyle ae^{t}+2be^{2t}=0}$

for all values of t. We need to show that ${\displaystyle a=0}$ and ${\displaystyle b=0}$. In order to do this, we subtract the first equation from the second, giving ${\displaystyle be^{2t}=0}$. Since ${\displaystyle e^{2t}}$ is not zero for some t, ${\displaystyle b=0}$. It follows that ${\displaystyle a=0}$ too. Therefore, according to the definition of linear independence, ${\displaystyle e^{t}}$ and ${\displaystyle e^{2t}}$ are linearly independent.