Linear independence

We will work with row vectors in ${\mathbb{R}}^n$. Consider a set of $m$ vectors. If one of these vectors can be expressed as a linear combination of the other vectors, these vectors are said to be linearly dependent. If none of these vectors can be expressed as a linear combination of the other vectors, they are said to be linearly independent. We show that the determinant of a matrix is zero if its row vectors are linearly dependent and vice versa.

Definition (Linear dependence)

We say that a finite sequence ${\mathbf{v}}_1,{\mathbf{v}}_2,\cdots ,{\mathbf{v}}_m$ of vectors is linearly dependent if there are real numbers ${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_m$ which are not all 0 (i.e., some are non-zero), such that

$$\sum _{i=1}^m{\lambda }_i{\mathbf{v}}_i=0.$$

Why this terminology? Suppose ${\lambda }_1\ne 0$. Then we can rearrange the above equation into

$${\mathbf{v}}_1=-(1/{\lambda }_1)({\lambda }_2{\mathbf{v}}_2+\cdots +{\lambda }_m{\mathbf{v}}_m).$$

That is, ${\mathbf{v}}_1$ can be expressed as a linear combination of the other vectors. In this sense, the vector ${\mathbf{v}}_1$ depends on the other vectors.

Example. $(2,-1,0),(-3,2,1),$ and $(7,-4,-1)$ are linearly dependent because

$$2(2,-1,0)-1\cdot (-3,2,1)-1\cdot (7,-4,-1)=(0,0,0).$$ □

Definition (Linear independence)

If a finite sequence ${\mathbf{v}}_1,{\mathbf{v}}_2,\cdots ,{\mathbf{v}}_m$ of vectors is not linearly dependent, we say that it is linearly independent.

Remark. Therefore, ${\mathbf{v}}_1,{\mathbf{v}}_2,\cdots ,{\mathbf{v}}_m$ are linearly independent if and only if

$$\sum _{i=1}^m{\lambda }_i{\mathbf{v}}_i=0$$

implies $${\lambda }_1={\lambda }_2=\cdots ={\lambda }_m=0.$$ □

Example. ${\mathbf{e}}_1=(1,0)$ and ${\mathbf{e}}_2=(0,1)$ are linearly independent vectors. (Verify this!) □

Theorem

Suppose that ${\mathbf{v}}_1,{\mathbf{v}}_2,\cdots ,{\mathbf{v}}_m$ are linearly independent. If there is a vector $\mathbf{u}$ such that

$$\mathbf{u}=\sum _{i=1}^m{\beta }_i{\mathbf{v}}_i$$

for some ${\beta }_1,{\beta }_2,\cdots ,{\beta }_m$, then this representation is unique.

Remark. What this uniqueness means is the following. If $\mathbf{u}$ can be represented in terms of another linear combination of ${\mathbf{v}}_i$, say,

$$\mathbf{u}=\sum _{i=1}^m{\gamma }_i{\mathbf{v}}_i,$$

then, we have ${\beta }_1={\gamma }_1,{\beta }_2={\gamma }_2,\cdots ,{\beta }_m={\gamma }_m.$ □

Proof. Suppose there is an alternative representation

$$\mathbf{u}=\sum _{i=1}^m{\gamma }_i{\mathbf{v}}_i.$$

Then subtracting both sides, we have

$$\mathbf{0}=\mathbf{u}-\mathbf{u}=\sum _{i=1}^m({\beta }_i-{\gamma }_i){\mathbf{v}}_i.$$

Since ${\mathbf{v}}_1,{\mathbf{v}}_2,\cdots ,{\mathbf{v}}_m$ are linearly independent by assumption, it follows that ${\beta }_i-{\gamma }_i=0$, and hence ${\beta }_i={\gamma }_i$ for all $i=1,2,\cdots ,m$ as required. ■

Linear independence and matrix determinant

Theorem

Let $$A=\left(\begin{array}{c}{\mathbf{a}}_1\\ {\mathbf{a}}_2\\ ⋮\\ {\mathbf{a}}_n\end{array}\right)$$

be an $n\times n$ matrix where ${\mathbf{a}}_1,{\mathbf{a}}_2,\cdots ,{\mathbf{a}}_n$ are $n$-dimensional row vectors. If these row vectors are linearly dependent, then $detA=0$.

Proof. Since the row vectors are linearly dependent, at least one of them can be expressed as a linear combination of the other vectors. Without the loss of generality, assume ${\mathbf{a}}_n=\sum _{i=1}^{n-1}{\lambda }_i{\mathbf{a}}_i$ with not all ${\lambda }_1,\cdots ,{\lambda }_{n-1}$ being equal to 0.

Then, by the properties of determinants, we have $$\begin{array}{rcl}\left|\begin{array}{c}{\mathbf{a}}_1\\ {\mathbf{a}}_2\\ ⋮\\ {\mathbf{a}}_{n-1}\\ {\mathbf{a}}_n\end{array}\right|& =& \left|\begin{array}{c}{\mathbf{a}}_1\\ {\mathbf{a}}_2\\ ⋮\\ {\mathbf{a}}_{n-1}\\ {\lambda }_1{\mathbf{a}}_1+\cdots +{\lambda }_{n-1}{\mathbf{a}}_{n-1}\end{array}\right|\\ & =& \sum _{i=1}^{n-1}\left|\begin{array}{c}{\mathbf{a}}_1\\ {\mathbf{a}}_2\\ ⋮\\ {\mathbf{a}}_{n-1}\\ {\lambda }_i{\mathbf{a}}_i\end{array}\right|\\ & =& \sum _{i=1}^{n-1}{\lambda }_i\left|\begin{array}{c}{\mathbf{a}}_1\\ {\mathbf{a}}_2\\ ⋮\\ {\mathbf{a}}_{n-1}\\ {\mathbf{a}}_i\end{array}\right|.\end{array}$$

In the last sum, each determinant contains two rows of the same vectors (i.e., ${\mathbf{a}}_i$ is equal to one of ${\mathbf{a}}_1,\cdots ,{\mathbf{a}}_{n-1}$). Therefore, by the property of determinants, all the terms are 0. ■

See also: Review More on determinants for the properties of determinants.

Example. Consider the matrix determinant $$\left|\begin{array}{ccc}2& -1& 0\\ -3& 2& 1\\ 7& -4& -1\end{array}\right|.$$ We know that the row vectors are linearly dependent (see the example above), and hence, the determinant is zero. □

Corollary 

Let

$$A=\left(\begin{array}{c}{\mathbf{a}}_1\\ {\mathbf{a}}_2\\ ⋮\\ {\mathbf{a}}_n\end{array}\right)$$

be an $n\times n$ matrix where ${\mathbf{a}}_1,{\mathbf{a}}_2,\cdots ,{\mathbf{a}}_n$ are $n$-dimensional row vectors. If $detA\ne 0$, then these row vectors are linearly independent.

Proof. The contrapositive of the above theorem. ■

The converse is also true: If the row vectors are linearly independent, then $detA\ne 0$. However, proving this is beyond the scope of this module. See some textbooks on Linear Algebra (We need to define the notion of the rank of a matrix).