Question

If two non-zero vectors are linearly dependent, then each of them is a scalar multiple of the other.

Short answer

The above statement is true.

If two non-zero vectors are linearly dependent, then each of them is a scalar multiple of the other.

Step-by-step solution

Definition of linearly dependent vectors

Let V be a vector space over a field K. Let$v_1,v_2,v_3,...,v_n\in V$ , then if there exist scalars $a_1,a_2,a_3,...,a_n\in K$, not all of them 0, such that

$a_1{v}_1+a_2{v}_{}+...a_n{v}_n=0$

Then the vectors $v_1,v_2,v_3,...,v_n$are called linearly dependent vectors.

Otherwise, linearly independent vectors.

To prove the given statement

Let$\overrightarrow{u},\overrightarrow{v}\in V$ are two non-zero vectors that are linearly dependent. Then, we have two scalars $a,b\in K$, not all of them 0, such that

$a\overrightarrow{u}+b\overrightarrow{v}=0$

Since both scalars$a,b\in K$ cannot be zero as vectors$\overrightarrow{u},\overrightarrow{v}\in V$ are linearly dependent, so let us assume$a\ne 0.$

$$\begin{gathered} \Rightarrow a\overrightarrow{u}=-b\overrightarrow{v} \\ \Rightarrow \overrightarrow{u}=-\frac{b}{a}\overrightarrow{v} \end{gathered}$$

This shows that$\overrightarrow{u}$ is a scalar multiple of$\overrightarrow{v}.$

Final Answer

Since $\overrightarrow{u},\overrightarrow{v}\in V$are two non-zero vectors that are linearly dependent, then we can write

$a\overrightarrow{u}+b\overrightarrow{v}=0$

role="math" localid="1664192943751" $$\begin{gathered} \Rightarrow a\overrightarrow{u}=-b\overrightarrow{v} \\ \Rightarrow \overrightarrow{u}=-\frac{b}{a}\overrightarrow{v} \end{gathered}$$

Thus, if two non-zero vectors are linearly dependent, then each of them is a scalar multiple of the other.