Linear Combinations of a Set of Vectors

# Linear Combinations of a Set of Vectors

 Definition: Let $V = \{ \mathbf{v_1}, \mathbf{v_2}, ..., \mathbf{v_m} \}$ be a set of $m$-vectors, and let $k_1, k_2, ..., k_m \in \mathbb{F}$. A Linear Combination of the set of vectors in $V$ is a vector in the form $\mathbf{u} = k_1\mathbf{v_1} + k_2\mathbf{v_2} + ... + k_n\mathbf{v_n}$ .

For example, consider the vectors $\mathbf{v_1} = (1, 2, 3)$ and $\mathbf{v_2} = (2, 5, 4)$. Any vector $\mathbf{u}$ can be called a linear combination of $v_1$ and $v_2$ if there exists scalars $k_1$ and $k_2$ such that $\mathbf{u} = k_1(1, 2, 3) + k_2(2, 5, 4)$.

For example, if we let $k_1 = 2$ and $k_2 = 3$, we obtain that $\mathbf{u} = 2(1, 2, 3) + 3(2, 5, 4) = (2, 4, 6) + (6, 15, 12) = (8, 19, 18)$. Therefore, the vector $(8, 19, 18)$ is a linear combination of $v_1$ and $v_2$.

Furthermore, recall the Standard Unit Vectors for $\mathbb{R}^3$ which are $\vec{i} = (1, 0, 0)$, $\vec{j} = (0, 1, 0)$ and $\vec{k} = (0, 0, 1)$. Recall that we could write any vector in $\mathbb{R}^3$ from these vectors, for example, $(2, 3, 4) = 2\vec{i} + 3\vec{j} + 4\vec{k}$. We can therefore say that any vector in $\mathbf{R}^3$ is a linear combination of the standard unit vectors for $\mathbb{R}^3$.

We will now look at an example to show that a specific vector is a linear combination of other vectors.

## Example 1

Suppose that $\mathbf{v_1} = (1, 4, 9)$ and $\mathbf{v_2} = (2, 3, 5)$. Show whether or not the vector $\mathbf{u} = (1, 2, 3)$ is a linear combination of $\mathbf{v_1}$ and $\mathbf{v_2}$.

We note that if $\mathbf{u}$ is a linear combination of $\mathbf{v_1}$ and $\mathbf{v_2}$, then there exists scalars $k_1$ and $k_2$ such that $\mathbf{u} = k_1\mathbf{v_1} + k_2\mathbf{v_1} = k_1(1, 4, 9) + k_2(2, 3, 5) = (1, 2, 3)$. We thus get the following system of linear equations:

(1)
\begin{align} k_1 + 2k_2 = 1 \\ 4k_1 + 3k_2 = 2 \\ 9k_1 + 5k_2 = 3 \end{align}

If we attempt to solve this system, we see that there exists no solution for $k_1$ and $k_2$ and therefore, $\mathbf{u}$ is not a linear combination of $\mathbf{v_1}$ and $\mathbf{v_2}$. The diagram below illustrates the three lines in the system and how there does not exist a point where all three lines intersect each other. ## Example 2

Suppose that $\mathbf{v_1} = (2, 3, 5)$ and $\mathbf{v_2} = (8, 13, 21)$. Show whether or not the vector $\mathbf{u} = (1, 1, 2)$ is a linear combination of $\mathbf{v_1}$ and $\mathbf{v_2}$.

Remember that if $\mathbf{u}$ is a linear combination of the set of vectors $V = \{ \mathbf{v_1}, \mathbf{v_2} \}$, then for $k_1, k_2 \in \mathbb{F}$:

(2)
\begin{equation} k_1(2, 3, 5) + k_2(8, 13, 21) = (1, 1, 2) \end{equation}

Which is equivalent to the following system of linear equations:

(3)
\begin{align} 2k_1 + 8k_2 = 1 \\ 3k_1 + 13k_2 = 1 \\ 5k_1 + 21k_2 = 2 \end{align}

In this example we have an exact solution, namely $k_1 = 2.5$ and $k_2 = -0.5$ and so $\mathbf{u}$ is a linear combination of $\mathbf{v_1}$ and $\mathbf{v_2}$.