# Linear Algebra: Linear Combinations

## Linear Combinations

A linear combination of a set of vectors $v_1, \ldots, v_n$ is a weighted sum of those vectors. Mathematically, a linear combination is the sum

$$\sum\limits_{i=1}^{n} a_i v_i$$

where each $a_i \in F$.

For example, if $u=[2, -1]$ and $v = [0, 4]$, then $w = [4, 2]$ is a linear combination of $u$ and $v$. Specifically, $w = 2u + v$.

## Problems

1. Let $u, v, w, q \in \mathbb{R}^4$, where $u = [1, 0, 0, 0]$, $v = [0, 1, 0, 1]$, $w = [0, 0, 1, 0]$, and $q = [4, -2, 0, -2]$. Find a linear combination of $u$, $v$, and $w$ that equals $q$.

Solving for the coefficients is a matter of using a little vector algebra:

$q = a_1u + a_2v + a_3w \\ [4, -2, 0, -2] = a_1 [1, 0, 0, 0] + a_2 [0, 1, 0, 1] + a_3 [0, 0, 1, 0] \\ [4, -2, 0, -2] = [a_1, 0, 0, 0] + [0, a_2, 0, a_2] + [0, 0, a_3, 0] \\ [4, -2, 0, -2] = [a_1, a_2, a_3, a_2] \\$

Lining up the vector components shows us that $a_1 = 4$, $a_2 = -2$, and $a_3 = 0$. We can now write $q$ as a linear combination of $u$, $v$, and $w$:

$q = 4u - 2v + 0w \\$

2. Determine whether $[4, 3]$ is a linear combination of $[2, 1]$, $[0, -1]$, where all are vectors in $\mathbb{R^2}$.

$[4, 3] = a_1[2, 1] + a_2 [0, -1] \\ [4, 3] = [2a_1, 1a_1] + [0a_2, -1 a_2] \\ [4, 3] = [2a_1, a_1] + [0, -a_2] \\ [4, 3] = [2a_1, a_1 - a_2] \\$

We have two equations and two unknowns, so we can solve for the solution:

$4 = 2a_1 \\ 2 = a_1 \\$

$3 = a_1 - a_2 \\ 3 = 2 - a_2 \\ 1 = -a_2 \\ -1 = a_2 \\$

Looks like we have a linear combination on our hands:

$[4, 3] = 2[2, 1] - [0, -1]$

3. Determine whether $[2, i, 3]$ is a linear combination of $[1, i, i]$ and $[1, 0, 1 + i]$, where all three vectors are in $\mathbb{C}^3$.

$[2, i, 3] = a_1[1, i, i] + a_2[1, 0, 1 + i] \\ [2, i, 3] = [a_1, ia_1, ia_1] + [a_2, 0, (1 + i)a_2] \\ [2, i, 3] = [a_1 + a_2, ia_1, ia_1 + (1 + i)a_2] \\$

There are three equations and two unknowns, so there's a chance this might not work out so well:

$i = ia_1 \\ 1 = a_1$

$2 = a_1 + a_2 \\ 2 = 1 + a_2 \\ 1 = a_2 \\$

$3 = ia_1 + (1+i)a_2 \\ 3 = i1 + (1+i)1 \\ 3 = 1+2i\\$

Oops, looks like that last computation didn't check out. Looks like $[2, i, 3]$ is not a linear combination of $[1, i, i]$ and $[1, 0, 1 + i]$.