Linear Algebra: Linear Combinations

Linear Combinations


A linear combination of a list of vector $v_1, \ldots, v_n$ in a vector space $V$ over a field $F$ is a vector of the form

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

where each $a_i \in F$.

In plain English, a linear combination of vectors is just some weighted sum of those vectors. For example, if $u=\langle 2, -1 \rangle$ and $v = \langle 0, 4 \rangle$, then $w = \langle 4, 2 \rangle$ is a linear combination of $u$ and $v$. Specifically, $w = 2u + v$.


Problems

  1. Let $u, v, w, q \in \mathbb{R}^4$, where $u = \langle 1, 0, 0, 0 \rangle$, $v = \langle 0, 1,  0, 1 \rangle$, $w = \langle 0, 0, 1, 0 \rangle$, and $q = \langle 4, 2, 0, -2 \rangle$. 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 \\ \langle 4, -2, 0, -2 \rangle = a_1\langle 1, 0, 0, 0 \rangle + a_2\langle 0, 1, 0, 1 \rangle + a_3\langle 0, 0, 1, 0 \rangle \\ \langle 4, -2, 0, -2 \rangle = \langle a_1, 0, 0, 0 \rangle + \langle 0, a_2, 0, a_2 \rangle + \langle 0, 0, a_3, 0 \rangle \\ \langle 4, -2, 0, -2 \rangle = \langle a_1, a_2, a_3, a_2 \rangle \\ $

    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 \\ $

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  2. For the following vectors in $\mathbb{R}^2$, determine whether $\langle 4, 3 \rangle$ is a linear combination of $\langle 2, 1 \rangle$, $\langle 0, -1\rangle$.

    $ \langle 4, 3 \rangle = a_1\langle 2, 1 \rangle + a_2\langle 0, -1 \rangle \\ \langle 4, 3 \rangle = \langle 2a_1, 1a_1 \rangle + \langle 0a_2, -1 a_2 \rangle \\ \langle 4, 3 \rangle = \langle 2a_1, a_1 \rangle + \langle 0, -a_2 \rangle \\ \langle 4, 3 \rangle = \langle 2a_1, a_1 - a_2 \rangle \\ $

    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:

    $\langle 4, 3 \rangle = 2\langle 2, 1\rangle - \langle 0, -1\rangle$

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  3. For the following vectors in $\mathbb{C}^3$ over $\mathbb{C}$, determine whether $\langle 2, i, 3 \rangle$ is a linear combination of $\langle 1, i, i \rangle$ and $\langle 1, 0, 1 + i\rangle$.

    $ \langle 2, i, 3 \rangle = a_1\langle 1, i, i \rangle + a_2\langle 1, 0, 1 + i \rangle \\ \langle 2, i, 3 \rangle = \langle a_1, ia_1, ia_1 \rangle + \langle a_2, 0, (1 + i)a_2 \rangle \\ \langle 2, i, 3 \rangle = \langle a_1 + a_2, ia_1, ia_1 + (1 + i)a_2 \rangle \\ $

    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 $\langle 2, i, 3\rangle$ is not a linear combination of $\langle 1, i, i\rangle$ and $\langle 1, 0, 1 + i \rangle$

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