Given $n$ vectors $\vectorlist{u}{n}$ from $\complex{m}$ and $n$ scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_n$, their linear combination is the vector \begin{equation*} \lincombo{\alpha}{u}{n} \end{equation*}
So this definition takes an equal number of scalars and vectors, combines them using our two new operations (scalar multiplication and vector addition) and creates a single brand-new vector, of the same size as the original vectors. When a definition or theorem employs a linear combination, think about the nature of the objects that go into its creation (lists of scalars and vectors), and the type of object that results (a single vector). Computationally, a linear combination is pretty easy.
Our next two examples are key ones, and a discussion about decompositions is timely. Have a look at Proof Technique DC before studying the next two examples.
With any discussion of Archetype A or Archetype B we should be sure to contrast with the other.
There is a lot going on in the last two examples. Come back to them in a while and make some connections with the intervening material. For now, we will summarize and explain some of this behavior with a theorem.
Denote the columns of the $m\times n$ matrix $A$ as the vectors $\vectorlist{A}{n}$. Then $\vect{x}\in\complex{n}$ is a solution to the linear system of equations $\linearsystem{A}{\vect{b}}$ if and only if $\vect{b}$ equals the linear combination of the columns of $A$ formed with the entries of $\vect{x}$, \begin{equation*} \vectorentry{\vect{x}}{1}\vect{A}_1+ \vectorentry{\vect{x}}{2}\vect{A}_2+ \vectorentry{\vect{x}}{3}\vect{A}_3+ \cdots+ \vectorentry{\vect{x}}{n}\vect{A}_n = \vect{b} \end{equation*}
In other words, this theorem tells us that solutions to systems of equations are linear combinations of the $n$ column vectors of the coefficient matrix ($\vect{A}_j$) which yield the constant vector $\vect{b}$. Or said another way, a solution to a system of equations $\linearsystem{A}{\vect{b}}$ is an answer to the question “How can I form the vector $\vect{b}$ as a linear combination of the columns of $A$?” Look through the archetypes that are systems of equations and examine a few of the advertised solutions. In each case use the solution to form a linear combination of the columns of the coefficient matrix and verify that the result equals the constant vector (see Exercise LC.C21).