Matrix Representation of a Vector

Vectors as matrices. A vector with three rectangular components can equally well be written as a row matrix $[x\ \ y\ \ z]$ or a column matrix $\begin{pmatrix}x\\y\\z\end{pmatrix}$. Concretely, for $\vec A=a_1\hat i+a_2\hat j+a_3\hat k$: $\vec A\ \longleftrightarrow\ \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}.$

Why this matters. Vector addition and scalar multiplication then read exactly like matrix addition and scalar multiplication, component by component. If $\vec A=a_1\hat i+a_2\hat j+a_3\hat k$ and $\vec B=b_1\hat i+b_2\hat j+b_3\hat k$: $\vec A+\vec B=\begin{pmatrix}a_1+b_1\\a_2+b_2\\a_3+b_3\end{pmatrix}=(a_1+b_1)\hat i+(a_2+b_2)\hat j+(a_3+b_3)\hat k,$ and for a scalar $k$: $k\vec A=\begin{pmatrix}ka_1\\ka_2\\ka_3\end{pmatrix}=ka_1\hat i+ka_2\hat j+ka_3\hat k.$

Reading the effect of the scalar $k$. For $k>1$, $k\vec A$ is a magnification of $\vec A$; for $0<k<1$, it is a contraction; and $k=0$ collapses $\vec A$ to the zero vector $\vec 0=0\hat i+0\hat j+0\hat k$.

Component-wise properties (all provable from the vector-addition and scalar-multiplication laws already established). For $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$, $\vec b=b_1\hat i+b_2\hat j+b_3\hat k$, and scalar $m$:

1. $\vec a+\vec b=(a_1+b_1)\hat i+(a_2+b_2)\hat j+(a_3+b_3)\hat k$
2. $\vec a-\vec b=(a_1-b_1)\hat i+(a_2-b_2)\hat j+(a_3-b_3)\hat k$
3. $m\vec a=ma_1\hat i+ma_2\hat j+ma_3\hat k$ …