Basis and coordinates

A basis of the plane is an ordered pair of vectors $b_1, b_2$ that do not lie on one line. Every vector $x$ can then be written in exactly one way as $x = c_1b_1 + c_2b_2$, and the numbers $c_1, c_2$ are the coordinates of $x$ in the basis $B = (b_1, b_2)$, written $[x]_B$.

The left picture is the standard grid, with $b_1$ and $b_2$ as ink arrows, the slanted grid they make, and $x$ in blue. The right picture is the B-coordinate chart, where $b_1$ and $b_2$ are the sides of the unit square and $x$ sits at its coordinates. Drag $b_1$ or $b_2$ and $x$ does not move on the left. Only the grid and the numbers change, because coordinates are a reading taken with a ruler, not a property of the arrow.

To read the coordinates, push $x$ back to each axis of $B$ along the other basis vector. Moving parallel to $b_2$, it reaches the $b_1$ axis at $c_1b_1$. Moving parallel to $b_1$, it reaches the $b_2$ axis at $c_2b_2$. These are the two yellow segments, and with the dashed lines from $x$ they form a parallelogram whose diagonal is $x$. Because $b_1$ and $b_2$ are not on one line, each push meets its axis in exactly one point, so the coordinates exist and are unique.

Stretch $b_1$ to $k\,b_1$ with $k \ne 0$, and keep $b_2$. Since $x = (c_1/k)(k\,b_1) + c_2b_2$, the first coordinate becomes $c_1/k$ and the second stays the same: a ruler twice as long gives half the reading. Turning a basis vector follows no such rule, and both coordinates can change.

Put $b_1$ and $b_2$ side by side as the columns of a matrix $P$. Then $x = c_1b_1 + c_2b_2$ says $x = P\,[x]_B$: $P$ turns coordinates in $B$ into standard ones, and going the other way needs $P^{-1}$.

The B-coordinate chart is the whole plane moved by $P^{-1}$. It sends $b_1$ and $b_2$ to $(1, 0)$ and $(0, 1)$, so the grid of $B$ is square there and our standard grid can become slanted. The standard basis vectors $\hat\imath$ and $\hat\jmath$ land on the columns of $P^{-1}$. The chart is drawn in B-coordinates. It keeps all lengths and angles only when $b_1$ and $b_2$ are perpendicular unit vectors.

A tempting shortcut is to drop perpendiculars from $x$ to the lines of $b_1$ and $b_2$, and read where they land. Perpendicular feet draws this reading. It gives the coordinates when $b_1 \perp b_2$, and for perpendicular unit vectors each reading is simply the dot product $x\cdot b_i$. In a slanted basis it is wrong for every $x$ except 0.

Two vectors on one line are not a basis. Then $\det P = 0$ and the grid collapses onto that line, or onto O if both vectors are zero. If $x$ lies in the collapsed grid, it has endless readings: with $b_2 = 2b_1 = (2, 2)$ and $x = (3, 3)$, every $(c_1, c_2)$ with $c_1 + 2c_2 = 3$ works. If $x$ lies outside it, it has none.