3.1 The Image of the Unit Square

A $2 times 2$ matrix with entries $a, b, c, d$ moves the point $(x, y)$ to $(a x + b y, c x + d y)$ . One point at a time, that is just arithmetic. The geometry appears when you move every point of a shape at once and look at what arrives. Our shape is the unit square: all points $(u, v)$ with $0 less than or equal to u less than or equal to 1$ and $0 less than or equal to v less than or equal to 1$ .

Follow the two special corners first. The corner $𝐞 sub 1 equals open parenthesis 1, 0 close parenthesis$ is sent to $(a, c)$ — the first column of the matrix. The corner $𝐞 sub 2 equals open parenthesis 0, 1 close parenthesis$ is sent to $(b, d)$ — the second column. And a general point of the square, $(u, v) = u 𝐞_{1} + v 𝐞_{2}$ , is sent to $u (a, c) + v (b, d)$ : the same combination of the columns that built the point from the corners. The figure below draws the arriving set live, with the four matrix entries on sliders.

A flat colored parallelogram lying in the xy-plane, seen from above: the image of the unit square under a 2-by-2 matrix. Its two edges from the origin are drawn as a red segment (the first matrix column) and a blue segment (the second column). Four sliders a, b, c, d set the matrix entries; dragging them shears, rotates, stretches, or flattens the parallelogram live. — The image of the unit square under the matrix with entries $a, b, c, d$ , drawn flat in the plane. The red edge is the image of $𝐞 sub 1$ — the column $(a, c)$ — and the blue edge is the image of $𝐞 sub 2$ — the column $(b, d)$ . The sliders start at the identity: $a equals 1$ , $b equals 0$ , $c equals 0$ , $d equals 1$ .

Explore the figure

At the identity defaults, the image is the unit square: red edge along the $x$ -axis, blue edge along the $y$ -axis. Confirm the two colored edges really are the columns $open parenthesis 1, 0 close parenthesis$ and $open parenthesis 0, 1 close parenthesis$ .
Drag $b$ to $0.5$ and stop. The red edge has not moved — the first column is untouched — but the blue edge tips over and the square shears into a parallelogram. Every slider controls one coordinate of one edge; convince yourself which.
Predict before you drag: with $a equals 1$ , $b equals 0.5$ , $c equals 0.5$ , $d equals 1.5$ , where will the far corner $open parenthesis 1, 1 close parenthesis$ land? It must arrive at the sum of the two columns. Set the sliders and check.
Now aim the two columns the same way: $a equals 1$ , $c equals 0.5$ , $b equals 2$ , $d equals 1$ , so the blue edge $open parenthesis 2, 1 close parenthesis$ is exactly twice the red edge $open parenthesis 1, 0.5 close parenthesis$ . What is left of the parallelogram, and why should the last chapter's word dependent come to mind?

The columns are the whole story

The computation behind everything you just saw is one line. Since $(u, v) = u 𝐞_{1} + v 𝐞_{2}$ , and the transformation respects adding and scaling,

$T (u, v) = u (a, c) + v (b, d)$

The image of the unit square is the set of all combinations $u (a, c) + v (b, d)$ with coefficients between $0$ and $1$ — a parallelogram whose edges from the origin are the two columns. Let the coefficients run over all real numbers instead and you get the image of the entire plane: the span of the columns, in exactly the sense of Chapter 2. When the columns are independent, that span is the whole plane and the square becomes an honest parallelogram; when they are dependent, as in your final exploration step, the span collapses to a line, the parallelogram collapses to a segment, and infinitely many different input points are forced to share each output point.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.