3.2 Morphing the Identity

The previous section treated a matrix as a destination: the square is here, its image is there. But a matrix is also somewhere you can travel gradually. Fix the matrix $A$ that sends $(u, v)$ to $(u + 0.8v, \; 0.6u + 0.5v)$ — columns $(1, 0.6)$ and $(0.8, 0.5)$ — and blend it with the identity map using a dial $s$ :

$T_s(u, v) = (1 - s)(u, v) + s\,(u + 0.8v, \; 0.6u + 0.5v)$

At $s = 0$ this is the identity, leaving every point in place. At $s = 1$ it is $A$ itself. In between it is a perfectly legitimate linear transformation in its own right — multiplying out gives $T_s(u, v) = (u + 0.8s\,v, \; 0.6s\,u + (1 - 0.5s)v)$ , the matrix with columns $(1, 0.6s)$ and $(0.8s, 1 - 0.5s)$ . The figure animates the whole family.

A colored square with a pale amber outline, and a light gray outline of the unit square, seen face-on in one upright plane against a gray horizontal axis and a gray vertical axis with numbered ticks. The gray outline never moves; the colored square is controlled by a slider named s. With s at 0 the colored square exactly fills the gray outline; dragging s toward 1 continuously shears and tilts it into the thin parallelogram that the matrix A produces, its far corner travelling in a straight line, so the reader watches the identity deform into A. — The unit square carried by the interpolated map $T_s$ , seen face-on, with a gray outline of the unit square that never moves drawn in front of it. At $s = 0$ the colored square exactly fills the gray outline; dragging $s$ to $1$ deforms it continuously into the image parallelogram of $A$ .

Explore the figure

At $s = 0$ , confirm the colored square coincides with the gray reference outline in front of it.
Drag $s$ slowly from $0$ to $1$ and watch one landmark: the far corner starts at $(1, 1)$ and arrives at the sum of the columns of $A$ , which is $(1.8, 1.1)$ . Does it travel along a curve or a straight line?
Watch the shape itself near the end of the slide. The parallelogram gets thin — by $s = 1$ its two edges, the columns $(1, 0.6)$ and $(0.8, 0.5)$ , point almost the same way. Compare $(0.8, 0.5)$ with $0.8 \cdot (1, 0.6) = (0.8, 0.48)$ .
Run $s$ back and forth across the last stretch of the slider. Does the square ever collapse to a segment completely, the way the dependent columns did in the previous section — or does it stop just short?

Straight-line motion, and a near collapse

Why straight lines? For a fixed point $\mathbf{p}$ , its position at dial value $s$ is $(1 - s)\mathbf{p} + s\,A\mathbf{p}$ — a weighted average of the start $\mathbf{p}$ and the destination $A\mathbf{p}$ . As $s$ runs from $0$ to $1$ , that weighted average slides along the segment joining the two, at constant speed. The whole square deforms coherently because every one of its points is running its own straight errand at once.

The thinning is quantitative. For a parallelogram with edge vectors $(x_1, y_1)$ and $(x_2, y_2)$ , the area is $|x_1 y_2 - x_2 y_1|$ , which for a matrix with columns $(a, c)$ and $(b, d)$ reads $|ad - bc|$ — and this book has that formula already, from Section 0.8 rather than from geometry in general. Lay the two edges in the floor of space as $(x_1, y_1, 0)$ and $(x_2, y_2, 0)$ ; their cross product is $(0, 0, x_1 y_2 - x_2 y_1)$ , standing straight up, and Section 0.8's result that a cross product's length is the area of the parallelogram spanned is exactly the claim being made here. Applied to the columns of $T_s$ , the quantity inside the absolute value is

$(1)(1 - 0.5s) - (0.8s)(0.6s) = 1 - 0.5s - 0.48s^2$

1 - \frac{1}{2} \cdot s - \frac{12}{25} \cdot s^{2}

Area along the morph ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The area quantity for the whole family at once: a computer algebra system expands the cross-product formula on the columns of $T_s$ and returns $1 - \tfrac{1}{2}s - \tfrac{12}{25}s^2$ — the same polynomial as above, with $\tfrac{1}{2} = 0.5$ and $\tfrac{12}{25} = 0.48$ in fraction form.

Example 1 How much area survives the trip?

Evaluate the area of the image parallelogram at $s = 0$ , $s = 0.5$ , and $s = 1$ , and determine whether the area ever reaches zero for $s$ between $0$ and $1$ .

Show solution

At $s = 0$ : $1 - 0 - 0 = 1$ , the unit square. At $s = 0.5$ : $1 - 0.25 - 0.48(0.25) = 1 - 0.25 - 0.12 = 0.63$ . At $s = 1$ : $1 - 0.5 - 0.48 = 0.02$ — only two percent of the area is left, which is why the parallelogram looks like a sliver. Setting $1 - 0.5s - 0.48s^2 = 0$ and solving the quadratic gives $s = \big({-0.5 + \sqrt{0.25 + 1.92}}\big)/0.96 = (-0.5 + \sqrt{2.17})/0.96 \approx 1.014$ . The collapse point exists, but it sits just past the end of the slider: on $[0, 1]$ the area stays positive, bottoming out at $0.02$ . The square is never quite destroyed — but $A$ comes within a hair of being the kind of matrix that flattens the plane.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Chapters 1–6 follow the topic sequence of Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL). Chapter 0's scope (its topics, their order, and the idea each figure shows) follows the Vectors chapter of Ximera's Linear Algebra: An Interactive Introduction. The prose, worked examples and figures were written for this book, which is not affiliated with or endorsed by the authors of either work. License: CC-BY-NC-SA-4.0.