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 $(u + 0.8 v, 0.6 u + 0.5 v)$ — columns $open parenthesis 1, 0.6 close parenthesis$ and $open parenthesis 0.8, 0.5 close parenthesis$ — and blend it with the identity map using a dial $s$ :

$T_{s} (u, v) = (1 - s) (u, v) + s (u + 0.8 v, 0.6 u + 0.5 v)$

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

Two flat squares in the xy-plane seen from above: a faint gray unit square that never moves, and a colored copy controlled by a slider named s. At s = 0 the colored copy sits exactly on the gray square; dragging s toward 1 continuously shears and tilts it into the parallelogram that the matrix A produces, so the reader watches the identity deform into A. — The unit square carried by the interpolated map $T_{s}$ , over a faint gray copy of the unit square that never moves. At $s equals 0$ the colored square sits exactly on the gray one; dragging $s$ to $1$ deforms it continuously into the image parallelogram of $A$ .

Explore the figure

At $s equals 0$ , confirm the colored square coincides with the gray reference square underneath it.
Drag $s$ slowly from $0$ to $1$ and watch one landmark: the far corner starts at $open parenthesis 1, 1 close parenthesis$ and arrives at the sum of the columns of $A$ , which is $open parenthesis 1.8, 1.1 close parenthesis$ . 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 $s equals 1$ its two edges, the columns $open parenthesis 1, 0.6 close parenthesis$ and $open parenthesis 0.8, 0.5 close parenthesis$ , point almost the same way. Compare $open parenthesis 0.8, 0.5 close parenthesis$ 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 $𝐩$ , its position at dial value $s$ is $(1 - s) 𝐩 + s A 𝐩$ — a weighted average of the start $𝐩$ and the destination $A 𝐩$ . 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 $open parenthesis x sub 1, y sub 1 close parenthesis$ and $open parenthesis x sub 2, y sub 2 close parenthesis$ , the area is $| x sub 1 y sub 2 minus x sub 2 y sub 1 |$ — the cross-product formula from geometry. Applied to the columns of $T_{s}$ , the quantity inside the absolute value is

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

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.