Shears
Hold the bottom edge of a square down and push its top edge sideways. The square leans over into a
slanted parallelogram — the bottom hasn't moved at all, the top has slid the furthest, and every
height in between has slid by an amount in proportion to how high up it is. That sliding, height-
dependent push is a shear, and it's hiding in more places than you'd guess: it's
why italic letters lean the way they do, it's the effect behind those "leaning tower"
tourist photos where someone appears to be pushing a famous tower over, and it's exactly what a
deck of cards does when you fan it out across a table.
A horizontal shear leaves \hat{\imath} exactly where it
is but tips \hat{\jmath} over to the side:
\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}.
The entry k is how far the top leans for every unit of height. Squares
become slanted parallelograms, yet — and this is the surprise that makes shears worth knowing —
the area never changes. Sliding doesn't add or remove any space, it only skews it.
Building the matrix: track the basis vectors
Just as with scaling and reflection, the whole matrix falls out of one question: where do
\hat{\imath}=(1,0) and \hat{\jmath}=(0,1)
land? A pure shear is defined by pinning one of them and tilting only the other.
For a horizontal shear, \hat{\imath} sits on the
x-axis already, right where the sliding happens — it doesn't move, so it
stays at (1,0). That's the first column. Meanwhile
\hat{\jmath}=(0,1) sits one full unit up, so it picks up the maximum
sideways nudge: it tilts over to (k, 1). That's the
second column. Side by side:
\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}.
The 1s on the diagonal say "nothing is being stretched" — lengths along
each axis survive unchanged. The single off-diagonal k is the entire
story of the shear: it says "as you go up by 1 in
y, drift sideways by k in
x." A point twice as high drifts twice as far, which is exactly what
"proportional to height" means.
A vertical shear is the mirror image of this idea: now
\hat{\jmath}=(0,1) is the one sitting on its own axis (the
y-axis) and stays fixed, while \hat{\imath}=(1,0)
— sitting one full unit out along x — tilts upward to
(1, k). The columns become
\begin{bmatrix} 1 & 0 \\ k & 1 \end{bmatrix},
a matrix that shifts y by an amount proportional to
x, sliding things up and down instead of left and right.
Notice the recipe is identical in both cases, just with the roles of the two basis vectors swapped:
pick one basis vector to fix in place — whichever one already sits on the axis that stays
still — and tilt the other one by an amount that lands exactly k away
from where it started. A pure shear never fixes neither vector, and it never tilts both.
Worked example 1: shear a unit square
Take the unit square with corners (0,0), (1,0),
(1,1), (0,1), and apply the horizontal shear
with k = 2, so H = \begin{bmatrix} 1 & 2 \\ 0 & 1
\end{bmatrix}. Multiply it into each corner in turn:
H\begin{bmatrix}0\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}, \quad H\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}, \quad H\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}, \quad H\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}.
The bottom two corners — height y=0 — didn't move an inch, exactly as
promised: zero height means zero sideways drift. The top two corners, at height
y=1, both slid right by k=2. The result is a
parallelogram sitting on the same base, the same height, just leaning over —
vertical sides have become slanted sides running parallel to each other, while the top and bottom
edges stayed horizontal and the same length as before.
Worked example 2: area survives, even though the shape doesn't
That parallelogram from example 1 has base 1 and height
1, and the base-times-height formula for a parallelogram's area only
cares about the base and the perpendicular height — not how far the top has slid sideways. So its
area is 1 \times 1 = 1, exactly the same as the original unit square's
area, no matter what k is. Try it with
k = 100: an extremely slanted sliver of a parallelogram, but its base and
height haven't changed, so its area is still 1.
This is exactly what it means to say a shear's
determinant
is 1: the determinant is the factor by which a transformation scales
area, and a shear's factor is always exactly 1, for every value of
k. Compare that with scaling, where changing a diagonal entry always
changes the area — a shear is the rare transformation that can distort a shape as violently as you
like while leaving its area completely alone.
Worked example 3: a vertical shear, for contrast
Apply the vertical shear V = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}
(that's k=1) to the same unit square. This time
\hat{\jmath}=(0,1) stays put, and \hat{\imath}=(1,0)
tilts upward:
V\begin{bmatrix}0\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}, \quad V\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}, \quad V\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix}, \quad V\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}.
Now it's the corners on the y-axis (x=0) that
didn't move, and the corner furthest out at x=1 that slid the most — but
upward this time, not sideways. Left and right edges stay vertical and the same length;
top and bottom edges tilt. It's the same idea as the horizontal shear rotated ninety degrees in
which direction does the sliding, not a different kind of transformation.
Check the area again: the resulting parallelogram still has base 1 and
(perpendicular) height 1, so its area is still exactly
1. Whichever direction a shear slides in, area-preservation isn't a
special property of the horizontal case — it's built into the very shape of a shear matrix, since
the diagonal is always 1, 1 and only one off-diagonal entry is ever
non-zero.
Lean it over, live
Slide k below and watch the grid tilt. Horizontal lines stay exactly
where they are; everything above the x-axis drifts sideways in
proportion to its height, so vertical lines tilt into parallel diagonals while staying parallel to
each other. Watch the two coloured arrows: \hat{\imath} is pinned to
(1,0) the entire time; only \hat{\jmath}
swings over.
Watch the corner label as you slide: it always reads "area unchanged," at
k=0 and at k=2 and everywhere in between —
the single fact that makes a shear worth its own name instead of just being lumped in with
"distortions."
Try dragging k all the way to one extreme, then straight through
k=0 (no shear at all — the grid is perfectly square) and out to the other
extreme. Because \hat{\imath} never moves, the grid always pivots around
the same fixed line, the x-axis — it never spins around a point the way a
rotation would. That single fixed line, with everything else sliding past it in proportion to
distance, is the visual signature that tells you "shear" at a glance, in any picture, before you've
even seen the matrix.
Two mistakes are extremely common the first time you meet shears:
-
A shear can look like a rotation on a single shape, but it isn't one. Shear a
square by a small k and, glancing quickly, it can resemble a slightly
rotated square. The resemblance stops the moment you look closer: a rotation keeps every length
and every angle exactly the same and keeps area fixed automatically because nothing stretches at
all; a shear does change most angles and side lengths (only area survives, and only by
a coincidence of geometry, not because nothing moved). The two also behave completely differently
under repetition — apply the same small rotation four times and you're back roughly where you
started; apply the same shear four times and the slant just keeps compounding, further and
further, with no return.
-
Only one basis vector moves in a pure shear — never both. A very common error is
to imagine the whole square tilting from both ends, like a book falling open, sliding
x by height and y by width at the
same time. That's not a shear at all — check the matrix: a genuine shear matrix has exactly one
off-diagonal entry non-zero (\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}
or \begin{bmatrix} 1 & 0 \\ k & 1 \end{bmatrix}), never both at once.
One basis vector is always left exactly where it started.
Because they're built the same way. An italic typeface is very often literally the upright letters
put through a horizontal shear — every point of every letterform slides sideways by an amount
proportional to how high up it sits, exactly like the matrix on this page. That's why italics keep
their letterforms recognisable (curves are still curves, dots are still dots — only positions
shifted) while leaning at a single consistent angle across an entire alphabet: one shared value of
k applied to every letter.
Engineers borrowed the same word for a very physical idea. "Shear stress" in
materials science describes forces that slide layers of a material sideways past each other — the
top of a bolt pushed one way while the bottom is held still, or two tectonic plates grinding past
one another along a fault line. The geometry is exactly this page's picture: one edge fixed, another
edge sliding parallel to it, and everything in between sliding proportionally. When a material
finally snaps under this kind of force, engineers call the failure a shear failure
— the same shear, doing damage instead of typesetting.
Area-preserving, and oddly important
Because a shear keeps area, its
determinant
is 1. Shears might look like a curiosity, but they're the elementary
moves behind row
operations — "add a multiple of one row to another," the workhorse step of solving a
linear system, is geometrically nothing but a shear. Every time an equation gets solved by
elimination, the plane (or space, or higher-dimensional space) is quietly being sheared, over and
over, until the system falls into a shape simple enough to read off the answer. Master shears and
you've quietly understood the geometry behind how equations get solved.