Shear Transformations and Their Standard Matrices

What a Shear Changes

A shear changes one coordinate of each point in proportion to the other coordinate, while leaving that other coordinate fixed. In a horizontal shear, the height stays the same and the horizontal displacement depends on height. In a vertical shear, the horizontal position stays the same and the vertical displacement depends on horizontal position. The parameter $k\in\mathbb{R}$ controls the amount and direction of the shear: $k=0$ makes no change, while changing the sign of $k$ reverses the direction of the displacement.

Let $\mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{R}^2$. A horizontal shear with parameter $k$ sends $(x,y)$ to $(x+ky,y)$; a vertical shear with parameter $k$ sends $(x,y)$ to $(x,y+kx)$. The coordinate descriptions specify the effect on every point. To find the standard matrices, it is enough to determine what each transformation does to the standard basis vectors $\mathbf{e}_1=\begin{bmatrix}1\\0\end{bmatrix}$ and $\mathbf{e}_2=\begin{bmatrix}0\\1\end{bmatrix}$.

Definition: For $k\in\mathbb{R}$, the horizontal shear $H_k:\mathbb{R}^2\to\mathbb{R}^2$ is $H_k(x,y)=(x+ky,y)$. The vertical shear $V_k:\mathbb{R}^2\to\mathbb{R}^2$ is $V_k(x,y)=(x,y+kx)$. Their standard matrices are the matrices $\mathbf{A}$ for which $H_k(\mathbf{x})=\mathbf{A}\mathbf{x}$ or $V_k(\mathbf{x})=\mathbf{A}\mathbf{x}$, respectively.

Building the Standard Matrices

For a horizontal shear, $\mathbf{e}_1$ has height zero, so it does not move. The vector $\mathbf{e}_2$ has height one, so it moves horizontally by $k$. For a vertical shear, $\mathbf{e}_1$ moves vertically by $k$, while $\mathbf{e}_2$ does not move. The columns of a standard matrix are the images of $\mathbf{e}_1$ and $\mathbf{e}_2$, in that order.

H_k(\mathbf{e}_1)=\begin{bmatrix}1\\0\end{bmatrix}, \qquad H_k(\mathbf{e}_2)=\begin{bmatrix}k\\1\end{bmatrix}, \qquad \mathbf{H}_k= \begin{bmatrix}1&k\\0&1\end{bmatrix}.

V_k(\mathbf{e}_1)=\begin{bmatrix}1\\k\end{bmatrix}, \qquad V_k(\mathbf{e}_2)=\begin{bmatrix}0\\1\end{bmatrix}, \qquad \mathbf{V}_k= \begin{bmatrix}1&0\\k&1\end{bmatrix}.

These matrices produce the coordinate formulas directly: multiplying $\mathbf{H}_k$ by $\begin{bmatrix}x\\y\end{bmatrix}$ gives $\begin{bmatrix}x+ky\\y\end{bmatrix}$, while multiplying $\mathbf{V}_k$ by that vector gives $\begin{bmatrix}x\\kx+y\end{bmatrix}$. The distinction between the two matrices is not merely a matter of naming: the entry $k$ is in the first row for a horizontal shear and in the second row for a vertical shear.

Theorem: For each $k\in\mathbb{R}$, the standard matrices of $H_k$ and $V_k$ are $\mathbf{H}_k=\begin{bmatrix}1&k\\0&1\end{bmatrix}$ and $\mathbf{V}_k=\begin{bmatrix}1&0\\k&1\end{bmatrix}$, respectively.

Proof. By the definition of the maps, $H_k(\mathbf{e}_1)=\begin{bmatrix}1\\0\end{bmatrix}$ and $H_k(\mathbf{e}_2)=\begin{bmatrix}k\\1\end{bmatrix}$. Since $\mathbf{e}_1,\mathbf{e}_2$ are the standard basis, every $\mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}$ can be written as $x\mathbf{e}_1+y\mathbf{e}_2$. The coordinate formula gives $H_k(\mathbf{x})=\begin{bmatrix}x+ky\\y\end{bmatrix}=xH_k(\mathbf{e}_1)+yH_k(\mathbf{e}_2)=\begin{bmatrix}1&k\\0&1\end{bmatrix}\mathbf{x}$. For $V_k$, its basis-vector images are $V_k(\mathbf{e}_1)=\begin{bmatrix}1\\k\end{bmatrix}$ and $V_k(\mathbf{e}_2)=\begin{bmatrix}0\\1\end{bmatrix}$. Thus $V_k(\mathbf{x})=\begin{bmatrix}x\\kx+y\end{bmatrix}=xV_k(\mathbf{e}_1)+yV_k(\mathbf{e}_2)=\begin{bmatrix}1&0\\k&1\end{bmatrix}\mathbf{x}$. These equations establish the stated standard matrices. $\square$

Computing Images of Vectors

Once the matrix is built, applying a shear is a matrix-vector multiplication. The coordinate formula is a useful independent check: for a horizontal shear, add $ky$ to the original $x$-coordinate and keep $y$; for a vertical shear, keep $x$ and add $kx$ to $y$.

Worked Example: A Horizontal Shear

Apply the horizontal shear with parameter $k=2$ to $\mathbf{x}=\begin{bmatrix}3\\-1\end{bmatrix}$.

The matrix is $\mathbf{H}_2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$. Multiplying gives

\mathbf{H}_2\mathbf{x} = \begin{bmatrix}1&2\\0&1\end{bmatrix} \begin{bmatrix}3\\-1\end{bmatrix} = \begin{bmatrix}1(3)+2(-1)\\0(3)+1(-1)\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix}.

Check: the original height is $-1$, so the horizontal displacement is $2(-1)=-2$. The new horizontal coordinate is $3-2=1$, and the height remains $-1$, agreeing with the matrix product.

Worked Example: A Vertical Shear

Apply the vertical shear with parameter $k=-3$ to $\mathbf{x}=\begin{bmatrix}-2\\4\end{bmatrix}$.

Here $\mathbf{V}_{-3}=\begin{bmatrix}1&0\\-3&1\end{bmatrix}$, so

\mathbf{V}_{-3}\mathbf{x} = \begin{bmatrix}1&0\\-3&1\end{bmatrix} \begin{bmatrix}-2\\4\end{bmatrix} = \begin{bmatrix}1(-2)+0(4)\\-3(-2)+1(4)\end{bmatrix} = \begin{bmatrix}-2\\10\end{bmatrix}.

Check: the horizontal coordinate remains $-2$. The vertical displacement is $k x=(-3)(-2)=6$, so the new vertical coordinate is $4+6=10$, as computed.

Images of Lines Through the Origin

A line through the origin is described by a nonzero direction vector $\mathbf{u}=\begin{bmatrix}a\\b\end{bmatrix}$: it consists of all vectors $t\mathbf{u}$ as $t$ ranges over $\mathbb{R}$. The image of a set under a map means the set of all outputs obtained from its elements. Because a shear is linear, a line's image can be found by applying the shear to its direction vector. The resulting direction vector cannot be zero: a horizontal shear sends $(a,b)$ to $(a+kb,b)$, which can be zero only if $b=0$ and $a=0$; a vertical shear sends $(a,b)$ to $(a,b+ka)$, which can be zero only if $a=0$ and $b=0$. Both would contradict that $\mathbf{u}\ne\mathbf{0}$.

Theorem: Let $k\in\mathbb{R}$, and let $L=\{t\mathbf{u}:t\in\mathbb{R}\}$ be a line through the origin with $\mathbf{u}=\begin{bmatrix}a\\b\end{bmatrix}\ne\mathbf{0}$. Then $$ H_k(L)=\left\{t\begin{bmatrix}a+kb\\b\end{bmatrix}:t\in\mathbb{R}\right\}, \qquad V_k(L)=\left\{t\begin{bmatrix}a\\b+ka\end{bmatrix}:t\in\mathbb{R}\right\}. $$ In each case the image is a line through the origin.

Proof. Every vector in $L$ has the form $t\mathbf{u}$. For the horizontal shear, direct substitution gives $H_k(t\mathbf{u})=H_k(ta,tb)=(ta+ktb,tb)=t(a+kb,b)$. As $t$ ranges over $\mathbb{R}$, these outputs are exactly the vectors listed for $H_k(L)$. The vector $(a+kb,b)$ is nonzero: if it were zero, then $b=0$ and $a+kb=a=0$, contradicting $\mathbf{u}\ne\mathbf{0}$. Its scalar multiples therefore form a line through the origin. For the vertical shear, $V_k(t\mathbf{u})=V_k(ta,tb)=(ta,tb+kta)=t(a,b+ka)$. These outputs are exactly the vectors listed for $V_k(L)$. If $(a,b+ka)=(0,0)$, then $a=0$, which makes $b+ka=b=0$, again contradicting $\mathbf{u}\ne\mathbf{0}$. Its scalar multiples also form a line through the origin. $\square$

For a nonvertical line written as $y=mx$, a direction vector is $\begin{bmatrix}1\\m\end{bmatrix}$. Under $H_k$, that direction becomes $\begin{bmatrix}1+km\\m\end{bmatrix}$. If $1+km\ne0$, the image line has slope $m/(1+km)$. If $1+km=0$, the first coordinate of the direction is zero and its second coordinate is nonzero, so the image is the vertical line $x=0$. Under $V_k$, the direction becomes $\begin{bmatrix}1\\m+k\end{bmatrix}$, so the image has slope $m+k$. A vertical line has direction $\begin{bmatrix}0\\1\end{bmatrix}$: a horizontal shear sends it to direction $\begin{bmatrix}k\\1\end{bmatrix}$, while a vertical shear leaves that direction unchanged.

Worked Example: Shearing a Line Through the Origin

Find the image under the horizontal shear with parameter $k=2$ of the line $y=3x$, and verify the result using a point on the line.

A direction vector for $y=3x$ is $\mathbf{u}=\begin{bmatrix}1\\3\end{bmatrix}$. Apply the shear matrix to that direction:

\mathbf{H}_2\mathbf{u} = \begin{bmatrix}1&2\\0&1\end{bmatrix} \begin{bmatrix}1\\3\end{bmatrix} = \begin{bmatrix}1+6\\3\end{bmatrix} = \begin{bmatrix}7\\3\end{bmatrix}.

Therefore the image line has direction $\begin{bmatrix}7\\3\end{bmatrix}$, so its slope is $3/7$ and its equation is $y=\tfrac{3}{7}x$. The same slope follows from the formula $m/(1+km)=3/(1+2\cdot3)=3/7$.

Check: the point $\begin{bmatrix}1\\3\end{bmatrix}$ lies on the original line. Its image is $\begin{bmatrix}7\\3\end{bmatrix}$, which satisfies $y=\tfrac{3}{7}x$, since $3=\tfrac{3}{7}\cdot7$. More generally, an original point $\begin{bmatrix}t\\3t\end{bmatrix}$ maps to $\begin{bmatrix}7t\\3t\end{bmatrix}$, which satisfies the same image-line equation for every $t\in\mathbb{R}$.

Common Pitfall: Putting the Shear in the Wrong Row

A frequent error is to use the vertical-shear matrix for a horizontal shear, or vice versa. For example, if $k=2$, the incorrect matrix $\begin{bmatrix}1&0\\2&1\end{bmatrix}$ is sometimes used for a horizontal shear. Applied to $\mathbf{e}_2$, it gives $\begin{bmatrix}0\\1\end{bmatrix}$, so it leaves $\mathbf{e}_2$ fixed; that is not the required horizontal effect, which moves $\mathbf{e}_2$ to $\begin{bmatrix}2\\1\end{bmatrix}$.

The correction is to build the columns from the basis-vector images in order. For a horizontal shear with $k=2$, $H_2(\mathbf{e}_1)=\begin{bmatrix}1\\0\end{bmatrix}$ and $H_2(\mathbf{e}_2)=\begin{bmatrix}2\\1\end{bmatrix}$, so the correct matrix is $\mathbf{H}_2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$. Its columns give the required images, and its product with $\begin{bmatrix}x\\y\end{bmatrix}$ is $\begin{bmatrix}x+2y\\y\end{bmatrix}$. The quickest check is to test both standard basis vectors before using a proposed matrix.

Similarly, for $V_2$, the images are $V_2(\mathbf{e}_1)=\begin{bmatrix}1\\2\end{bmatrix}$ and $V_2(\mathbf{e}_2)=\begin{bmatrix}0\\1\end{bmatrix}$, giving $\mathbf{V}_2=\begin{bmatrix}1&0\\2&1\end{bmatrix}$. Keeping the coordinate effect, basis-vector images, and matrix columns aligned prevents this row-placement error.

Check Your Understanding

For each question, use the coordinate effect or the standard matrix, and check your calculations against the defining formulas.

Write the standard matrix for a horizontal shear with parameter $k=-2$, and state the images of $\mathbf{e}_1$ and $\mathbf{e}_2$.
Find the image of $\begin{bmatrix}2\\-3\end{bmatrix}$ under the vertical shear with parameter $k=4$.
Under the horizontal shear with parameter $k=1$, what is the image line of $y=2x$?
Under the horizontal shear with parameter $k=-1$, what is the image of the line $y=x$?
Explain why the image of a line through the origin under a vertical shear is still a line through the origin.

Shear Transformations and Their Standard Matrices

What You'll Learn

What a Shear Changes

Building the Standard Matrices