[ALGEBRA] Geometric examples

Matrix maps, scaling, reflections, rotations, dot products, projections and a translation non-example.

This page reads linear maps geometrically, through the points, lines and directions they move in $\mathbb R^2$ and $\mathbb R^3$ . It first shows that linear maps between these spaces are exactly matrix maps, then works through scaling, reflections, rotations and projections.

[DEFINITION]

Linear map

Let $V$ and $W$ be vector spaces over the same field $F$ . A function $T:V\to W$ is a linear map if

T(u+v)=T(u)+T(v)\qquad\text{for all }u,v\in V,

T(\omega u)=\omega T(u)\qquad\text{for all }\omega\in F,\ u\in V.

Matrix maps and standard coordinates

Multiplication by a fixed matrix is a linear map. In standard coordinates, the columns of the matrix are the images of the standard basis vectors.

[THEOREM]

Matrix maps are linear

For every $m\times n$ matrix $A$ , the map

T_A:\mathbb R^n\to\mathbb R^m,\qquad T_A(x)=Ax

is linear.

[1]:Linear map
[2]:Proof
[3]:Reading a matrix map in coordinates
[4]:The matrix of a linear map on standard coordinates
[5]:Scaling in $\mathbb R^2$
4 more

[PROOF]

For $x,y\in\mathbb R^n$ and $\lambda\in\mathbb R$ ,

T_A(x+y)=A(x+y)=Ax+Ay=T_A(x)+T_A(y),

and

T_A(\lambda x)=A(\lambda x)=\lambda Ax=\lambda T_A(x).

Thus $T_A$ satisfies both conditions in the definition of a linear map.

[1]:Matrix maps are linear

[EXAMPLE]

Reading a matrix map in coordinates

For

A= \begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23} \end{pmatrix},

the associated map $T_A:\mathbb R^3\to\mathbb R^2$ is

T_A\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{pmatrix} a_{11}x_1+a_{12}x_2+a_{13}x_3\\ a_{21}x_1+a_{22}x_2+a_{23}x_3 \end{pmatrix}.

The matrix columns give the images of the standard basis vectors, so the coordinate formula is determined by those three images.

[routine]

[1]:Matrix maps are linear

The converse also holds: by the basis-data principle, every linear map $\mathbb R^n\to\mathbb R^m$ is a matrix map.

[THEOREM]

The matrix of a linear map on standard coordinates

Let $T:\mathbb R^n\to\mathbb R^m$ be a linear map and let $e_1,\ldots,e_n$ be the standard basis vectors of $\mathbb R^n$ . If $A$ is the $m\times n$ matrix whose $j$ th column is $T(e_j)$ , then

T(x)=Ax

for every $x\in\mathbb R^n$ . Thus every linear map from $\mathbb R^n$ to $\mathbb R^m$ is represented by a matrix.

[1]:Linear map
[2]:Preservation of finite linear combinations
[3]:Matrix maps are linear
[4]:Proof
[5]:Constructing the matrix from a formula

[PROOF]

Write $x=x_1e_1+\cdots+x_ne_n$ . Preservation of finite linear combinations gives

T(x)=x_1T(e_1)+\cdots+x_nT(e_n).

These vectors are exactly the columns of $A$ , so the right-hand side is $Ax$ .

[1]:The matrix of a linear map on standard coordinates

[EXAMPLE]

Constructing the matrix from a formula

Suppose

T\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{pmatrix}x_2+x_3\\x_1+x_2\end{pmatrix}.

Evaluating at the standard basis vectors gives

T\begin{pmatrix}1\\0\\0\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix}, \qquad T\begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}1\\1\end{pmatrix}, \qquad T\begin{pmatrix}0\\0\\1\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}.

Therefore

A= \begin{pmatrix} 0&1&1\\ 1&1&0 \end{pmatrix},

whose columns reproduce the given coordinate formula.

[routine]

[1]:The matrix of a linear map on standard coordinates

Matrix representations extends this construction to arbitrary bases. Changing the basis changes the entries of the matrix, but not the underlying map.

Scaling and reflections

Diagonal matrices show how separate coordinate directions can be stretched or reversed.

[EXAMPLE]

Scaling in $\mathbb R^2$

For fixed positive factors $\lambda_x,\lambda_y$ , the map

\begin{pmatrix}x\\y\end{pmatrix} \longmapsto \begin{pmatrix}\lambda_xx\\\lambda_yy\end{pmatrix}

has matrix

\begin{pmatrix} \lambda_x&0\\ 0&\lambda_y \end{pmatrix}.

The two coordinate directions are scaled independently. Each axis is sent to itself: $(x,0)$ is sent to $(\lambda_xx,0)$ and $(0,y)$ to $(0,\lambda_yy)$ . Any other line through the origin is sent to a line through the origin, but its slope can change unless $\lambda_x=\lambda_y$ .

[routine]

[1]:Matrix maps are linear

[EXAMPLE]

Reflections in $\mathbb R^2$

Reflection in the $x$ -axis and reflection in the $y$ -axis are respectively

\begin{pmatrix}x\\y\end{pmatrix} \longmapsto \begin{pmatrix}x\\-y\end{pmatrix}, \qquad \begin{pmatrix}x\\y\end{pmatrix} \longmapsto \begin{pmatrix}-x\\y\end{pmatrix},

with matrices

\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad \begin{pmatrix}-1&0\\0&1\end{pmatrix}.

[routine]

[1]:Matrix maps are linear

Rotations

A rotation turns every vector through the same angle about the origin and it is linear.

[EXAMPLE]

Rotation about the origin

Write $(x,y)=(r\cos\varphi,r\sin\varphi)$ . A counterclockwise rotation through $\theta$ sends the angle $\varphi$ to $\varphi+\theta$ . The angle-addition identities give

\begin{pmatrix}x\\y\end{pmatrix} \longmapsto \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}.

Explicitly,

r\cos(\varphi+\theta)=r\cos\varphi\cos\theta-r\sin\varphi\sin\theta =x\cos\theta-y\sin\theta,

and similarly the second coordinate is $x\sin\theta+y\cos\theta$ . So the signs and the placement of $\sin\theta$ in the matrix are forced by these identities, not a convention to memorise.

The first column is the image of $(1,0)^T$ , namely $(\cos\theta,\sin\theta)^T$ and the second is the image of $(0,1)^T$ , namely $(-\sin\theta,\cos\theta)^T$ .

Under the identification $\mathbb C\cong\mathbb R^2$ , write $z=x+iy=re^{i\varphi}$ . Multiplication by $e^{i\theta}$ gives

e^{i\theta}z =re^{i(\varphi+\theta)} =(x\cos\theta-y\sin\theta) +i(x\sin\theta+y\cos\theta),

so the real and imaginary parts are the two coordinates produced by the rotation matrix. Polar form and the matrix describe the same rotation.

[routine]

[1]:Matrix maps are linear

[EXAMPLE]

Reflection in a line through the origin

More generally, reflection in a line through the origin making angle $\phi$ with the positive $x$ -axis has matrix

\begin{pmatrix} \cos(2\phi)&\sin(2\phi)\\ \sin(2\phi)&-\cos(2\phi) \end{pmatrix}.

Setting $\phi=0$ gives reflection in the $x$ -axis, while $\phi=\pi/2$ gives reflection in the $y$ -axis.

To derive the formula, observe that a vector at angle $\theta$ reflects to angle $2\phi-\theta$ . Applying the cosine and sine subtraction identities to this new angle produces the two rows of the matrix.

[routine]

[1]:Matrix maps are linear

Dot products and projections

The codomain need not have the same dimension as the domain. A dot product produces a scalar, while projection produces a vector on a fixed line.

[EXAMPLE]

A dot-product map

Fix $u\in\mathbb R^n$ and define

T_u:\mathbb R^n\to\mathbb R,\qquad T_u(x)=u\cdot x.

Bilinearity of the dot product gives

T_u(x+y)=u\cdot(x+y)=u\cdot x+u\cdot y=T_u(x)+T_u(y),

T_u(\lambda x)=u\cdot(\lambda x)=\lambda(u\cdot x)=\lambda T_u(x).

Thus taking the dot product with a fixed vector is a linear map.

[routine]

[1]:Linear map

[EXAMPLE]

Projection onto a line

For $u\ne0$ , the projection of $x\in\mathbb R^n$ onto the line spanned by $u$ is

\operatorname{proj}_u x=\frac{x\cdot u}{u\cdot u}\,u.

To see this, write the projected vector as $cu$ . The residual $x-cu$ must be perpendicular to $u$ and solving $(x-cu)\cdot u=0$ gives $c=(x\cdot u)/(u\cdot u)$ . The condition $u\ne0$ makes the denominator positive.

The projection map $P_u:\mathbb R^n\to\mathbb R^n$ is therefore

P_u(x)=\frac{x\cdot u}{u\cdot u}\,u.

Since $u\cdot u$ is a fixed non-zero scalar, bilinearity gives

P_u(x+y)=P_u(x)+P_u(y), \qquad P_u(\lambda x)=\lambda P_u(x).

Hence projection onto a fixed line through the origin is a linear map.

For orthogonal projection onto a subspace $H$ , every vector in $H$ is fixed and every vector in $H^\perp$ is sent to zero. When those subspaces are non-zero, they are eigenspaces for eigenvalues $1$ and $0$ respectively; this links projection geometry to eigenvalues.

[routine]

[1]:Linear map

Translation looks like these geometric maps but is not linear. For a fixed non-zero vector $a$ , the map $T(x)=x+a$ satisfies $T(0)=a\ne0$ and, in general,

T(\lambda x)=\lambda x+a\ne\lambda x+\lambda a=\lambda T(x).

Kernel, image and rank turns next to the vectors a map sends to $0$ and the outputs it can produce.

Geometric examples

Linear map

On this page