Examples from Geometric examples

Reading a matrix map in coordinates

For

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

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

Constructing the matrix from a formula

Suppose

Evaluating at the standard basis vectors gives

Therefore

whose columns reproduce the given coordinate formula.

Scaling in $\mathbb R^2$

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

has matrix

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 $\lambda_x=\lambda_y$.

Reflections in $\mathbb R^2$

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

with matrices

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

Explicitly,

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

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

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

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.

A dot-product map

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

Bilinearity of the dot product gives

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

Projection onto a line

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

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

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

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.