Linear Transformations

Each map below is linear, each is given by its matrix, and each is worth recognising on sight.

Scaling by $k$. . Stretches everything away from the origin by a factor $k$. For $k=0$ it is the zero map, whose kernel is all of ${\mathbb{R}}^2$ and whose image is $\{0\}$: nullity 2, rank 0. For $k=1$ it is the identity: nullity 0, rank 2.

Reflection in the $x$-axis. , sending $(x,y)$ to $(x,-y)$. Determinant $-1$, kernel trivial, image everything. Applying it twice gives the identity.

Rotation by $\theta$. . The columns are the images of $(1,0)$ and $(0,1)$, which land at angle $\theta$ from where they started. Determinant ${\cos }^2\theta +{\sin }^2\theta =1$, so nothing is collapsed and nothing is scaled.

Projection onto the $x$-axis. , sending $(x,y)$ to $(x,0)$. The kernel is the $y$-axis and the image is the $x$-axis: nullity 1, rank 1, summing to 2. Applying it twice changes nothing after the first time, which is what makes it a projection.

Shear. , sending $(x,y)$ to $(x+sy,y)$. Horizontal lines slide by an amount proportional to their height. Determinant 1, so areas are preserved even though the picture is distorted. A reminder that the determinant measures area scaling, not distortion.

The determinant is zero exactly when the kernel is nontrivial, and in the plane the three rank values 0, 1 and 2 are the only ones available, collapse everything, collapse a line, collapse nothing. Rank–nullity accounts for each row: the entries in the last two columns sum to 2 throughout.

Two that are not on the list. Translation by a fixed vector is not linear, since it moves the origin. "Reflection in the line $y=1$" is not linear for the same reason; only reflections in lines through the origin qualify.

A shift. $S(x,y)=(x+1,y)$. Immediately $S(0,0)=(1,0)\ne 0$, so it is not linear. Additivity fails too: $S(1,2)+S(-3,4)=(2,2)+(-2,4)=(0,6)$, while $S(-2,6)=(-1,6)$. Any map with a constant term fails this way, including $x\mapsto mx+c$ for $c\ne 0$. The schoolroom "linear function" is affine, not linear.

A squaring. $Q(x,y)=(x^2,y)$. Here $Q(0,0)=(0,0)$, so the origin test passes and gives no information. Homogeneity is what fails: $Q(2\cdot (1,1))=Q(2,2)=(4,2)$, while $2\cdot Q(1,1)=2(1,1)=(2,2)$. The map is nonlinear in a way the origin never reveals, which is why the origin test rules out rather than rules in.

A norm. $N(x,y)=\sqrt{x^2+y^2}$, as a map ${\mathbb{R}}^2\to \mathbb{R}$. It fixes the origin and it satisfies $N(av)=|a|N(v)$, but $|a|$, not $a$. Taking $a=-1$: $N(-v)=N(v)$ while $-N(v)$ is negative for $v\ne 0$. Additivity fails too, since $N(1,0)+N(0,1)=2$ while $N(1,1)=\sqrt{2}$.

Transposition, which is linear. $T(M)=M^{\mathsf{T}}$ on $2\times 2$ matrices. It looks like a rearrangement rather than an algebraic operation, and it is linear: $(M+N{)}^{\mathsf{T}}=M^{\mathsf{T}}+N^{\mathsf{T}}$ and $(aM{)}^{\mathsf{T}}=a{M}^{\mathsf{T}}$. Its kernel is $\{0\}$, since only the zero matrix transposes to zero, so by rank–nullity its image is all of $M_{2\times 2}$, nullity $0$, rank $4$, summing to $\dim M_{2\times 2}=4$.

Linearity is not about looking like a formula with no exponents. It is two equations that either hold for all inputs or do not, and the cases divide by how they fail: $S$ at the origin, $Q$ under scaling only, $N$ under negative scaling and addition both. A map is linear when it respects the operations, whatever it looks like, which is why transposition qualifies and squaring does not.

A map given by a first-degree formula is linear, so $x\mapsto mx+c$ and maps that add a constant vector qualify, since their graphs are straight lines.