Worked examples — Linear transformations — definition, kernel, image

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The scenario matrix

Every linear-map problem lands in one of these cells. Read this table like a checklist — each row is a shape of behaviour. Recall $n=\dim V$ (input dimension) and $m=\dim W$ (output dimension), just defined above.

#	Case class	What's special	Nullity / Rank	Example
A	Injective, not onto ($n<m$: fewer input than output dims)	maps up a dimension, crushes nothing	null $=0$, rank $=n$	Ex 1
B	Onto, not injective ($n>m$: more input than output dims)	maps down, must crush something	null $>0$, rank $=m$	Ex 2
C	Bijective (square, $n=m$, invertible)	perfect one-to-one, both directions	null $=0$, rank $=n$	Ex 3
D	The zero map (degenerate)	crushes everything	null $=n$, rank $=0$	Ex 4
E	Projection (idempotent, kernel = a line/plane)	keeps a subspace, kills the rest	null $>0$, rank $<n$	Ex 5
F	Rank-deficient square ($n=m$ but $\det =0$)	square but NOT invertible	null $>0$, rank $<n$	Ex 6
G	Non-${\mathbb{R}}^n$ space (polynomials, the derivative)	vectors aren't columns	null $=1$, rank $=n-1$	Ex 7
H	Word problem (real world)	translate first, then compute	depends	Ex 8
I	Exam twist (find $k$ so a vector is reachable)	image-membership test	—	Ex 9

The Injective surjective bijective properties are just cells A, B, C read off the nullity/rank.

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Example 1 — Cell A: injective, not onto ($n=2<m=3$)

1. Kernel: set the output to zero: $x=0$, $y=0$, $x+y=0$. The first two already force $x=y=0$. Why this step? Kernel = inputs sent to $0$; solve $T(x,y)=(0,0,0)$ directly.
2. So $\ker T=\{(0,0)\}$, nullity $=0$ → $T$ is injective. Why this step? Injective $⟺\ker T=\{0\}$ (parent's rule, no exceptions).
3. Image: the matrix is $A=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right)$; columns $(1,0,1)$ and $(0,1,1)$. They aren't multiples of each other, so they're independent → rank $=2$. Why this step? For $T(x)=Ax$, the image is the column space; its dimension is the number of independent columns.
4. Since rank $=2<3=\dim {\mathbb{R}}^3$, the image is a plane inside 3D, not all of it → not onto.

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Example 2 — Cell B: onto, not injective ($n=3>m=2$)

1. Kernel: $x-z=0$ and $y+z=0$, so $x=z$ and $y=-z$. Let $z=t$: $(t,-t,t)=t(1,-1,1)$. Why this step? Solve $T=0$; one free parameter $t$ remains.
2. $\ker T=\mathrm{span}\{(1,-1,1)\}$, nullity $=1$. A nonzero kernel → not injective. Why this step? One free variable = a whole line of inputs crushed to $0$.
3. Image: matrix $A=\left(\begin{array}{ccc}1& 0& -1\\ 0& 1& 1\end{array}\right)$; columns $(1,0),(0,1),(-1,1)$. The first two already span all of ${\mathbb{R}}^2$ → rank $=2$, image $={\mathbb{R}}^2$ → onto. Why this step? Two independent columns fill the 2D output space.

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Example 3 — Cell C: bijective (invertible, $n=m=2$)

1. Matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$, determinant $=2\cdot 1-1\cdot 1=1\ne 0$. Why this step? A nonzero determinant means the columns are independent — no collapse.
2. Kernel: $2x+y=0$, $x+y=0$. Subtract: $x=0$, then $y=0$. So $\ker T=\{0\}$, nullity $=0$, injective. Why this step? Solve $T=0$; the unique solution being $0$ confirms $\det \ne 0$.
3. Image: rank $=2$ (two independent columns) $=\dim {\mathbb{R}}^2$ → onto. Injective + onto = bijective.

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Example 4 — Cell D: the zero map (degenerate)

1. Kernel: every input gives $0$, so $\ker T={\mathbb{R}}^2$, nullity $=2$. Why this step? Kernel is the set sent to $0$ — here that's the whole space.
2. Image: the only output is $(0,0)$, so $\mathrm{im}T=\{0\}$, rank $=0$. Why this step? Nothing else can come out.
3. Not injective (kernel $\ne \{0\}$), not onto (image $\ne {\mathbb{R}}^2$).

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Example 5 — Cell E: projection (idempotent)

1. Kernel: $T(x,y)=(0,0)$ needs $x=0$; $y$ is free. So $\ker T=\{(0,y)\}=\mathrm{span}\{(0,1)\}$, the $y$-axis, nullity $=1$. Why this step? Points directly above/below origin cast their shadow onto $0$.
2. Image: outputs are $(x,0)$ = the whole $x$-axis, rank $=1$. Why this step? Every shadow lands on the $x$-axis; nothing above it is reachable.
3. Idempotent check: $T(T(x,y))=T(x,0)=(x,0)=T(x,y)$. Projecting twice = projecting once. ✓ Why this step? A projection fixes what's already in its image (that's the definition of idempotent above).

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Example 6 — Cell F: rank-deficient square ($n=m=2$, $\det =0$)

1. $\det A=1\cdot 4-2\cdot 2=0$. Not invertible. Why this step? Zero determinant flags dependent columns → a collapse.
2. Kernel: $x+2y=0$ (both rows give this), so $x=-2y$. $(-2y,y)=y(-2,1)$. $\ker T=\mathrm{span}\{(-2,1)\}$, nullity $=1$. Why this step? The two equations are secretly one equation → one free variable.
3. Image: columns $(1,2)$ and $(2,4)$; the second is $2\times$ the first → only one independent column. $\mathrm{im}T=\mathrm{span}\{(1,2)\}$, rank $=1$ — a line, not the plane. Why this step? Dependent columns span a lower-dimensional space.

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Example 7 — Cell G: the derivative on polynomials

1. Kernel: $D(a+bx+c{x}^2)=b+2cx=0$ forces $b=0$ and $c=0$; $a$ is free. So $\ker D=\{a\cdot 1\}=$ constants, nullity $=1$. Why this step? Only constant functions have zero slope everywhere.
2. Image: $D(a+bx+c{x}^2)=b+2cx$, which is any polynomial of degree $\le 1$ = $\mathrm{span}\{1,x\}$, rank $=2$. Why this step? Differentiating drops the degree by one, so outputs live in $P_1$.

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Example 8 — Cell H: word problem

1. Write $T(a,b,c)=(a+b+c,2a+b+3c)$, matrix $A=\left(\begin{array}{ccc}1& 1& 1\\ 2& 1& 3\end{array}\right)$. Why this step? Read the coefficients straight off the two recipe formulas.
2. (i) Kernel: $a+b+c=0$ and $2a+b+3c=0$. Subtract the first from the second: $a+2c=0\Rightarrow a=-2c$. Then $b=-a-c=2c-c=c$. So $(a,b,c)=(-2c,c,c)=c(-2,1,1)$. Why this step? Eliminate one variable to expose the free parameter.
3. $\ker T=\mathrm{span}\{(-2,1,1)\}$: blends along this direction cancel to zero volume and zero sugar (algebraically). Nullity $=1$.
4. (ii) Image: columns $(1,2)$ and $(1,1)$ are independent → rank $=2$ = $\dim {\mathbb{R}}^2$. Every (volume, sugar) pair is reachable → onto.

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Example 9 — Cell I: exam twist (image membership)

1. We need $(x+y,x-y,2x)=(3,1,k)$ for some $x,y$. From the first two: $x+y=3$, $x-y=1$ → add: $2x=4$, $x=2$, then $y=1$. Why this step? Solve the two "free" coordinates first; they pin down $x,y$ uniquely.
2. The third coordinate is forced: $2x=2\cdot 2=4$. So we need $k=4$. Why this step? Once $x$ is fixed, the third output can't be chosen independently — the plane's constraint.
3. Therefore $(3,1,k)\in \mathrm{im}T⟺\boxed{k=4}$.

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Active recall

Recall Which cell is each map? (hide answers)

${\mathbb{R}}^4\to {\mathbb{R}}^2$, onto ::: Cell B — onto, not injective (nullity $=2$). $T(v)=0$ ::: Cell D — the zero map (nullity = dim $V$, rank $0$). $2\times 2$ with $\det =0$ ::: Cell F — rank-deficient square. Derivative on $P_3$ ::: Cell G — non-${\mathbb{R}}^n$; nullity $1$, rank $3$. ${\mathbb{R}}^2\to {\mathbb{R}}^5$ injective ::: Cell A — injective, not onto. What does idempotent mean? ::: $T\circ T=T$ — applying twice equals applying once (projections).

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Connections

• Column space and null space — every "image" here is a column space, every "kernel" a null space.
• Rank–Nullity Theorem — the running verify for all nine examples.
• Injective surjective bijective — cells A/B/C are these three properties read off nullity & rank.
• Eigenvalues and eigenvectors — a zero determinant (Cell F) means $0$ is an eigenvalue; its eigenvectors form the kernel.
• Change of basis — turns the abstract derivative map (Ex 7) into a concrete matrix.
• Matrix multiplication — composing two linear maps multiplies their matrices.