Linear transformations — definition, kernel, image: Exercises

Level 1 — Recognition

Exercise 1.1

Which of these maps $T : R^{2} \to R^{2}$ are linear?

(a) $T (x, y) = (3 x, x - y)$
(b) $T (x, y) = (x + 2, y)$
(c) $T (x, y) = (x y, 0)$

Recall Solution

What "linear" demands: the map must send $0 \to 0$ , and it must respect adding and scaling (see the definition). The fastest first filter is the free sanity check $T (0) = 0$ .

(a) $T (0, 0) = (0, 0)$ ✓. Now check homogeneity and additivity — but notice every output is a sum of constant-multiples of $x$ and $y$ , i.e. degree-1 with no constants. $T (c (x, y)) = (3 c x, c x - cy) = c (3 x, x - y) = c T (x, y)$ ✓, and expanding a sum splits cleanly. Linear. ✓

(b) $T (0, 0) = (2, 0) \neq = (0, 0)$ . Why it fails: the $+ 2$ is a constant shift (affine), and shifting moves the origin. Not linear.

(c) $T (0, 0) = (0, 0)$ ✓ — but this passes the cheap test and still fails. Try homogeneity: $T (c (x, y)) = (c x \cdot cy, 0) = (c^{2} x y, 0)$ , whereas $c T (x, y) = (c x y, 0)$ . These differ ( $c^{2} \neq = c$ for e.g. $c = 2$ ). Not linear — the product $x y$ is degree 2.

Exercise 1.2

For $T (x, y, z) = (x - z, y)$ , write down the matrix $A$ with $T (v) = A v$ .

Recall Solution

What we do: feed the standard basis vectors in and stack the outputs as columns — this is how a linear map becomes a matrix (parent note §1). $T (e_{1}) = T (1, 0, 0) = (1, 0)$ , $T (e_{2}) = T (0, 1, 0) = (0, 1)$ , $T (e_{3}) = T (0, 0, 1) = (- 1, 0)$ . Why columns: any $x = x_{1} e_{1} + x_{2} e_{2} + x_{3} e_{3}$ gives $T (x) = \sum x_{i} T (e_{i})$ , exactly matrix-times-vector. $A = (1001 - 1 0) .$ This connects to Matrix multiplication.

Level 2 — Application

Exercise 2.1

Find $ker T$ and its dimension (the nullity) for $T : R^{3} \to R^{2}, T (x, y, z) = (x + 2 y + z, 2 x + 4 y + 2 z) .$

Recall Solution

What the kernel is: all inputs squashed to $(0, 0)$ (parent §2). So solve $x + 2 y + z = 0, 2 x + 4 y + 2 z = 0.$ Why look before solving: the second equation is exactly twice the first, so it carries no new information — one honest equation, three unknowns. Set $x + 2 y + z = 0 \Rightarrow x = - 2 y - z$ . Two free choices $y, z$ : $(x, y, z) = (- 2 y - z, y, z) = y (- 2, 1, 0) + z (- 1, 0, 1) .$ $ker T = span {(- 2, 1, 0), (- 1, 0, 1)}, nullity = 2.$ These two vectors are independent (neither is a multiple of the other), so the dimension really is $2$ . See Column space and null space.

Exercise 2.2

For the same $T$ , find $im T$ and the rank.

Recall Solution

What the image is: everything $T$ can output = the column space of its matrix. Columns of $A = (122412)$ are $(1, 2), (2, 4), (1, 2)$ . Why this shortcut works: $A x = \sum x_{i} (column_{i})$ , so outputs are exactly combinations of columns. All three columns are multiples of $(1, 2)$ . So they span a single line: $im T = span {(1, 2)} \subseteq R^{2}, rank = 1.$ Check with Rank–Nullity Theorem: $dim V = 3$ , nullity $2$ + rank $1 = 3$ ✓.

Figure — Linear transformations — definition, kernel, image

Level 3 — Analysis

Exercise 3.1

A linear map $T : R^{4} \to R^{3}$ has $dim (ker T) = 2$ . Is $T$ injective? Is it surjective? Justify.

Recall Solution

Injective? $T$ is injective $⟺ ker T = {0}$ (parent §2). Here $dim (ker T) = 2 > 0$ , so the kernel contains nonzero vectors. Not injective — see Injective surjective bijective.

Surjective? Use Rank–Nullity Theorem: $dim V = 4 = nullity + rank = 2 + rank$ , so $rank = 2$ . The image sits in $W = R^{3}$ but has dimension only $2 < 3$ . Why that settles it: a 2-dimensional image cannot fill a 3-dimensional space. Not surjective.

Exercise 3.2

The projection $P : R^{3} \to R^{3}$ , $P (x, y, z) = (x, y, 0)$ , flattens space onto the $x y$ -plane. Find $ker P$ , $im P$ , and verify rank–nullity.

Recall Solution

Kernel — what gets crushed to $0$ : solve $(x, y, 0) = (0, 0, 0)$ , giving $x = 0, y = 0$ , $z$ free. $ker P = span {(0, 0, 1)} = the z -axis, nullity = 1.$ What it looks like: the whole vertical $z$ -axis is the shadow-line that lands on the origin — every point directly above/below the origin collapses onto it.

Image — what can come out: every output has third coordinate $0$ , and any $(a, b, 0)$ is reachable via $P (a, b, anything)$ . $im P = {(a, b, 0)} = the x y -plane, rank = 2.$ Rank–nullity: $dim V = 3 = 1 + 2$ ✓.

Level 4 — Synthesis

Exercise 4.1

Let $D : P_{2} \to P_{2}$ be differentiation on polynomials of degree $\leq 2$ , i.e. $D (a + b x + c x^{2}) = b + 2 c x$ . Using the basis ${1, x, x^{2}}$ , find the matrix of $D$ , then $ker D$ , $im D$ , and check rank–nullity.

Recall Solution

Why differentiation is linear: $\frac{d}{d x} (f + g) = f^{'} + g^{'}$ and $\frac{d}{d x} (c f) = c f^{'}$ — exactly additivity and homogeneity. So $D$ is a genuine linear transformation on the vector space of polynomials.

Matrix (via basis images): feed each basis polynomial in. $D (1) = 0$ , $D (x) = 1$ , $D (x^{2}) = 2 x$ . Writing outputs in coordinates $(a, b, c)$ for $a + b x + c x^{2}$ : $D (1) = (0, 0, 0), D (x) = (1, 0, 0), D (x^{2}) = (0, 2, 0) \Rightarrow M = 000100020 .$ Kernel: $D (p) = 0$ means $b + 2 c x = 0$ for all $x$ , forcing $b = 0, c = 0$ ; $a$ free. $ker D = span {1} = {constants}, nullity = 1.$ Why: the derivative of a constant is zero — constants are exactly the information a derivative loses. Image: outputs $b + 2 c x$ range over all polynomials of degree $\leq 1$ . $im D = span {1, x}, rank = 2.$ Rank–nullity: $dim P_{2} = 3 = 1 + 2$ ✓.

Exercise 4.2

Build a linear map $T : R^{3} \to R^{3}$ whose kernel is the line $span {(1, 1, 1)}$ and whose image is the plane $z = 0$ . Give an explicit matrix, or explain why impossible.

Recall Solution

Feasibility first (rank–nullity): we want nullity $= 1$ and rank $= 2$ ; $1 + 2 = 3 = dim R^{3}$ ✓, so it is possible. Strategy: we need a matrix $A$ with (i) $A (1, 1, 1)^{⊤} = 0$ and (ii) column space $= {z = 0}$ , i.e. all outputs have third coordinate $0$ → the whole bottom row of $A$ is zero. Try $A = 100 - 1 10 0 - 1 0 .$ Check kernel: $A (1, 1, 1)^{⊤} = (1 - 1, 1 - 1, 0) = (0, 0, 0)$ ✓. Solving $A x = 0$ : $x - y = 0, y - z = 0 \Rightarrow x = y = z$ , so $ker T = span {(1, 1, 1)}$ ✓, nullity 1. Check image: top two rows are independent, third row zero, so outputs are all $(a, b, 0)$ = the plane $z = 0$ ✓, rank 2. Done. (Many valid answers exist; connects to Change of basis if you build $T$ by choosing where basis vectors go.)

Level 5 — Mastery

Exercise 5.1

Let $T : V \to V$ be linear on a finite-dimensional space with $T \circ T = T$ (a projection, or idempotent map). Prove that $V = ker T \oplus im T$ — i.e. every $v$ splits uniquely as (kernel part) + (image part).

Recall Solution

What " $\oplus$ " means: two subspaces $A, B$ satisfy $V = A \oplus B$ when (1) $A \cap B = {0}$ and (2) $A + B = V$ (every vector is a sum of one from each). We must show both for $A = ker T$ , $B = im T$ .

Step 1 — spanning ( $A + B = V$ ). For any $v$ , write the clever identity $v = call it k (v - T (v)) + call it i T (v) .$ Why this split: $T (v)$ is manifestly in the image. Check $k \in ker T$ : apply $T$ and use $T \circ T = T$ : $T (k) = T (v) - T (T (v)) = T (v) - T (v) = 0. ✓$ So $v = k + i$ with $k \in ker T$ , $i \in im T$ . Every vector splits.

Step 2 — trivial intersection ( $A \cap B = {0}$ ). Take $w \in ker T \cap im T$ . Being in the image, $w = T (u)$ for some $u$ . Being in the kernel, $T (w) = 0$ . Then $0 = T (w) = T (T (u)) = T (u) = w .$ Why the middle equality: $T \circ T = T$ collapses $T (T (u))$ to $T (u)$ . Hence $w = 0$ , so the only shared vector is $0$ .

Both conditions hold, therefore $V = ker T \oplus im T$ . $■$ (This is why a projection cleanly separates "what it keeps" from "what it kills" — echoing Exercise 3.2. Linked idea: Eigenvalues and eigenvectors, since idempotents have eigenvalues only $0$ and $1$ .)

Exercise 5.2

Suppose $T : R^{n} \to R^{n}$ is linear and injective. Prove it is automatically surjective (hence bijective). Then give a counterexample showing this fails when the map goes $R^{n} \to R^{m}$ with $m \neq = n$ .

Recall Solution

Why the tool: rank–nullity. Injective means $ker T = {0}$ (parent §2), so nullity $= 0$ . Then $n = dim R^{n} = nullity 0 + rank \Rightarrow rank = n .$ The image is an $n$ -dimensional subspace of $R^{n}$ , and the only $n$ -dimensional subspace of an $n$ -dimensional space is the whole space. So $im T = R^{n}$ → surjective, hence bijective (see Injective surjective bijective).

Why it needs equal dimensions: the argument used $dim V = dim W = n$ . Break that and it collapses. Counterexample $T : R^{2} \to R^{3}$ , $T (x, y) = (x, y, 0), A = 100010 .$ $ker T = {0}$ (injective), yet $im T = {z = 0} \neq = R^{3}$ (rank 2 < 3), so not surjective. Injectivity does not force surjectivity when the spaces differ in size.

Recap

Recall One-line takeaways per level

L1 — $T (0) = 0$ is necessary, not sufficient; always test scaling. L2 — image = column space; repeated columns don't raise rank. L3 — decide inject/surject by comparing kernel to ${0}$ and rank to $dim W$ . L4 — kernel and image are chained by rank–nullity; construct with both in view. L5 — "injective ⇒ surjective" needs equal dimensions; projections split $V = ker \oplus im$ .