Linearity線型性せんけいせい and linear maps線型写像せんけいしゃぞう - Basic Exercises

date2026-07-14document_iddoc_067c66eb0599fbb0489998a59d592bd9description線型性を加法性と同次性から判定し、基底ベクトルの像と行列の列の関係を確認する基本演習である。prerequisites線型性の基本 / ベクトルの基本演算 / 線型結合と張る空間の基本type問題演習content_typeexercisestatusactive

mathlinear-algebraexerciselinear-maplinearity

data/lecture/math/linear-algebra/linearity-basics.lecture.n.md data/lecture/math/linear-algebra/linear-maps-and-matrices.lecture.n.md

1Exercise plan

To decide whether a map is a linear map線型写像せんけいしゃぞう, do not rely on a few numerical examples. Check additivity加法性かほうせい and homogeneity同次性どうじせい for arbitrary vectorsベクトル and scalarsスカラー.

To give a counterexample反例はんれい, one example where additivity or homogeneity fails is enough. To prove linearity, compute with arbitrary symbols.

2Problem 1

Define $T : R^{2} \to R^{2}$ by

T (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x + y \\ 2 x - y \end{matrix}) .

Show that $T$ is a linear map線型写像せんけいしゃぞう.

2.1Answer example

Let $u = (x_{1}, y_{1})^{T}$ and $v = (x_{2}, y_{2})^{T}$ . Then

T (u + v) = (\begin{matrix} x_{1} + x_{2} + y_{1} + y_{2} \\ 2 (x_{1} + x_{2}) - (y_{1} + y_{2}) \end{matrix}),

and

T (u) + T (v) = (\begin{matrix} x_{1} + y_{1} \\ 2 x_{1} - y_{1} \end{matrix}) + (\begin{matrix} x_{2} + y_{2} \\ 2 x_{2} - y_{2} \end{matrix}) = (\begin{matrix} x_{1} + x_{2} + y_{1} + y_{2} \\ 2 (x_{1} + x_{2}) - (y_{1} + y_{2}) \end{matrix}) .

Thus additivity holds. For a scalar $c$ ,

T (c u) = (\begin{matrix} c x_{1} + c y_{1} \\ 2 c x_{1} - c y_{1} \end{matrix}) = c (\begin{matrix} x_{1} + y_{1} \\ 2 x_{1} - y_{1} \end{matrix}) = c T (u),

so homogeneity holds. Therefore $T$ is linear.

2.2Explanation

The proof returns to the definition定義ていぎ of linearity. A formula that looks linear is not proved linear until additivity and homogeneity are checked for arbitrary inputs.

3Problem 2

Define $S : R^{2} \to R^{2}$ by

S (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x + 1 \\ y \end{matrix}) .

Show that $S$ is not a linear map.

3.1Answer example

A linear map must satisfy $S (0) = 0$ , but

S (\begin{matrix} 0 \\ 0 \end{matrix}) = (\begin{matrix} 1 \\ 0 \end{matrix}) \neq (\begin{matrix} 0 \\ 0 \end{matrix}) .

Therefore $S$ is not linear.

3.2Explanation

A translation平行移動へいこういどう sends lines to lines, but it does not keep the origin原点げんてん fixed. Checking $T (0) = 0$ is often the fastest way to find a counterexample.

4Problem 3

Define $R : R^{2} \to R^{2}$ by

R (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x^{2} \\ y \end{matrix}) .

Show that $R$ is not linear by showing failure of homogeneity同次性どうじせい.

4.1Answer example

Let $u = (1, 0)^{T}$ and $c = 2$ . Then

R (2 u) = R (\begin{matrix} 2 \\ 0 \end{matrix}) = (\begin{matrix} 4 \\ 0 \end{matrix}),

but

2 R (u) = 2 (\begin{matrix} 1 \\ 0 \end{matrix}) = (\begin{matrix} 2 \\ 0 \end{matrix}) .

Thus $R (2 u) \neq 2 R (u)$ , so homogeneity fails.

4.2Explanation

A nonlinear非線型ひせんけい term such as $x^{2}$ does not preserve scalar multiplication. Preserving the origin alone is not enough for linearity.

5Problem 4

A linear map $T : R^{3} \to R^{2}$ satisfies

T (e_{1}) = (\begin{matrix} 1 \\ 0 \end{matrix}), T (e_{2}) = (\begin{matrix} - 1 \\ 2 \end{matrix}), T (e_{3}) = (\begin{matrix} 3 \\ 1 \end{matrix}) .

Find the matrix of $T$ in the standard basis標準基底ひょうじゅんきてい, and compute $T (2, - 1, 4)^{T}$ .

5.1Answer example

The columns are the images of the standard basis vectors:

A = (\begin{matrix} 1 & - 1 & 3 \\ 0 & 2 & 1 \end{matrix}) .

Since $(2, - 1, 4)^{T} = 2 e_{1} - e_{2} + 4 e_{3}$ ,

T (2, - 1, 4)^{T} = 2 T (e_{1}) - T (e_{2}) + 4 T (e_{3}) = (\begin{matrix} 15 \\ 2 \end{matrix}) .

5.2Explanation

The columns of a matrix record the destinations of the basis vectors基底きていベクトル. Linearity lets every input be computed using the same coefficients on the images.

6Problem 5

Let $T : V \to W$ be linear and suppose $v_{2} = 3 v_{1}$ . State the relation between $T (v_{2})$ and $T (v_{1})$ .

6.1Answer example

By homogeneity,

T (v_{2}) = T (3 v_{1}) = 3 T (v_{1}) .

6.2Explanation

A linear map preserves linear-combination線型結合せんけいけつごう relations. Therefore, if the input side has a dependent relation, the output side keeps the same coefficients in the corresponding relation. The converse need not hold: even if the output vectors are dependent, the same relation need not have existed on the input side, because the map may collapse directions.

7Problem 6

Define $T : C \to C$ by $T (z) = \overline{z}$ . Show that $T$ is real-linear実線型じつせんけい when $C$ is a vector space over $R$ , but not complex-linear複素線型ふくそせんけい over $C$ .

7.1Answer example

For $z, w \in C$ ,

T (z + w) = \overline{z + w} = \overline{z} + \overline{w} = T (z) + T (w) .

If $a \in R$ , then

T (a z) = \overline{a z} = a \overline{z} = a T (z),

so $T$ is real-linear. But for the complex scalar $i$ ,

T (i \cdot 1) = \overline{i} = - i,

while

i T (1) = i .

Thus $T (i \cdot 1) \neq i T (1)$ , so $T$ is not complex-linear.

7.2Explanation

Linearity depends on which field体たい supplies the scalars. The same map may be linear over $R$ but not over $C$ .

8Related lectures

data/lecture/math/linear-algebra/linearity-basics.lecture.n.md data/lecture/math/linear-algebra/linear-maps-and-matrices.lecture.n.md data/lecture/math/linear-algebra/linear-combinations-and-spans.lecture.n.md data/lecture/math/linear-algebra/vector-spaces-and-bases.lecture.n.md

9Proof exercise: the kernel核かく and image像ぞう are subspaces部分空間ぶぶんくうかん

9.1Problem

Let $T : V \to W$ be a linear map. Prove that $\ker T$ is a subspace of $V$ and $Im T$ is a subspace of $W$ .

9.2Answer

If $u, v \in \ker T$ and $a, b$ are scalars, then by linearity

T (a u + b v) = a T (u) + b T (v) = 0.

Thus $a u + b v \in \ker T$ .

If $y_{1}, y_{2} \in Im T$ , write $y_{1} = T (v_{1})$ and $y_{2} = T (v_{2})$ . Then

a y_{1} + b y_{2} = T (a v_{1} + b v_{2}),

so $a y_{1} + b y_{2} \in Im T$ .

9.3Explanation

The kernel is the set of directions collapsed to 0, and the image is the reachable range. Both are closed under linear combinations.