Vectors and linear combinations線型結合せんけいけつごう - Basic Exercises

1Exercise plan

Read vectors not only as lists of components成分せいぶん, but also as points or arrows in space. A linear combination線型結合せんけいけつごう records how much of each given direction is mixed, and a span records the range reachable from those directions.

Use the same checkpoint in every problem: a linear combination線型結合せんけいけつごう is an equality of vectors, so the unknowns are the coefficients, not new vectors. Translate the vector equation into coordinate equations, solve those equations, and then interpret the result as membership in a span, failure of spanning, or a dependence relation.

Pay special attention to the zero vector and to non-unique coefficients. A nontrivial coefficient relation giving $0$0 proves linear dependence線型従属せんけいじゅうぞく; a system with no solution proves that the target vector is outside the span. These two conclusions answer different questions and should not be conflated.

2Problem 1

For

find $2u-3v$2u-3v and describe the geometric meaning.

2.1Answer example

This means moving twice in the direction of $u$u and three times in the opposite direction of $v$v.

2.2Explanation

Addition and scalar multiplication are componentwise. The result is one vector, constructed as a linear combination線型結合せんけいけつごう of $u$u and $v$v.

3Problem 2

Let

Decide whether $b$b is a linear combination of $a_1,a_2$a_1,a_2, and find coefficients if it is.

3.1Answer example

Set $x{a}_1+y{a}_2=b$xa_1+ya_2=b. Then

Solving gives $y=\frac{11}{5}$y=\frac{11}{5} and $x=\frac{12}{5}$x=\frac{12}{5}. Therefore

3.2Explanation

Membership in a span is equivalent to solvability of the coefficient equation. This is not just substitution; it is a test of whether $b$b lies in the space spanned by $a_1,a_2$a_1,a_2.

4Problem 3

Determine whether

are linearly independent一次独立いちじどくりつ or linearly dependent一次従属いちじじゅうぞく.

4.1Answer example

Set $c_1{w}_1+c_2{w}_2+c_3{w}_3=0$c_1w_1+c_2w_2+c_3w_3=0. This gives

The first two equations give $c_1=-2c_3$c_1=-2c_3 and $c_2=c_3$c_2=c_3, and the third then holds identically. Taking $c_3=1$c_3=1 gives

Thus the vectors are linearly dependent.

4.2Explanation

A nontrivial coefficient relation proves dependence. Here $w_3=2w_1-w_2$w_3=2w_1-w_2, so no new direction is added.

5Problem 4

Describe geometrically the space spanned by

5.1Answer example

Since $q=2p$q=2p,

which is a line through the origin.

5.2Explanation

The dimension of a span is determined by independent directions, not by the number of listed vectors. This viewpoint leads to rank階数かいすう.

6Problem 5

Decide whether each statement is true and explain why.

1. A vector set containing the zero vector is linearly dependent.
2. A one-vector set $\left\{u\right\}$\{u\} is linearly independent if $u\ne 0$u\ne0.
3. The empty set is linearly independent in the standard definition.

6.1Answer example

All are true.

6.2Explanation

If the zero vector is included, coefficient 1 on that vector and 0 on all others gives a nontrivial combination equal to 0. If $u\ne 0$u\ne0, then $cu=0$cu=0 implies $c=0$c=0. The empty set has no nontrivial coefficients that could give a counterexample, so it is defined as linearly independent.

7Supplementary problems: checks moved from the lecture

7.1Problem 6

For $u=(2,-1,3{)}^T$u=(2,-1,3)^T and $v=(0,4,1{)}^T$v=(0,4,1)^T, compute $u+v$u+v and $3u-2v$3u-2v.

7.2Answer example

$u+v=(2,3,4{)}^T$u+v=(2,3,4)^T and $3u-2v=(6,-11,7{)}^T$3u-2v=(6,-11,7)^T. This checks componentwise vector addition and scalar multiplication.

7.3Problem 7

Find $a,b$a,b such that $a(1,0{)}^T+b(0,1{)}^T=(5,-2{)}^T$a(1,0)^T+b(0,1)^T=(5,-2)^T.

7.4Answer example

$a=5$a=5 and $b=-2$b=-2. In the standard basis標準基底ひょうじゅんきてい, coefficients are the components.

7.5Problem 8

Use components to identify the direction represented by $2(1,1{)}^T-(3,-1{)}^T$2(1,1)^T-(3,-1)^T.

7.6Answer example

$2(1,1{)}^T-(3,-1{)}^T=(-1,3{)}^T$2(1,1)^T-(3,-1)^T=(-1,3)^T.

7.7Problem 9

Decide whether $(1,0{)}^T$(1,0)^T and $(1,1{)}^T$(1,1)^T span $R^2$\mathbb R^2.

7.8Answer example

$a(1,0{)}^T+b(1,1{)}^T=(a+b,b{)}^T$a(1,0)^T+b(1,1)^T=(a+b,b)^T. For any $(x,y{)}^T$(x,y)^T, take $b=y$b=y and $a=x-y$a=x-y. Thus they span $R^2$\mathbb R^2.

7.9Problem 10

Describe the span of $(1,2{)}^T$(1,2)^T and $(2,4{)}^T$(2,4)^T.

7.10Answer example

Since $(2,4{)}^T=2(1,2{)}^T$(2,4)^T=2(1,2)^T, the span is $\left\{t\right(1,2{)}^T\mid t\in \mathbb{R}\}$\{t(1,2)^T\mid t\in\mathbb R\}.

7.11Problem 11

Confirm that $(3,1{)}^T$(3,1)^T is a linear combination of $(1,0{)}^T,(0,1{)}^T$(1,0)^T,(0,1)^T.

7.12Answer example

$(3,1{)}^T=3(1,0{)}^T+1(0,1{)}^T$(3,1)^T=3(1,0)^T+1(0,1)^T.

8Supplementary problems: zero vector and boundary cases

8.1Problem 12

For $u=\left(\begin{array}{c}1\\ 2\end{array}\right)$u=\begin{pmatrix}1\\2\end{pmatrix}, find $0u$0u, $-u$-u, and $u+(-u)$u+(-u) and explain each.

8.2Answer example

$0u$0u is the zero vector, $-u$-u points opposite to $u$u, and $u+(-u)$u+(-u) returns to zero by the additive inverse.

This problem checks boundary cases in vector operations. The vector $0u$0u no longer preserves the direction of $u$u, while $-u$-u keeps the same line but reverses the direction.

8.3Problem 13

For $u=\left(\begin{array}{c}1\\ 2\end{array}\right)$u=\begin{pmatrix}1\\2\end{pmatrix}, explain why

are the same set.

8.4Answer example

Since $0=0u$0=0u and $2u$2u is a scalar multiple of $u$u, adding them does not enlarge the set of linear combinations. Therefore

This problem checks that adding redundant vectors to a generating set does not necessarily change the span. Increasing the number of listed vectors is different from increasing the number of independent directions.

9Related lectures

10Proof exercise: minimality of span

10.1Problem

Let $S=\{v_1,\dots ,v_k\}$S=\{v_1,\ldots,v_k\}. Prove that $\mathrm{span}\left(S\right)$\operatorname{span}(S) is the smallest subspace部分空間ぶぶんくうかん containing $S$S.

10.2Answer

Elements of $\mathrm{span}\left(S\right)$\operatorname{span}(S) are $x=\sum a_i{v}_i$x=\sum a_iv_i and $y=\sum b_i{v}_i$y=\sum b_iv_i. For scalars $c,d$c,d,

so the span is closed under linear combinations and is a subspace. If $U$U is any subspace containing $S$S, it contains every linear combination of elements of $S$S, so $\mathrm{span}\left(S\right)\subseteq U$\operatorname{span}(S)\subseteq U.

10.3Explanation

The span is the smallest linear world built from the given vectorsベクトル using only linear combinations.