Vector spaces, bases基底きてい, and rank階数かいすう - Basic Exercises

1Exercise plan

A basis基底きてい is a measuring system that represents every point of a space by unique coordinates. Rank階数かいすう is the dimension次元じげん that survives under a linear map線型写像せんけいしゃぞう and reaches the output side.

For basis and rank problems, first separate the two requirements: spanning and linear independence線型独立せんけいどくりつ. Having the right number of vectors is useful only after you know the ambient dimension and one of the two requirements. Row reduction is the common tool because pivots identify independent columns, free variables, and the rank at the same time.

When a problem asks for rank, nullity, or a basis of a subspace, keep track of which space the vectors live in. Pivot columns of the original matrix give a basis for the column space, nonzero rows of an echelon form give a basis for the row space, and free-variable solutions give a basis for the null space. These are related by rank-nullity but are not interchangeable lists of vectors.

2Problem 1

Confirm that

is a subspace部分空間ぶぶんくうかん of $R^3$\mathbb R^3.

2.1Answer example

The zero vector is included because $0+0+0=0$0+0+0=0. If $u,v\in W$u,v\in W, both component sums are 0, so the component sum of $u+v$u+v is 0. For $c\in \mathbb{R}$c\in\mathbb R, the component sum of $cu$cu is $c·0=0$c\cdot0=0. Hence $W$W is a subspace.

2.2Explanation

To check a subspace, test the zero vector, closure under sums, and closure under scalar multiples.

3Problem 2

Explain why

are not a basis基底きてい of $W$W.

3.1Answer example

$b_1$b_1 is not in $W$W because $1+1+0=2$1+1+0=2. Therefore $b_1,b_2$b_1,b_2 are not a basis of $W$W.

3.2Explanation

A basis vector must first belong to the target space. Check membership before linear independence or spanning.

4Problem 3

Show that

are a basis of $W$W, and find the coordinates of

with respect to this basis.

4.1Answer example

Both $u_1,u_2$u_1,u_2 have component sum 0, so they lie in $W$W. If $a{u}_1+b{u}_2=0$au_1+bu_2=0, then

so $a=0$a=0 and $b=0$b=0. They are linearly independent. Any $(x,y,z{)}^T\in W$(x,y,z)^T\in W satisfies $x=-y-z$x=-y-z, so

Thus they span $W$W and form a basis. For $w=a{u}_1+b{u}_2$w=au_1+bu_2, $-a=-1$-a=-1 and $-b=-2$-b=-2, so $a=1,b=2$a=1,b=2. The coordinate vector is

4.2Explanation

Check basis in the order membership, linear independence, and spanning. Coordinates are the coefficients used to mix basis vectors.

5Problem 4

For the linear map $T:R^3\to R^2$T:\mathbb R^3\to\mathbb R^2 represented by

find $\mathrm{rank}A$\operatorname{rank}A and $\dim \ker T$\dim\ker T.

5.1Answer example

The two rows are linearly independent, so there are two pivots:

By rank-nullity,

5.2Explanation

The input has 3 dimensions. Two dimensions survive in the image像ぞう, and one dimension collapses into the kernel核かく.

6Problem 5

For finite-dimensional spaces of the same dimension, explain the relation between a linear map $T:R^2\to R^2$T:\mathbb R^2\to\mathbb R^2 being invertible可逆かぎゃく and $\ker T=\left\{0\right\}$\ker T=\{0\}.

6.1Answer example

If $\ker T=\left\{0\right\}$\ker T=\{0\}, no nonzero difference of inputs is collapsed to zero, so $T$T is injective. For $R^2\to R^2$\mathbb R^2\to\mathbb R^2, injective implies surjective, so $T$T is invertible. Conversely, if $T$T is invertible and $T\left(x\right)=0$T(x)=0, then $x=T^{-1}\left(0\right)=0$x=T^{-1}(0)=0.

6.2Explanation

The equal-dimension assumption is essential; in unequal dimensions injectivity and surjectivity are not generally equivalent.

7Supplementary problems: column vectors and rank

7.1Problem 6

Decide whether $(1,0{)}^T,(2,0{)}^T$(1,0)^T,(2,0)^T are linearly independent.

7.2Answer example

They are linearly dependent because $(2,0{)}^T=2(1,0{)}^T$(2,0)^T=2(1,0)^T.

7.3Problem 7

Give the rank of $(1,0{)}^T,(0,1{)}^T,(1,1{)}^T$(1,0)^T,(0,1)^T,(1,1)^T.

7.4Answer example

$(1,0{)}^T,(0,1{)}^T$(1,0)^T,(0,1)^T are independent and $(1,1{)}^T=(1,0{)}^T+(0,1{)}^T$(1,1)^T=(1,0)^T+(0,1)^T, so the rank is 2.

7.5Problem 8

Explain why a column relation can be written as $Ac=0$Ac=0.

7.6Answer example

If $A=[a_1\cdots a_n]$A=[a_1\ \cdots\ a_n] and $c=(c_1,\dots ,c_n{)}^T$c=(c_1,\ldots,c_n)^T, then $Ac=c_1{a}_1+\cdots +c_n{a}_n$Ac=c_1a_1+\cdots+c_na_n. Thus a column relation is exactly $Ac=0$Ac=0.

7.7Problem 9

For

decide which are subspaces and state which condition fails when not.

7.8Answer example

$W_1$W_1 is not a subspace because it does not contain the zero vector. $W_2$W_2 is a subspace. $W_3$W_3 contains the zero vector but is not closed under scalar multiplication, since $(1,0,0{)}^T\in W_3$(1,0,0)^T\in W_3 while $(-1,0,0{)}^T\notin W_3$(-1,0,0)^T\notin W_3.

7.9Problem 10

Use row operations on

to find $\mathrm{rank}A$\operatorname{rank}A, a basis for the column space, and $\dim \ker A$\dim\ker A.

7.10Answer example

Pivots are in columns 1 and 2, so $\mathrm{rank}A=2$\operatorname{rank}A=2. A basis for the column space is given by the original columns

Since there are 3 unknowns, $\dim \ker A=3-2=1$\dim\ker A=3-2=1.

7.11Problem 11

Explain how the basis and dimension of $\mathbb{C}$\mathbb C change when viewed over $\mathbb{C}$\mathbb C versus over $\mathbb{R}$\mathbb R.

7.12Answer example

Over $\mathbb{C}$\mathbb C, a basis is $\left\{1\right\}$\{1\} and the dimension is 1. Over $\mathbb{R}$\mathbb R, a basis is $\{1,i\}$\{1,i\} and the dimension is 2, because $a+bi=a·1+b·i$a+bi=a\cdot1+b\cdot i with real coefficients.

7.13Problem 12

Represent $v=(3,1{)}^T$v=(3,1)^T in the standard basis $E$E and in

7.14Answer example

For the basis $B$B, solve $v=a(1,1{)}^T+b(1,-1{)}^T$v=a(1,1)^T+b(1,-1)^T. Then $a+b=3$a+b=3 and $a-b=1$a-b=1, so $a=2,b=1$a=2,b=1. Thus

The vector is unchanged, but its coordinates change.

7.15Problem 13

View $0_{2\times 3}$0_{2\times3} as the matrix of the zero map $R^3\to R^2$\mathbb R^3\to\mathbb R^2. Find its rank, column space, and kernel dimension.

7.16Answer example

All columns are zero, so $\mathrm{rank}{0}_{2\times 3}=0$\operatorname{rank}0_{2\times3}=0. The column space is $\left\{0\right\}\subset R^2$\{0\}\subset\mathbb R^2. Every $x\in R^3$x\in\mathbb R^3 maps to 0, so $\ker 0_{2\times 3}=R^3$\ker0_{2\times3}=\mathbb R^3 and $\dim \ker 0_{2\times 3}=3$\dim\ker0_{2\times3}=3. Thus rank plus nullity is $0+3=3$0+3=3.

7.17Problem 14

For

solve $Ac=0$Ac=0 and choose a basis for the column space.

7.18Answer example

Since $a_3=a_1+a_2$a_3=a_1+a_2,

Thus

solves $Ac=0$Ac=0. The columns are dependent, and $a_1,a_2$a_1,a_2 form a basis for the column space.

7.19Problem 15

For $A=\left(\begin{array}{cc}1& 1\\ 2& 2\end{array}\right)$A=\begin{pmatrix}1&1\\2&2\end{pmatrix}, decide solvability of $Ax=b$Ax=b for $b_1=(3,6{)}^T$b_1=(3,6)^T and $b_2=(3,7{)}^T$b_2=(3,7)^T by comparing ranks.

7.20Answer example

Since the second row of $A$A is twice the first, $\mathrm{rank}A=1$\operatorname{rank}A=1. For $b_1$b_1, the right side has the same relation, so $\mathrm{rank}(A\mid b_1)=1$\operatorname{rank}(A\mid b_1)=1 and solutions exist. Since there are 2 unknowns and rank 1, infinitely many solutions exist.

For $b_2$b_2,

so $\mathrm{rank}(A\mid b_2)=2\ne 1$\operatorname{rank}(A\mid b_2)=2\ne1 and no solution exists.

7.21Explanation

Equality of the ranks of the coefficient and augmented matrices determines existence; comparison with the number of unknowns determines uniqueness.

8Related lectures

9Proof exercise: uniqueness of basis coordinates and rank-nullity

9.1Problem

Prove that if $B=(v_1,\dots ,v_n)$B=(v_1,\ldots,v_n) is a basis, then coordinate representation is unique. Also explain rank-nullity:

9.2Answer

If $v=\sum x_i{v}_i=\sum y_i{v}_i$v=\sum x_iv_i=\sum y_iv_i, then

Since the $v_i$v_i are linearly independent, every $x_i-y_i=0$x_i-y_i=0, so $x_i=y_i$x_i=y_i.

For rank-nullity, extend a basis of $\ker T$\ker T to a basis of $V$V. The images of the added basis vectors form a basis of $\mathrm{Im}T$\operatorname{Im}T, splitting the dimension into kernel and image parts.

9.3Explanation

A basis makes coordinates unique. Rank-nullity separates those coordinates into collapsed and surviving directions.