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Wize University Linear Algebra Textbook > Vector Spaces and Subspaces

Basis and Dimension

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Basis and Dimension

Let VV be a vector space, and let B={v⃗1,…,v⃗n}B = \{\vec v_1,\ldots,\vec v_n\} be a set of vectors in VV. BB is a basis for VV if:
  • BB is linearly independent
  • BB is a generating set of VV (VV is the span of the vectors in BB)
In other words, a basis BB is the smallest set such that every vector in VV is a linear combination of the vectors in BB.
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Example 1
B={[100],[010],[001]}B=\left\{ \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix}, \begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix} \right\} is a basis for R3\mathbb{R}^3 because:
  • The vectors in BB are linearly independent
  • Every vector in R3\mathbb{R}^3 can be generated from the vectors in BB
Note: removing any vector would mean that the set is no longer a generating set, so it is as small as possible.
Example 2

B={[1000],[0100],[0010],[0001]}B=\left\{\left[\begin{array}{rr}1&0\\0&0\end{array}\right], \left[\begin{array}{rr}0&1\\0&0\end{array}\right], \left[\begin{array}{rr}0&0\\1&0\end{array}\right], \left[\begin{array}{rr}0&0\\0&1\end{array}\right]\right\} is a basis for M2×2M_{2\times 2}.
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Dimension

The dimension of a vector space VVis the number of vectors in its basis, which we denote dim(V)\text{dim}(V).
Example 1
B={[100],[010],[001]}B=\left\{ \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix}, \begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix} \right\} is a basis for R3\mathbb{R}^3, so dim(R3)=3\text{dim}(\reals^3) = 3.
Example 2
B={[1000],[0100],[0010],[0001]}B=\left\{\left[\begin{array}{rr}1&0\\0&0\end{array}\right], \left[\begin{array}{rr}0&1\\0&0\end{array}\right], \left[\begin{array}{rr}0&0\\1&0\end{array}\right], \left[\begin{array}{rr}0&0\\0&1\end{array}\right]\right\} is a basis for M2×2M_{2\times2}, so dim(M2×2)=4\text{dim}(M_{2 \times 2}) = 4.
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Example: Verifying a Basis

Let VV be the vector space {(x,y)  ∣  x,y∈R+}\{(x,y)\;|\;x,y\in\mathbb{R}^+\} with operations:
  • (x1,y1)+(x2,y2)=(x1x2, y1y2)(x_1,y_1)+(x_2,y_2)=(x_1x_2,\ y_1y_2)
  • k⋅(x,y)=(xk,yk)k\cdot(x,y)=(x^k,y^k)
Show that B={(10,1),(1,10)}B=\left\{ (10,1),(1,10) \right\} a basis for VV. [Hint: 0→=(1,1)\overrightarrow{0}=(1,1)]
Step 1
Show that the vectors in BB are linearly independent.
c1⋅(10,1)+c2⋅(1,10)=0⃗  ⟹  (10c1,1c1)+(1c2,10c2)=(1,1)  ⟹  (10c1,1)+(1,10c2)=(1,1)  ⟹  (10c1⋅1,1⋅10c2)=(1,1)  ⟹  (10c1,10c2)=(1,1)\begin{array}{rlrrl} c_1\cdot(10,1)+c_2\cdot(1,10)=\vec{0} \quad &\implies& (10^{c_1},1^{c_1})+(1^{c_2},10^{c_2})&=&(1,1)\\[0.5em] &\implies& (10^{c_1},1)+(1,10^{c_2})&=&(1,1)\\[0.5em] &\implies& (10^{c_1}\cdot 1,1\cdot 10^{c_2})&=&(1,1)\\[0.5em] &\implies& (10^{c_1},10^{c_2})&=&(1,1)\\[0.5em] \end{array}
So we need 10c1=110^{c_1}=1 and 10c2=110^{c_2}=1.
The only solution to these equations is c1=c2=0c_1=c_2=0, so BB is linearly independent.
Step 2
Show that BB is a generating set for VV.
We must show that every vector (x,y)(x,y) in VV can be written as a combination of the vectors in BB:
(x,y)=c1⋅(10,1)+c2⋅(1,10)=(10c1,1c1)+(1c2,10c2)=(10c1,1)+(1,10c2)=(10c1⋅1,1⋅10c2)=(10c1,10c2)\begin{array}{rcl} (x,y)&=&c_1\cdot(10,1)+c_2\cdot(1,10)\\[0.5em] &=&(10^{c_1},1^{c_1})+(1^{c_2},10^{c_2})\\[0.5em] &=&(10^{c_1},1)+(1,10^{c_2})\\[0.5em] &=&(10^{c_1}\cdot 1,1\cdot 10^{c_2})\\[0.5em] &=&(10^{c_1},10^{c_2}) \end{array}
From this we can see that:
x=10c1  ⟹  c1=log⁡10(x)y=10c2  ⟹  c2=log⁡10(y)\begin{array}{rcl} x=10^{c_1} &\implies& c_1=\log_{10}(x)\\[0.5em] y=10^{c_2}&\implies&c_2=\log_{10}(y) \end{array}
For every pair (x,y)∈V(x,y)\in V: (x,y)=log⁡(x)⋅(10,1)+log⁡(y)⋅(1,10)(x,y)=\log(x)\cdot(10,1)+\log(y)\cdot(1,10)
Note that x,y∈R+x,y \in \reals^+ so the logarithms are well-defined.

Since we can find appropriate coefficients for any values of xx and yy, BB is a generating set for VV.
Therefore, BB is a basis of VV.
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Example: Finding a Basis

Let VV be the vector space {(x,y)  ∣  x,y∈R+}\{(x,y)\;|\;x,y\in\mathbb{R}^+\} with operations:
  • (x1,y1)+(x2,y2)=(x1x2, y1y2)(x_1,y_1)+(x_2,y_2)=(x_1x_2,\ y_1y_2)
  • k⋅(x,y)=(xk,yk)k\cdot(x,y)=(x^k,y^k)
Consider the following subspace of VV: S={(x,y)∈V  ∣  x3y=1}S=\{(x,y)\in V\;|\;x^3y=1\}.
Find a basis for SS, and find the dimension of SS.
[Hint: for any real number x>0x>0 and any real number kk: xk=eln⁡(xk)=(ek)ln⁡(x){\displaystyle x^k=e^{\ln(x^k)}=(e^k)^{\ln(x)}}]

Step 1
To find a basis, first find a generating set for SS.
If x⃗=(x,y)∈S\vec x=(x,y)\in S is an arbitrary vector, then we know that: x3y=1    ⟹    y=1x3    ⟹    y=x−3x^3y=1 \;\implies\; y=\frac{1}{x^3} \;\implies\; y=x^{-3}
So x⃗=(x,y)=(x,x−3)\vec x=(x,y)=(x,x^{-3}).
Goal: Write x⃗\vec x as a linear combination of vectors in SS.
We will "factor out" the variable xx using the defined operations of VV along with the hint:
(x,x−3)=(x1,x−3)=((e1)ln⁡(x),(e−3)ln⁡(x))using the hint=ln⁡(x)⋅(e1,e−3)by the definition of scalar multiplication=ln⁡(x)⋅(e,e−3)\begin{array}{lcll} (x,x^{-3}) &=& (x^1,x^{-3})\\[0.5em] &=& \big((e^1)^{\ln(x)},(e^{-3})^{\ln(x)}\big)& \text{using the hint}\\[0.5em] &=& \ln(x)\cdot(e^1,e^{-3}) & \text{by the definition of scalar multiplication}\\[0.5em] &=&\ln(x)\cdot(e,e^{-3}) \end{array}
Note that (e,e−3)∈S(e,e^{-3})\in S because (e)3⏟x3(e−3)⏟y=1\underbrace{(e)^3}_{x^3} \underbrace{(e^{-3})}_{y}=1 (or simply: y=x−3y=x^{-3}).
We can generate any vector in SS as a scalar multiple of (e,e−3)(e,e^{-3}): (x,y)=(x,x−3)=ln⁡(x)⋅(e,e−3)(x,y)=(x,x^{-3})=\ln(x)\cdot(e,e^{-3})
Therefore, the generating set is B={(e,e−3)}B=\{(e,e^{-3})\}.
Step 2
BBis linearly independent because it contains only one vector (and it is not the zero vector).
Therefore, BB is a basis for SS. Since BB contains only one vector, SS is a 1-dimensional subspace of VV.

Practice: Finding a Basis

Let A=[1−11]A=\left[\begin{array}{rrr} 1&-1&1\\ \end{array}\right] and consider the special subspace of R3\mathbb{R}^3 called the null space:
N={x⃗∈R3  ∣  Ax⃗=0⃗}N=\{\vec x\in\mathbb{R}^3\;|\;A\vec x=\vec{0}\}
Find a basis for NN, and state dim(N)\text{dim}(N).
Extra Practice
Example: Linear Independence (pre-Matrix)
Show that for x⃗,y⃗∈R2\vec x,\vec y\in \mathbb{R}^2 , if Span{x⃗,y⃗}=R2Span\{\vec x,\vec y\}=\mathbb{R}^2, then x⃗,y⃗\vec x,\vec y are linearly independent.
Practice Question: Bases
Practice Question: Bases
For each of the following subspaces, find the dimension of the subspace by giving a specific example of a basis:
Q:=span{(0,0,0,0),(1,1,0,0),(0,0,1,1),(1,1,1,1)}⊆R4Q:=span\{(0,0,0,0), (1,1,0,0),(0,0,1,1),(1,1,1,1)\}\subseteq\mathbb{R}^4
Linear Independence and Basis
Let v⃗1,v⃗2,v⃗3,v⃗4\vec v_1,\vec v_2,\vec v_3, \vec v_4 be any four vectors in R4\mathbb{R}^4
Explain why the set C={v⃗1−v⃗2,v⃗2−v⃗3,v⃗3−v⃗4,v⃗4−v⃗1}C=\{\vec v_1-\vec v_2,\vec v_2-\vec v_3,\vec v_3-\vec v_4,\vec v_4-\vec v_1\} cannot be a basis for R4\mathbb{R}^4
133-WML2_Quiz-19.4F_eg_17
It is known that the set {v⃗1, v⃗2, … , v⃗n}\{\vv_1,\, \vv_2,\, \dots \,,\, \vv_n\} is a basis for Rn\mathbb{R^n}. If w⃗∈\vw \inRn\mathbb{R^n}, is the set {v⃗1, v⃗2, … , v⃗n, w⃗}\{\vv_1,\, \vv_2,\, \dots \,,\, \vv_n,\, \vw\} linearly in/dependent? Briefly justify your answer (using mathematical notation is preferable to words, if possible).
Which of the following is a basis for R2\mathbb{R}^2?
Show that a subset is a subspace
Let V={(x,y)  ∣  x∈R,y∈R+}V=\{(x,y)\;|\;x\in\mathbb{R},y\in\mathbb{R}^+\} be the vector space with operations:
(x1,y1)+(x2,y2)=(x1+x2,y1y2)(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1y_2)
k⋅(x,y)=(kx,yk)k\cdot(x,y)=(kx,y^k)
Which of the following is a basis for R2\mathbb{R}^2?
Which of the following is a basis for R2\mathbb{R}^2?
Consider the vector space M2,2(R)\mathcal{M}_{2, 2} (\mathbb{R}) . What is the dimension of this vector space?
Dimension of a subspace
In the vector space P2P_2, what is the dimension of the subspace that contains all polynomials of the form p(x)=a+bx+cx2p(x)=a+bx+cx^2, such that a=b+ca=b+c?
133 - FML 3 - 18.1W - e.g. 51
If it is known that the set of vectors {x⃗, y⃗, z⃗, w⃗}\bcb{\left\{ \vec{x},\, \vec{y},\, \vec{z},\, \vec{w} \right\}} form a basis for R4\bcb{\mathbb{R}^4}, show that the set {x⃗+kz⃗,y⃗, z⃗, w⃗}\bcb{\left\{ \vec{x} + k\vec{z}, \vec{y},\, \vec{z},\, \vec{w} \right\}} is also a basis, for any scalar k\bcb{k}.
Verify if the following vectors form a basis for R3\mathbb{R}^3:
{[112],[1−21],[2−11]}\left\{ \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix},\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} \right\}
133 - FML 3 - 18.1W - e.g. 42
Are the vectors [1310], [2−100], [021−1], [36−3−2]\bcb{\begin{bmatrix} 1 \\ 3 \\ 1 \\ 0 \end{bmatrix}, \, \begin{bmatrix} 2 \\ -1 \\ 0 \\ 0 \end{bmatrix}, \, \begin{bmatrix} 0 \\ 2 \\ 1 \\ -1 \end{bmatrix} , \, \begin{bmatrix} 3 \\ 6 \\ -3 \\ -2 \end{bmatrix} } a basis for R4\bcb{\mathbb{R}^4}?
133 - FML 3 - 18.1W - e.g. 44
Find a basis for the set of vectors {[10−1], [2−20], [0−22], [3−41]}\bcb{\left\{ \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix},\, \begin{bmatrix} 2 \\ -2 \\ 0 \end{bmatrix},\, \begin{bmatrix} 0 \\ -2 \\ 2 \end{bmatrix}, \, \begin{bmatrix} 3 \\ -4 \\ 1\end{bmatrix} \right\}}, and determine the dimension of their span.
Practice: Coordinates
Practice: Coordinates
Let B1={[−120],[3−52],[4−73]}B_1= \left\{ \begin{bmatrix} -1\\ 2 \\0\\ \end{bmatrix} , \begin{bmatrix} 3\\ -5\\ 2\\ \end{bmatrix} , \begin{bmatrix} 4\\ -7\\ 3\\ \end{bmatrix} \right\} and B2={[1−43],[52−2],[4−70]}B_2= \left\{ \begin{bmatrix} 1\\ -4 \\3\\ \end{bmatrix} , \begin{bmatrix} 5\\ 2\\ -2\\ \end{bmatrix} , \begin{bmatrix} 4\\ -7\\ 0\\ \end{bmatrix} \right\} be two bases of R3\mathbb{R}^3.
If [x⃗]B1=[−48−7][\vec x]_{\small B_1} = \begin{bmatrix} -4\\ 8\\ -7\\ \end{bmatrix}, find:
Which of the following are a basis for the vector space of polynomials with degree at most 3?
Let {e1,…,en}\{ e_1, \dots, e_n \} be a basis for an nn - dimensional vector space VV , and let L:V→WL : V\rightarrow W be a linear map, where WW is also a vector space. What conditions are necessary for {L(e1),…,L(en)}\{ L(e_1), \dots, L(e_n) \} to be a basis for WW ?
$\tkct{Mock F1}$ 19.4F_Mid_Builder_$\tkcth{8.6.}\tkco{x}$_$\key{original \#1.}\tkct{b}$__$\tkcth{Mock F}\tkct{?}$_$\tkct{no vid}$
If vectors u⃗\vu, v⃗\vv, x⃗\vx and y⃗\vec{y} are vectors in R7\mathbb{R}^7 such that u⃗=−2v⃗+0x⃗+4y⃗\vu = -2\vv + 0\vx +4\vec{y}, then a basis for span{u⃗, v⃗, x⃗, y⃗}\text{span}\big\{ \vu,\, \vv,\, \vx,\, \vec{y} \big\} is {u⃗, v⃗, y⃗}\big\{ \vu,\, \vv,\, \vec{y} \big\}.
Example: Linear Independence (pre-Matrix)
Example:
Show that for x⃗,y⃗∈R2\vec x,\vec y\in \mathbb{R}^2 , if Span{x⃗,y⃗}=R2Span\{\vec x,\vec y\}=\mathbb{R}^2, then x⃗,y⃗\vec x,\vec y are linearly independent.
Basis and Dimension
Practice: Finding a Basis
Consider the vector space M2×2M_{2\times2} with the standard component-wise operations of addition and scalar multiplication
The subset D={[abcd]∈M2×2  ∣  a+d=b+c}D=\left\{ \left[\begin{array}{cc} a&b\\ c&d \end{array}\right]\in M_{2\times2}\;\Big|\;a+d=b+c \right\} is a subspace of M2×2M_{2\times2}
Finding a basis and the dimension of a subspace
In the real vector space R3\mathbb{R}^3 with the standard component-wise operations of vector addition and scalar multiplication, consider the subspace:
W={(x,y,z)∈R3  ∣  y+3z=0}W=\{(x,y,z)\in\mathbb{R}^3\;|\;y+3z=0\}
Find a basis for WW, and use your answer to state the dimension of WW
Concept Clarifier
Determine whether the set T={(1,0,0),(0,−2,2),(1,1,1)}T=\{ (1,0,0),(0,-2,2), (1,1,1) \} is a basis of R3\mathbb{R}^3.
Basis of a Subspace
Find a basis for the following subspace of R3\mathbb{R}^3 and state its dimension:
S:={x⃗∈R3 | x⃗⋅v⃗=0 for every v⃗∈span{(1,−1,1),(1,1,0)}}S:=\{\vec x\in\mathbb{R}^3\text{ | }\vec x\cdot\vec v=0\text{ for every } \vec v\in span\{(1,-1,1),(1,1,0)\}\}
Example: Abstract Basis Proof
Suppose the set B={v1→,v2→,v3→}B=\{\overrightarrow{v_1},\overrightarrow{v_2},\overrightarrow{v_3}\} is a basis for a vector space VV
Define the vectors:
u1→=v1→−v2→u2→=v2→−v3→u3→=−v3→\overrightarrow{u_1}=\overrightarrow{v_1}-\overrightarrow{v_2} \hspace{1cm}\overrightarrow{u_2}=\overrightarrow{v_2}-\overrightarrow{v_3}\hspace{1cm}\overrightarrow{u_3}=-\overrightarrow{v_3}
Basis and Dimension
Example: Verifying a Basis
Let VV be the vector space {(x,y)  ∣  x,y∈R+}\{(x,y)\;|\;x,y\in\mathbb{R}^+\} with operations:
(x1,y1)+(x2,y2)=(x1x2, y1y2)(x_1,y_1)+(x_2,y_2)=(x_1x_2,\ y_1y_2)
Find a basis for a given space
Consider the vector space M2×2M_{2\times2} with the standard component-wise operations of addition and scalar multiplication
The subset D={[abcd]∈M2×2  ∣  a+d=b+c}D=\left\{ \left[\begin{array}{cc} a&b\\ c&d \end{array}\right]\in M_{2\times2}\;\Big|\;a+d=b+c \right\} is a subspace of M2×2M_{2\times2}
Find a basis for DD and the dimension of DD
Show that a set is a basis
Let A=[1−11]A=\left[\begin{array}{rrr} 1&-1&1\\ \end{array}\right] and consider the subspace of R3\mathbb{R}^3:
N={x→=(x,y,z)∈R3  ∣  A.x→=0→}N=\{\overrightarrow{x}=(x,y,z)\in\mathbb{R}^3\;|\;A.\overrightarrow{x}=\overrightarrow{0}\}
Show that B={(1,1,0),(−1,0,1)}B=\{(1,1,0), (-1,0,1)\} is a basis for NN
Consider the space of polynomials of degree 3 or lower, P3\mathcal{P}_3 . Give an example of a linearly independent set of vectors whose span is the entirety of P3\mathcal{P}_3 .
For which value(s) of c do [110],[2c1],[c43]\begin{bmatrix}1\\1\\0\end{bmatrix}, \begin{bmatrix}2\\c\\1\end{bmatrix}, \begin{bmatrix}c\\4\\3\end{bmatrix} form a basis for R3\mathbb{R}^3?
Which of the following sets are a basis for R3\mathbb{R}^3 ?