Subspaces

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Subspaces
A subspace is a mini vector space within a vector space consisting of vectors that behave "similarly".
Definition
Suppose WWW is a non-empty subset of a vector space VVV. Let w⃗1, w⃗2∈W\vec w_1,\ \vec w_2 \in Ww1​, w2​∈W and let k∈Rk \in \realsk∈R.
Then WWW is a subspace of VVV if:
WWW contains the zero vector of VVV: 0⃗∈W\vec 0 \in W0∈W
WWW is closed under vector addition: w⃗1+w⃗2∈W\vec w_1 + \vec w_2 \in Ww1​+w2​∈W
WWW is closed under scalar multiplication: k⋅w⃗1∈Wk\cdot \vec w_1 \in Wk⋅w1​∈W
Wize Tip
To prove that that a subset of a vector space is a subspace, you must show it satisfies all three conditions.
Shortcut: Showing that aw⃗1+bw⃗2a\vec w_1 + b \vec w_2aw1​+bw2​ is in WWW  is sufficient to prove all three conditions at once!
To show that a subset is not a subspace, you must provide an example where one condition fails.
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Example
Use the shortcut to show that W={⟨x,y⟩∈R2  ∣  y=2x}W=\{\lang x,y \rang \in \mathbb{R}^2\;|\;y=2x\}W={⟨x,y⟩∈R2∣y=2x} is a subspace of R2\reals^2R2.

WWW consists of vectors on a line through the origin (0⃗\vec 00) with vectors of the form ⟨x, 2x⟩\lang x,\ 2x \rang⟨x, 2x⟩.
Using the shortcut, we want to show that au⃗+bv⃗∈Wa \vec u +b \vec v \in Wau+bv∈W where u⃗=⟨u,2u⟩,  v⃗=⟨v,2v⟩\vec u=\lang u, 2u \rang, \ \ \vec v=\lang v, 2v \rangu=⟨u,2u⟩,  v=⟨v,2v⟩:
au⃗+bv⃗=a⟨u,2u⟩+b⟨v,2v⟩=⟨au,2au⟩+⟨bv,2bv⟩=⟨au+bv, 2(au+bv)⟩∈W\begin{aligned}
a \vec u +b \vec v 
&= a \lang u,2u \rang + b \lang v,2v \rang\\[0.5em]
&= \lang au, 2au \rang + \lang bv, 2bv \rang\\[0.5em]
&= \lang au + bv,\  2(au + bv) \rang \in W
\end{aligned}au+bv​=a⟨u,2u⟩+b⟨v,2v⟩=⟨au,2au⟩+⟨bv,2bv⟩=⟨au+bv, 2(au+bv)⟩∈W​
Here, we conclude that ⟨au+bv, 2(au+bv)⟩\lang au + bv,\  2(au + bv) \rang⟨au+bv, 2(au+bv)⟩ must be in WWW since it is a vector of the form ⟨x,2x⟩\lang x , 2x \rang⟨x,2x⟩.
Therefore, WWW is a subspace of R2\reals^2R2.

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Example: Subspaces
Let VVV be the vector space {(x,y)  ∣  x,y∈R+}\{(x,y)\;|\;x,y\in\mathbb{R}^+\}{(x,y)∣x,y∈R+} with operations:
(x1,y1)+(x2,y2)=(x1x2, y1y2)(x_1,y_1)+(x_2,y_2)=(x_1x_2,\ y_1y_2)(x1​,y1​)+(x2​,y2​)=(x1​x2​, y1​y2​)
k⋅(x,y)=(xk,yk)k\cdot(x,y)=(x^k,y^k)k⋅(x,y)=(xk,yk)
Recall that in this vector space: 0⃗=(1,1)\vec{0}=(1,1)0=(1,1)
Consider the following subset of VVV:
W={(x,y)∈V  ∣  x=y}W=\{(x,y)\in V\;|\;x=y\}W={(x,y)∈V∣x=y}
Show that WWW is a subspace of VVV.

According to the definition of WWW, vectors in WWW must have equal components.
Wize Tip
 Find actual vectors in WWWto get a feel for the space.
For example: w1→=(6,6)∈W\overrightarrow{w_1}=(6,6)\in Ww1​​=(6,6)∈W, and w2→=(2,2)∈W\overrightarrow{w_2}=(2,2)\in Ww2​​=(2,2)∈W.
Then notice that: w1→+w2→=(6,6)+(2,2)=(6⋅2,6⋅2)=(12,12)∈W\overrightarrow{w_1}+\overrightarrow{w_2}=(6,6)+(2,2)=(6\cdot2,6\cdot2)=(12,12)\in Ww1​​+w2​​=(6,6)+(2,2)=(6⋅2,6⋅2)=(12,12)∈W because the components are equal.
Using a scalar k=2k=2k=2, for example:
2⋅w1→=2⋅(6,6)=(62,62)=(36,36)∈W2\cdot\overrightarrow{w_1}=2\cdot(6,6)=(6^2,6^2)=(36,36)\in W2⋅w1​​=2⋅(6,6)=(62,62)=(36,36)∈W because 36=3636=3636=36.
Watch Out!
This is not a formal proof!
We need to show that this is true for every vector in WWW and for every scalar kkk.

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Formal Proof
Step 1
Show that 0⃗∈W\vec{0}\in W0∈W:
In VVV the zero vector is 0⃗=(1,1)∈W\vec{0}=(1,1) \in W0=(1,1)∈W because 1=11=11=1.
Therefore, 0⃗∈W\vec 0 \in W0∈W.
Step 2
Let  x⃗, y⃗∈W\vec{x},\ \vec{y} \in Wx, y​∈W be arbitrary vectors, with x⃗=(x,x)\vec {x}=(x,x)x=(x,x) and y⃗=(y,y)\vec{y}=(y,y)y​=(y,y) (in WWW, components must be the same!).
We must show that x⃗+y⃗∈W\vec x+\vec y \in Wx+y​∈W:
x⃗+y⃗=(x,x)+(y,y)=(xy,xy)∈W\vec x + \vec y = (x,x)+(y,y) = (xy,xy) \in Wx+y​=(x,x)+(y,y)=(xy,xy)∈W since the components are equal.
Therefore, WWW is closed under vector addition (the addition of any two vectors is still in WWW).
Step 3
Once again, let x⃗=(x,x)∈W\vec x=(x,x) \in Wx=(x,x)∈W be an arbitrary vector.
We must show that k⋅x⃗∈Wk\cdot \vec x \in Wk⋅x∈W:
k⋅x⃗=k⋅(x,x)=(xk,xk)∈Wk\cdot \vec x=k\cdot(x,x)=(x^k,x^k) \in Wk⋅x=k⋅(x,x)=(xk,xk)∈W since the components are equal.
Therefore, WWW is closed under scalar multiplication (multiplying a vector in WWW by a scalar results in another vector in WWW).
Thus, WWW is a subspace of VVV.
Note: We could have also used the shortcut method to prove WWW is a subspace.

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Practice: Subspaces
Let M2×2M_{2\times2}M2×2​ be the vector space whose vectors are 2×22\times 22×2 matrices with real number entries, and with the standard operations of matrix addition and scalar multiplication.

Consider the following subset of M2×2M_{2\times2}M2×2​:
W={[abcd]∈M2×2 ∣ a+d=b+c}W=\left\{\left[\begin{array}{rr}
a&b\\
c&d\\
\end{array}\right]\in M_{2\times2}\ \Big |\ a+d=b+c\right\}W={[ac​bd​]∈M2×2​ ​ a+d=b+c}
Show that WWW is a subspace of M2×2M_{2\times2}M2×2​.

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Extra Practice

Span

Practice: Spans are Subspaces
Suppose that v⃗1,…,v⃗k∈Rn\vec v_1, \dots, \vec v_k\in\reals^nv1​,…,vk​∈Rn. Prove that U=span⁡({v⃗1,...,v⃗k})U=\operatorname{span}(\{\vec v_1, ...,\vec v_k\})U=span({v1​,...,vk​}) is a subspace of Rn\mathbb{R}^nRn.

Practice: Subspaces
Let M2×2M_{2\times2}M2×2​ be the vector space whose vectors are 2×22\times 22×2 matrices with real number entries, and with the standard operations of matrix addition and scalar multiplication.
Consider the following subset of M2×2M_{2\times2}M2×2​:

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}^+\}V={(x,y)∣x∈R,y∈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)(x1​,y1​)+(x2​,y2​)=(x1​+x2​,y1​y2​)
k⋅(x,y)=(kx,yk)k\cdot(x,y)=(kx,y^k)k⋅(x,y)=(kx,yk)

19.4F_Mid_Builder_$\tkcth{8.6.}\tkco{x}$_$\key{original \#1.}\tkct{d}$

The vectors [−71000]\begin{bmatrix} -7 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}​−71000​​, [−10310]\begin{bmatrix}
  -1 \\ 0 \\ 3 \\ 1 \\ 0 
\end{bmatrix}​−10310​​, [20−501]\begin{bmatrix}
  2 \\ 0 \\ -5 \\ 0 \\ 1 
\end{bmatrix}​20−501​​ are a basis for some subspace of R5\mathbb{R}^5R5. 

Show that U={(abcd)∈M2×2(R)  ∣  a+d=b+c}U=\Big\{ \begin{pmatrix} a&b\\ c&d \end{pmatrix}\in M_{2\times 2}(\mathbb{R})\;|\; a+d=b+c\Big\}U={(ac​bd​)∈M2×2​(R)∣a+d=b+c} is a subspace of M2×2(R)M_{2\times2}(\mathbb{R})M2×2​(R) over R\mathbb{R}R. Find a basis for UUU and find its dimension.
Show that W={p(x)∈P2(R)  ∣  p(−1)=0}W=\Big\{ p(x)\in P_2(\mathbb{R})\;|\; p(-1)=0\Big\}W={p(x)∈P2​(R)∣p(−1)=0} is a subspace of P2(R)P_2(\mathbb{R})P2​(R) over R\mathbb{R}R. Find a basis for WWWand find its dimension.

Dimension of a subspace

In the vector space P2P_2P2​, what is the dimension of the subspace that contains all polynomials of the form p(x)=a+bx+cx2p(x)=a+bx+cx^2p(x)=a+bx+cx2, such that a=b+ca=b+ca=b+c?

Practice Question: Dimension

For subspaces U,VU,VU,Vof Rn\mathbb{R}^nRn, prove that dim(U+V)=dim(U)+dim(V)−dim(U∩V)dim(U+V) = dim(U)+dim(V)-dim(U\cap V)dim(U+V)=dim(U)+dim(V)−dim(U∩V)

Bases of Subspaces

Suppose that U,VU,VU,V are subspaces of R4\mathbb{R}^4R4 , with dim(U)=2,dim(V)=3dim(U)=2,dim(V)=3dim(U)=2,dim(V)=3 .  Show that there is a nonzero vector in R4\mathbb{R}^4R4 which is in both  UUU and VVV .

Basis of a Subspace

Find a basis for the following subspace of M2×2M_{2\times2}M2×2​ and state its dimension
Q={[abcd]∈M2×2  ∣  a+b=0,c=2d}Q=\left\{\left[\begin{array}{rr}
a&b\\
c&d
\end{array}\right]\in M_{2\times2}\;\Big|\; a+b=0, c=2d\right\}Q={[ac​bd​]∈M2×2​​a+b=0,c=2d}

Basis of a Subspace

Find a basis for the following subspace of R3\mathbb{R}^3R3 and state its dimension
S={x⃗∈R3  ∣   [1−2−1]⋅x⃗=0⃗}S=\left\{\vec x\in\mathbb{R}^3\;\Big|\;\ \left[\begin{array}{rrr}
1&-2&-1\\
\end{array}\right]\cdot\vec x=\vec 0\right\}S={x∈R3​ [1​−2​−1​]⋅x=0}

Practice Question: Subspaces

For the following subset WWW of R4\mathbb{R}^4R4, select the statements which are true:
W={w⃗∈R4  ∣  Aw⃗=5w⃗}W=\{\vec w\in \mathbb{R}^4\;|\;A\vec w = 5\vec w\}W={w∈R4∣Aw=5w} where AAA is a 4×44\times 44×4 matrix

Practice Question: Subspaces

Which of the following subsets are subspaces?
U={(x,y,z)∈R3  ∣  x=2y,z=1}U=\{(x,y,z)\in \mathbb{R}^3\;|\;x=2y,z=1\}U={(x,y,z)∈R3∣x=2y,z=1}
W={w⃗∈R4  ∣  Aw⃗=5w⃗}W=\{\vec w \in \mathbb{R}^4\;|\;A\vec w = 5 \vec w\}W={w∈R4∣Aw=5w} where AAA is a 4×44\times 44×4 matrix

Practice Question: Subspaces

For the following subset UUU of R3\mathbb{R}^3R3, select the statements which are true:
U={(x,y,z)∈R3  ∣  x=2y,z=1}U=\{(x,y,z)\in \mathbb{R}^3\;|\;x=2y,z=1\}U={(x,y,z)∈R3∣x=2y,z=1}

Practice Question: Subspaces

For the following subset WWW of R3\mathbb{R}^3R3, select the statements which are true:
W={(a,b,c)∈R3  ∣  c<0}W=\{(a,b,c)\in \mathbb{R}^3\;|\;c<0\}W={(a,b,c)∈R3∣c<0}

Give an example of a non-empty subset of R2\mathbb{R}^2R2 that is closed under scalar multiplication, but is not a subspace.

Let VVV be a real vector space. Given two subspaces, UUU and WWW , is their union ever also a subspace? If so, what is the condition required for their union to be a subspace?

Wize University Linear Algebra Textbook > Vector Spaces and Subspaces

Subspaces

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Subspaces

Definition

Example: Subspaces

Formal Proof

Practice: Subspaces

Extra Practice

Span

Subspaces

Show that a subset is a subspace

19.4F_Mid_Builder_$\tkcth{8.6.}\tkco{x}$_$\key{original \#1.}\tkct{d}$

Dimension of a subspace

Practice Question: Dimension

Bases of Subspaces

Basis of a Subspace

Basis of a Subspace

Practice Question: Subspaces

Practice Question: Subspaces

Practice Question: Subspaces

Practice Question: Subspaces