Wize High School Algebra II Textbook (Common Core) > Vectors (Extension Topic)

Vector Properties

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Vector Properties
You may be asked to prove some of the following properties, or to use these properties to prove other statements.

Let u⃗, v⃗, w⃗ ∈Rn\vec{u},\ \vec{v},\ \vec{w}\ \in \reals^nu, v, w ∈Rn be three vectors.
Let c, d∈Rc,\ d \in \realsc, d∈R be two scalars.
u⃗+v⃗=v⃗+u⃗\vec{u}+\vec{v}=\vec{v}+\vec{u}u+v=v+u
(u⃗+v⃗)+w⃗=u⃗+(v⃗+w⃗)\left(\vec{u}+\vec{v}\right)+\vec{w}=\vec{u}+\left(\vec{v}+\vec{w}\right)(u+v)+w=u+(v+w)
There exists a zero vector 0⃗=⟨0,0,...,0⟩∈Rn\vec{0} = \lang 0,0,...,0 \rang \in \reals^n0=⟨0,0,...,0⟩∈Rn such that v⃗+0⃗=v⃗\vec{v}+\vec{0}=\vec{v}v+0=v 
There exists a negative vector −v⃗-\vec{v}−v  such that v⃗+(−v⃗)=0⃗\vec{v}+\left(-\vec{v}\right)=\vec{0}v+(−v)=0 
(cd)v⃗=c(dv⃗)\left(cd\right)\vec{v}=c\left(d\vec{v}\right)(cd)v=c(dv)
(c+d)v⃗=cv⃗+dv⃗\left(c+d\right)\vec{v}=c\vec{v}+d\vec{v}(c+d)v=cv+dv
c(v⃗+u⃗)=cv⃗+cu⃗c\left(\vec{v}+\vec{u}\right)=c\vec{v}+c\vec{u}c(v+u)=cv+cu
1v⃗=v⃗1\vec{v}=\vec{v}1v=v
0v⃗=0⃗0\vec{v}=\vec{0}0v=0
Wize Tip
To prove a statement involving vectors, it is often helpful to write out the components of each vector. 
u⃗=[u1u2⋮un],v⃗=[v1v2⋮vn]\vec u =
\begin{bmatrix}
  u_1\\ u_2\\ \vdots\\ u_n
\end{bmatrix}, \quad
\vec v =
\begin{bmatrix}
  v_1\\ v_2\\ \vdots\\ v_n
\end{bmatrix}u=​u1​u2​⋮un​​​,v=​v1​v2​⋮vn​​​

To show that a statement is NOT always true, find one counter example (keep it simple with an example from R2\reals^2R2)

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Example: Proving Vector Properties
Given vectors u⃗, v⃗∈Rn\vec u, \ \vec v \in \reals^nu, v∈Rn and scalar c∈Rc \in \realsc∈R, prove the following property:
c(u⃗+v⃗)=cu⃗+cv⃗c(\vec u + \vec v) = c \vec u + c \vec vc(u+v)=cu+cv

Begin by writing out the components of u⃗\vec uu and v⃗\vec{v}v:
u⃗=[u1u2⋮un],v⃗=[v1v2⋮vn]\vec u =
\begin{bmatrix}
  u_1\\ u_2\\ \vdots\\ u_n
\end{bmatrix}, \quad
\vec v =
\begin{bmatrix}
  v_1\\ v_2\\ \vdots\\ v_n
\end{bmatrix}u=​u1​u2​⋮un​​​,v=​v1​v2​⋮vn​​​
Then:
c(u⃗+v⃗)=c([u1u2⋮un]+[v1v2⋮vn])=c[u1+v1u2+v2⋮un+vn]=   [c(u1+v1)c(u2+v2)⋮c(un+vn)]=   [cu1+cv1cu2+cv2⋮cun+cvn]=c[u1u2⋮un]+c[v1v2⋮vn]=cu⃗+cv⃗\begin{aligned}
c(\vec u + \vec v)
= 
c
&
\left(
\begin{bmatrix}
  u_1\\ u_2\\ \vdots\\ u_n
\end{bmatrix}
+
\begin{bmatrix}
  v_1\\ v_2\\ \vdots\\ v_n
\end{bmatrix}
\right)\\
=
c
&\begin{bmatrix}
  u_1 +v_1\\ u_2+v_2\\ \vdots\\ u_n+v_n
\end{bmatrix}\\
=\ \ \ 
&\begin{bmatrix}
  c(u_1 +v_1)\\ c(u_2+v_2)\\ \vdots\\ c(u_n+v_n)
\end{bmatrix}\\
=\ \ \ 
&\begin{bmatrix}
  cu_1 +cv_1\\ cu_2+cv_2\\ \vdots\\ cu_n+cv_n
\end{bmatrix}\\
= 
c
&\begin{bmatrix}
  u_1\\ u_2\\ \vdots\\ u_n
\end{bmatrix}
+
c
\begin{bmatrix}
  v_1\\ v_2\\ \vdots\\ v_n
\end{bmatrix}\\
= c &\vec u + c \vec v
\end{aligned}c(u+v)=c=c=   =   =c=c​​​u1​u2​⋮un​​​+​v1​v2​⋮vn​​​​​u1​+v1​u2​+v2​⋮un​+vn​​​​c(u1​+v1​)c(u2​+v2​)⋮c(un​+vn​)​​​cu1​+cv1​cu2​+cv2​⋮cun​+cvn​​​​u1​u2​⋮un​​​+c​v1​v2​⋮vn​​​u+cv​

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Example: Proving Vector Properties
Determine whether the following statement is always true.
If it is true, prove it. If it is not always true, provide a counterexample.

For any vectors u⃗, v⃗, w⃗∈Rn\vec u, \ \vec v, \ \vec w \in \reals^nu, v, w∈Rn,
u⃗−(v⃗−w⃗)=(u⃗−v⃗)−w⃗\vec u - (\vec v - \vec w) = (\vec u - \vec v) - \vec wu−(v−w)=(u−v)−w

Not always true.

Intuitively, we might think something isn't quite right when we try to distribute the negative on the LHS:
u⃗−(v⃗−w⃗)=u⃗−v⃗+w⃗ (???)\vec u - (\vec v - \vec w) = \vec u - \vec v + \vec w \qquad \text{ (???)}u−(v−w)=u−v+w (???)
This leads us to think of a counterexample. Keep it simple with vectors in R2\reals^2R2.

Let  u⃗=[11],v⃗=[22],w⃗=[33]\vec u =
\begin{bmatrix}
  1\\1
\end{bmatrix}, \quad
\vec v =
\begin{bmatrix}
  2\\2
\end{bmatrix}, \quad
\vec w =
\begin{bmatrix}
  3\\3
\end{bmatrix}u=[11​],v=[22​],w=[33​].

LHS
u⃗−(v⃗−w⃗)=[11]−([22]−[33])=[11]−[−1−1]=[22]\vec u - (\vec v - \vec w)
= 
\begin{bmatrix}
  1\\1
\end{bmatrix}
-
\left(
\begin{bmatrix}
  2\\2
\end{bmatrix} -
\begin{bmatrix}
  3\\3
\end{bmatrix}
\right)
= 
\begin{bmatrix}
  1\\1
\end{bmatrix}
-
\begin{bmatrix}
  -1\\-1
\end{bmatrix}
=
\begin{bmatrix}
  2\\2
\end{bmatrix}u−(v−w)=[11​]−([22​]−[33​])=[11​]−[−1−1​]=[22​]

RHS
(u⃗−v⃗)−w⃗=([11]−[22])−[33]=[−1−1]−[33]=[−4−4](\vec u - \vec v) - \vec w
= 
\left(
\begin{bmatrix}
  1\\1
\end{bmatrix}
-
\begin{bmatrix}
  2\\2
\end{bmatrix}
\right)
-
\begin{bmatrix}
  3\\3
\end{bmatrix}
= 
\begin{bmatrix}
  -1\\-1
\end{bmatrix}
-
\begin{bmatrix}
  3\\3
\end{bmatrix}
=
\begin{bmatrix}
  -4\\-4
\end{bmatrix}(u−v)−w=([11​]−[22​])−[33​]=[−1−1​]−[33​]=[−4−4​]

Since LHS≠RHS\text{LHS} \neq \text{RHS}LHS=RHS, this counterexample implies that the statement is not true in general.

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Practice: Vector Properties
Prove the following vector property:
u⃗+v⃗=v⃗+u⃗\vec{u}+\vec{v}=\vec{v}+\vec{u}u+v=v+u

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