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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 u, v, w Rn\vec{u},\ \vec{v},\ \vec{w}\ \in \reals^n be three vectors.
Let c, dRc,\ d \in \reals be two scalars.
  • u+v=v+u\vec{u}+\vec{v}=\vec{v}+\vec{u}
  • (u+v)+w=u+(v+w)\left(\vec{u}+\vec{v}\right)+\vec{w}=\vec{u}+\left(\vec{v}+\vec{w}\right)
  • There exists a zero vector 0=0,0,...,0Rn\vec{0} = \lang 0,0,...,0 \rang \in \reals^n such that v+0=v\vec{v}+\vec{0}=\vec{v}
  • There exists a negative vector v-\vec{v} such that v+(v)=0\vec{v}+\left(-\vec{v}\right)=\vec{0}
  • (cd)v=c(dv)\left(cd\right)\vec{v}=c\left(d\vec{v}\right)
  • (c+d)v=cv+dv\left(c+d\right)\vec{v}=c\vec{v}+d\vec{v}
  • c(v+u)=cv+cuc\left(\vec{v}+\vec{u}\right)=c\vec{v}+c\vec{u}
  • 1v=v1\vec{v}=\vec{v}
  • 0v=00\vec{v}=\vec{0}
Wize Tip
  • To prove a statement involving vectors, it is often helpful to write out the components of each vector.
u=[u1u2un],v=[v1v2vn]\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}

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

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Example: Proving Vector Properties

Given vectors u, vRn\vec u, \ \vec v \in \reals^n and scalar cRc \in \reals, prove the following property:
c(u+v)=cu+cvc(\vec u + \vec v) = c \vec u + c \vec v

Begin by writing out the components of u\vec u and v\vec{v}:
u=[u1u2un],v=[v1v2vn]\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}
Then:
c(u+v)=c([u1u2un]+[v1v2vn])=c[u1+v1u2+v2un+vn]=   [c(u1+v1)c(u2+v2)c(un+vn)]=   [cu1+cv1cu2+cv2cun+cvn]=c[u1u2un]+c[v1v2vn]=cu+cv\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}
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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 u, v, wRn\vec u, \ \vec v, \ \vec w \in \reals^n,
u(vw)=(uv)w\vec u - (\vec v - \vec w) = (\vec u - \vec v) - \vec w

Not always true.

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

Let u=[11],v=[22],w=[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}.

LHS
u(vw)=[11]([22][33])=[11][11]=[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}

RHS
(uv)w=([11][22])[33]=[11][33]=[44](\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}

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

checklist
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Practice: Vector Properties

Prove the following vector property:
u+v=v+u\vec{u}+\vec{v}=\vec{v}+\vec{u}

Extra Practice