Houjun Liu

NUS-MATH530 Plane and 1.B

Equation of a Plane

We want to determine all points on the plane formed by two vectors.

Let’s take two vectors \(\vec{u} \in V\) and \(\vec{v} \in V\). The orthogonal vector to the both of them (i.e. the normal direction of the plane) is:

\begin{equation} \vec{u}\times \vec{v} \end{equation}

by the definition of the cross product.

The points on the plane, therefore, have to be orthogonal themselves to this normal vector. This means that the dot product of the candidate vector against these vectors should be \(0\):

\begin{equation} (\vec{u} \times \vec{v}) \cdot \begin{pmatrix} x_{1} \\ \dots \\ x_{n} \end{pmatrix} = 0 \end{equation}

This forms the final equation for a plane given two vectors in \(\mathbb{F}^{n}\).

A.B Exercises

Double Negative

We desire that \(-(-v)=v \forall v \in V\)

By distributivity in vector spaces, and the fact that \(0v=0\), we have that:

\begin{equation} v+(-1)v = (1-1)v = 0v = 0 \end{equation}

Therefore, \((-1)v=-v\).

We now have:

\begin{equation} -(-v) = -((-1)v) \end{equation}

The scalar multiple of \(v\), by definition, is also \(\in V\) if \(v \in V\). Therefore, it itself holds that:

\begin{equation} (-1)((-1)v) \end{equation}

By associativity:

\begin{equation} (-1\cdot -1)v \end{equation}


\begin{equation} (-1\cdot -1)v = (1v) = v\ \blacksquare \end{equation}

One of it is zero

If \(a \in \mathbb{F}\), \(v \in V\), and \(av=0\), we desire that \(a=0\) or \(v=0\). We perform casework.

Case 1: \(a=0\) – we are done.

Case 2: \(a \neq 0\): As \(a \in \mathbb{F}\), and \(a \neq 0\), \(\exists \frac{1}{a}: a\cdot \frac{1}{a}=1\).


\begin{align} &av = 0 \\ \Rightarrow\ & \frac{1}{a}av = \frac{1}{a} 0 \\ \Rightarrow\ & 1v = \frac{1}{a} 0 \\ \Rightarrow\ & 1v = 0 \\ \Rightarrow\ & v=0\ \blacksquare \end{align}

Existence and Uniqueness Given Equation

Given \(v,w \in V\), we desire a unique \(x\in V: v+3x=w\).

Let’s first check existence. Take the expression:

\begin{equation} n = \frac{1}{3} (w-v) \end{equation}

As both \(v,w \in V\), subtraction (addition) and scalar multiplication are defined. Therefore, \(\forall w,v \in V\), we can construct such an \(n\).

Supplying the expression into \(v+3x\) for the definition of \(x\):

\begin{align} v+3x &= v+3\qty(\frac{1}{3}(w-v)) \\ &= v+(w-v) \\ &= v+w-v \\ &= v-v+w \\ &= 0+w \\ &= w \end{align}

by distributivity, associativity, and commutativity in vector spaces, yielding \(w\) as desired.

Now let’s check uniqueness.

Suppose \(\exists x_1, x_2: v+3x_1=w\) and \(v+3x_2=w\).

By transitivity:

\begin{equation} v+3x_1=v+3x_2 \end{equation}

Applying \(-v\) to both sides:

\begin{equation} 3x_1=3x_2 \end{equation}

Finally, applying \(\frac{1}{3}\) to both sides:

\begin{equation} x_{1}= x_2 \end{equation}

Therefore, there only exists one unique \(x\) which satisfies the expression. \(\blacksquare\)

Empty Set is Not a Vector Space

The empty set is not a vector space as it doesn’t have an additive identity. \(\blacksquare\)

Additive Inverse is also Zero Multiplication

We first take the additive inverse expression:

\begin{equation} \forall v \in V, \exists -v: v+(-v) = 0 \end{equation}

Take now:

\begin{equation} 0v \end{equation}

We have that:

\begin{align} 0v &= (0+0)v \\ &= 0v + 0v \end{align}

By distributivity.

As \(0v \in V\), \(\exists -0v: 0v+(-0v)=0\).

\begin{align} &0v = 0v+0v \\ \Rightarrow\ & 0v-0v = 0v+0v-0v \\ \Rightarrow\ & 0 = 0v \end{align}

as desired. Now, we will start from this condition and work out way backwards.

Note that the statement for additive inverse condition is that:

\begin{equation} \forall v \in V, \exists -v: v+(-v) = 0 \end{equation}

Let us begin with the expression that:

\begin{equation} 0=0v \end{equation}

We have that:

\begin{equation} 0=(1-1)v \end{equation}

Then, we have by distributivity:

\begin{equation} 0 = v + (-1)v \end{equation}

scalar multiplication is defined on a vector space. Therefore, we have \(-1v\) to construct such an additive inverse \(\forall v \in V\). \(\blacksquare\)

Weird Vector Space

All operations are defined as given.

Take scalars \(t_1, t_2 \in \mathbb{R}\).

\begin{equation} (t_1-t_2)\infty = \infty \end{equation}

Yet, if we follow the rules of distribution:

\begin{equation} (t_1 -t_2)\infty = \infty -\infty =0 \end{equation}

Therefore, distribution doesn’t hold on this new structure. It is not a vector space. \(\blacksquare\)