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 \(0\):

Prove but \(T\) is diagonalizable if and only if the matrix of \(T\) is similar to a diagonal matrix.

Try 2.

Given similarity:

So we have that:

\begin{equation} D = S^{-1} A S \end{equation}

where, \(D\) is diagonal. We apply \(S\) to both sides to yield:

\begin{equation} SD = AS \end{equation}

Now, note that \(S\) is invertible. This means that its column s are linearly independent (as it is an operator, which means it is injective, and hence has a zero null space; that indicates that the dimension of its range is that of the whole space: indicating its columns vectors are spanning; there is \(dim\ V\) such columns, so it is a basis and hence linearly independent).

Two Variables

Let’s begin with the equations:

\begin{equation} \begin{cases} 2x+y = 3 \\ x - y = 0 \end{cases} \end{equation}

We will first change this into a matrix equation:

\begin{equation} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix} \end{equation}

We need to find, then, the inverse of:

\begin{equation} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \end{equation}

Suppose \(\mathbb{F} = \mathbb{R}\), and \(V \neq \{0\}\). Replace the positivity condition with the condition that \(\langle v,v \rangle > 0\) for some \(v \in V\). Show that this change in definition does not change the set of functions from \(V \times V\) to \(\mathbb{R}\) that are inner products on \(V\).

We hope to show that \(\langle v,v \rangle >0\) for some \(v \in V\) implies that \(\langle v,v \rangle \geq 0\) for all \(v \in V\) in real vector spaces.

Proof: identity of a group is unique

Assume for contradiction that there exists two identities \(e_1\) and \(e_2\) which are identities of the group \(A\). Take also an \(a \in A\).

Given both \(e_1\) and \(e_2\) are identities, we have that:

\begin{equation} a * e_1 = a \end{equation}

as well as

\begin{equation} a * e_2 = a \end{equation}

Therefore, we have by the transitive property that:

\begin{equation} a * e_1 = a*e_2 \end{equation}