# Some notations

$S_n:=\left\{\mathbf a\in\mathbb R^{n}\,\middle|\,\sum_ja_j=1\right\}.$ $\mathbf1:=\left(\begin{matrix}1\\\vdots\\1\end{matrix}\right).$ $\mathbf v_1:=\left(\begin{matrix} \\\mathbf v\\\\1 \end{matrix}\right).$ $\mathbf M_1:=\left(\begin{matrix} \\&\mathbf M&\\\\&\mathbf1^{\mathrm T} \end{matrix}\right).$

# Introduction to barycentric coordinates

Let $\mathbf v_j$ be the vertices of a simplex in $\mathbb R^{n-1}$, then any point $\mathbf r\in\mathbb R^{n-1}$ can be expressed by a tuple $\boldsymbol\lambda\in S_n$ such that $\mathbf r=\sum_j\lambda_j\mathbf v_j$.

If regarding $\mathbf V$ as a $\left(n-1\right)\times n$ matrix whose $j$th column is $\mathbf v_j$, then we have $\mathbf r=\mathbf V\boldsymbol\lambda$.

Along with the normalization condition $\sum_j\lambda_j=1$ or $\mathbf1^{\mathrm T}\boldsymbol\lambda=1$, we have $\mathbf r_1=\mathbf V_1 \boldsymbol\lambda$, so

$\begin{equation} \boldsymbol\lambda=\mathbf V_1^{-1} \mathbf r_1. \label{as Cartesian} \end{equation}$

Usually, due to the convenience, we select the center of the Cartesian coordinate system so properly that $\sum_j\mathbf v_j=\mathbf0$ or

$\begin{equation} \mathbf V\mathbf1=\mathbf0. \label{barycenter zero} \end{equation}$

# The research object

We are going to show that the equation

$\begin{equation} \boldsymbol\lambda^{\mathrm T}\boldsymbol\lambda=1 \label{research object} \end{equation}$

depicts a hyperellipsoid whose center is $\mathbf0$ and its tangent hyperplane at $\mathbf v_j$ is parallel to the hyperplane that passes all $\mathbf v_k$ that $k\ne j$.

We are going to rewrite Formula \ref{research object} in the form of a quadric of $\mathbf r$.

Substitute Formula \ref{as Cartesian} into \ref{research object}, and then we can derive that

$\begin{equation} 1=\boldsymbol\lambda^{\mathrm T}\boldsymbol\lambda =\left(\mathbf V_1^{-1} \mathbf r_1\right)^{\mathrm T} \left(\mathbf V_1^{-1} \mathbf r_1\right) =\mathbf r_1^{\mathrm T} \left(\left(\mathbf V_1^{-1} \right)^{\mathrm T}\mathbf V_1^{-1} \right)\mathbf r_1. \label{r quadric} \end{equation}$

Let

$\begin{equation} \mathbf Q:=\left(\mathbf V_1^{-1} \right)^{\mathrm T}\mathbf V_1^{-1} =\left(\mathbf V_1 \mathbf V_1^{\mathrm T}\right)^{-1}, \label{Q def} \end{equation}$

and substitute Formula \ref{Q def} into \ref{r quadric}, and then we can derive the quadric of $\mathbf r_1$

$\begin{equation} \mathbf r_1^{\mathrm T}\mathbf Q\mathbf r_1=1. \end{equation}$

Note that besides $\mathbf r$, there is a $1$ in $\mathbf r_1$, so the quadric is a $2$nd-degree polynomial of $\mathbf r$, including quadratic terms, linear terms and a constant term.

In order to show that the center of the quadric is a hyperellipsoid whose center is $\mathbf0$, we need to prove that the coefficients of the linear terms are all $0$, and the determinant of the coefficients is positive.

# Proving that the center of the quadric is $\mathbf0$

Note that $\mathbf Q=\left(\mathbf V_1\mathbf V_1^{\mathrm T}\right)^{-1}$, so

$\begin{equation} \mathbf Q^{-1}= \left(\begin{matrix} \\&\mathbf V&\\\\1&\cdots&1 \end{matrix}\right) \left(\begin{matrix} &&&1\\&\mathbf V^{\mathrm T}&&\vdots\\&&&1 \end{matrix}\right)= \left(\begin{matrix} \\&\mathbf V\mathbf V^{\mathrm T}&&\mathbf V\mathbf1 \\\\&\mathbf1^{\mathrm T}\mathbf V^{\mathrm T}&&n \end{matrix}\right). \label{Q^-1} \end{equation}$

Substitute Formula \ref{barycenter zero} into \ref{Q^-1}, and then we can derive that

$\mathbf Q=\left(\begin{matrix} &&&0\\&\mathbf V\mathbf V^{\mathrm T}&&\vdots \\&&&0\\0&\cdots&0&n \end{matrix}\right)^{-1}= \left(\begin{matrix} &&&0\\&\mathbf W&&\vdots\\&&&0\\0&\cdots&0&\frac1n \end{matrix}\right),$

where $\mathbf W:=\left(\mathbf V\mathbf V^{\mathrm T}\right)^{-1}$, so

$\mathbf r_1^{\mathrm T}\mathbf Q\mathbf r_1= \mathbf r^{\mathrm T}\mathbf W\mathbf r+\frac1n.$

The linear terms are all $0$, so the center of the quadric is $\mathbf0$.

# Proving that the quadric is a hyperellipsoid

We need to show that determinant of the coefficients matrix is positive.

Because $\mathbf Q= \left(\mathbf V_1^{-1}\right)^{\mathrm T}\mathbf V_1^{-1}$, we have

$\left|\mathbf Q\right|= \left|\mathbf V_1^{-1}\right|^2>0.$

# Proving that the its tangent hyperplane at $\mathbf v_j$ is parallel to $P_j$

Here $P_j$ is defined as the hyperplane that passes all $\mathbf v_k$ that $k\ne j$.

The equation of the quadric is $F\!\left(\mathbf r\right)=0$, where the quadratic function

$F\!\left(\mathbf r\right):=\mathbf r^{\mathrm T}\mathbf W\mathbf r +\frac1n-1.$

According to geometry, the normal vector of the quadric at $\mathbf v_j$ is the gradient of $F$ at $\mathbf v_j$, which is

$\boldsymbol\nu_j:= \left.\frac{\partial F\!\left(\mathbf r\right)}{\partial\mathbf r}\right| _{\mathbf r=\mathbf v_j}= 2\mathbf W\mathbf v_j.$

Now consider the normal vector $\mathbf m_j$ of $P_j$. Assume that

$P_j:n\mathbf m_j^{\mathrm T}\mathbf r+2=0.$

The equation of $P_j$ should holds when $\mathbf r=\mathbf v_k$ for all $k\ne j$, so we can derive $n-1$ linear equations with respect to $\mathbf m_j$

$\begin{equation} \forall k\ne j:n\mathbf m_j^{\mathrm T}\mathbf v_k+2=0. \label{equations for m} \end{equation}$

If we can show that

$\begin{equation} \mathbf m_j=\boldsymbol\nu_j=2\mathbf W\mathbf v_j \label{solution for m} \end{equation}$

is the solution to Formula \ref{equations for m}, then we can say that the two hyperplane are parallel. Thus, we need to verify the equations derived from substituting Formula \ref{solution for m} into \ref{equations for m}

$\forall k\ne j:n\mathbf v_j^{\mathrm T}\mathbf W\mathbf v_k+1=0,$

which is to say that the $n\times n$ matrix

$\mathbf P:=\mathbf V^{\mathrm T}\mathbf W\mathbf V=\ \mathbf V^{\mathrm T}\left(\mathbf V\mathbf V^{\mathrm T}\right)^{-1} \mathbf V$

is such a matrix that all of its components except those on its diagonal are $-\frac1n$.

According to conclusions in matrix analysis, if we regard $\mathbf V^{\mathrm T}$ as $n-1$ $n$-dimensional vectors, then $\mathbf P$ is an orthogonal projection in $\mathbb R^n$ to the linear subspace whose basis is the $n-1$ vectors.

Note that with Formula \ref{barycenter zero}, we can say that the subspace is just a hyperplane whose normal vector is $\mathbf1$. With the conclusion, we can easily write out the form of $\mathbf P$ because we just need to write out one set of its basis $\mathbf B$. Writing out $\mathbf B$ only requires finding out $n-1$ linearly independent vectors that are perpendicular to $\mathbf1$. For example,

$\mathbf B:=\left(\begin{matrix} n-1&-1&-1&\cdots&-1\\-1&n-1&-1&\cdots&-1 \\-1&-1&n-1&\cdots&-1\\\vdots&\vdots&\vdots&\ddots&\vdots \\-1&-1&-1&\cdots&n-1\\-1&-1&-1&\cdots&-1 \end{matrix}\right).$

Then, we have

$\mathbf P=\mathbf B\left(\mathbf B^{\mathrm T}\mathbf B\right)^{-1} \mathbf B^{\mathrm T}.$

After some calculation, we can derive that the components of $\mathbf P$ are $1-\frac1n$ on the diagonal and $-\frac1n$ elsewhere, which is what we want to show.

We have proved that the tangent hyperplane of the quadric at $\mathbf v_j$ is parallel to $P_j$. 