# Some notations

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

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

Usually, due to the convenience, we select the center of the Cartesian coordinate system so properly that $\sum_j\mathbf v_j=\mathbf0$ or $$\mathbf V\mathbf1=\mathbf0. \label{barycenter zero}$$

# The research object

We are going to show that the equation $$\boldsymbol\lambda^{\mathrm T}\boldsymbol\lambda=1 \label{research object}$$ 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 $$1=\boldsymbol\lambda^{\mathrm T}\boldsymbol\lambda =\left(\mathbf V_1^{-1} \mathbf r_1\right)^{\mathrm T} \left(\mathbf V_1^{-1} \mathbf r_1\right) =\mathbf r_1^{\mathrm T} \left(\left(\mathbf V_1^{-1} \right)^{\mathrm T}\mathbf V_1^{-1} \right)\mathbf r_1. \label{r quadric}$$ Let $$\mathbf Q:=\left(\mathbf V_1^{-1} \right)^{\mathrm T}\mathbf V_1^{-1} =\left(\mathbf V_1 \mathbf V_1^{\mathrm T}\right)^{-1}, \label{Q def}$$ and substitute Formula \ref{Q def} into \ref{r quadric}, and then we can derive the quadric of $\mathbf r_1$ $$\mathbf r_1^{\mathrm T}\mathbf Q\mathbf r_1=1.$$ 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 coefficients of the suqare terms are all 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 $$\mathbf Q^{-1}= \left(\begin{matrix} \\&\mathbf V&\\\\1&\cdots&1 \end{matrix}\right) \left(\begin{matrix} &&&1\\&\mathbf V^{\mathrm T}&&\vdots\\&&&1 \end{matrix}\right)= \left(\begin{matrix} \\&\mathbf V\mathbf V^{\mathrm T}&&\mathbf V\mathbf1 \\\\&\mathbf1^{\mathrm T}\mathbf V^{\mathrm T}&&n \end{matrix}\right). \label{Q^-1}$$ Substitute Formula \ref{barycenter zero} into \ref{Q^-1}, and then we can derive that \begin{equation*} \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), \end{equation*} where $\mathbf W:=\left(\mathbf V\mathbf V^{\mathrm T}\right)^{-1}$, so \begin{equation*} \mathbf r_1^{\mathrm T}\mathbf Q\mathbf r_1= \mathbf r^{\mathrm T}\mathbf W\mathbf r+\frac1n. \end{equation*} 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 the square terms are all positive. In other words, the components on the diagonal of $\mathbf W$ are all positive.

Because $\mathbf Q= \left(\mathbf V_1^{-1}\right)^{\mathrm T}\mathbf V_1^{-1}$, we have \begin{equation*} \left(\mathbf Q\right)_{j,j}= \sum_k\left(\mathbf V_1^{-1}\right)_{j,k}^2>0. \end{equation*} Therefore, obviously, the components on the diagonal of $\mathbf W$ are all positive.

# 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 \begin{equation*} F\left(\mathbf r\right):=\mathbf r^{\mathrm T}\mathbf W\mathbf r +\frac1n-1. \end{equation*} According to geometry, the normal vector of the quadric at $\mathbf v_j$ is the gradient of $F$ at $\mathbf v_j$, which is \begin{equation*} \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. \end{equation*}

Now consider the normal vector $\mathbf m_j$ of $P_j$. Assume that \begin{equation*} P_j:n\mathbf m_j^{\mathrm T}\mathbf r+2=0. \end{equation*} 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 $\mathbf m_j$ $$\forall k\ne j:n\mathbf m_j^{\mathrm T}\mathbf v_k+2=0. \label{equations for m}$$

If we can show that $$\mathbf m_j=\boldsymbol\nu_j=2\mathbf W\mathbf v_j \label{solution for m}$$ 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} \begin{equation*} \forall k\ne j:n\mathbf v_j^{\mathrm T}\mathbf W\mathbf v_k+1=0, \end{equation*} which is to say that the $n\times n$ matrix \begin{equation*} \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 \end{equation*} 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, \begin{equation*} \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). \end{equation*}

Then, we have \begin{equation*} \mathbf P=\mathbf B\left(\mathbf B^{\mathrm T}\mathbf B\right)^{-1} \mathbf B^{\mathrm T}. \end{equation*} 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 $P_j$.