## Some notations

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

## 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 $\boldsymbol\lambda=\mathbf V_1^{-1} \mathbf r_1.$$\p{1}$ Usually, due to the convenience, we select the center of the Cartesian coordinate system so properly that $\sum_j\mathbf v_j=\mathbf0$ or $\mathbf V\mathbf1=\mathbf0.$$\p{2}$

## The research object

We are going to show that the equation $\boldsymbol\lambda^{\mathrm T}\boldsymbol\lambda=1$$\p{3}$ 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 3 in the form of a quadric of $\mathbf r$.

Substitute Formula 1 into 3, 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.$$\p{4}$ Let $\mathbf Q\coloneqq\left(\mathbf V_1^{-1} \right)^{\mathrm T}\mathbf V_1^{-1} =\left(\mathbf V_1 \mathbf V_1^{\mathrm T}\right)^{-1},$$\p{5}$ and substitute Formula 5 into 4, 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 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 $\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).$$\p{6}$ Substitute Formula 2 into 6, 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 $\mathbf W\coloneqq\left(\mathbf V\mathbf V^{\mathrm T}\right)^{-1}$, 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 $F\!\left(\mathbf r\right)\coloneqq\mathbf r^{\mathrm T}\mathbf W\mathbf r +\frac1n-1.$ According to geometry, the normal vector of the quadric at $\mathbf v_j$ is the gradient of $F$ at $\mathbf v_j$, which is $\boldsymbol\nu_j\coloneqq \left.\frac{\partial F\!\left(\mathbf r\right)}{\partial\mathbf r}\right| _{\mathbf r=\mathbf v_j}= 2\mathbf W\mathbf v_j.$

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 $\mathbf m_j$ $\forall k\ne j:n\mathbf m_j^{\mathrm T}\mathbf v_k+2=0.$$\p{7}$

If we can show that $\mathbf m_j=\boldsymbol\nu_j=2\mathbf W\mathbf v_j$$\p{8}$ is the solution to Formula 7, then we can say that the two hyperplane are parallel. Thus, we need to verify the equations derived from substituting Formula 8 into 7 $\forall k\ne j:n\mathbf v_j^{\mathrm T}\mathbf W\mathbf v_k+1=0,$ which is to say that the $n\times n$ matrix $\mathbf P\coloneqq\mathbf V^{\mathrm T}\mathbf W\mathbf V= \mathbf V^{\mathrm T}\left(\mathbf V\mathbf V^{\mathrm T}\right)^{-1} \mathbf V$ 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 2, 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, $\mathbf B\coloneqq\left(\begin{matrix} n-1&-1&-1&\cdots&-1\\-1&n-1&-1&\cdots&-1 \\-1&-1&n-1&\cdots&-1\\\vdots&\vdots&\vdots&\ddots&\vdots \\-1&-1&-1&\cdots&n-1\\-1&-1&-1&\cdots&-1 \end{matrix}\right).$ 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 $P_j$.