\begin{lemma}
Suppose we have a matrix $A$ of size $k \times k$. Then suppose we have a matrix $B$ such that the absolute value of the elements of $B$ are equal to the absolute value of the elements in $A$, however, the $j^{th}$ diagonal of B is multiplied by -1 when $j$ is odd.

Then $det(A) = (-1)^k \cdot det(B)$

For example, if $$A = \begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{bmatrix}$$, then $$B = \begin{bmatrix}
-a_{11} & a_{12} & -a_{13}\\
a_{21} & -a_{22} & a_{23}\\
-a_{31} & a_{32} & -a_{33}
\end{align*}$$.
and $det(A) = (-1)^3 \cdot det(B)$.

Proof.
Say we have a square matrix $N$ of size $n$. Where the $m^{th}$ diagonal of $N$ is non-positive.

Case 1, $n$ is even.

Then $n = 2p$, $p \in \mathbb{Z}$
$N = \begin{bmatrix}
-a_{11} & a_{12} & \dots & a_{1n}\\
a_{21} & -a_{22} & \dots & -a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & -a_{n2} & \dots & -a_{nn}
\end{bmatrix}$.

Multiplying the odd rows of by -1 gives us
$N' = \begin{bmatrix}
a_{11} & -a_{12} & \dots & -a_{1n}\\
a_{21} & -a_{22} & \dots & -a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & -a_{n2} & \dots & -a_{nn}
\end{bmatrix}$.

Since $N$ has $p$ odd rows, $det(N) = (-1)^p \cdot det(N')$.

Then multiplying the even columns of $N'$ by -1 gives us
$N'' = \begin{bmatrix}
a_{11} & a_{12} & \dots & a_{1n}\\
a_{21} & a_{22} & \dots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \dots & a_{nn}
\end{bmatrix}$.

Since $N'$ has $p$ even columns, $det(N') = (-1)^p \cdot det(N'')$.

Then, $det(N) = (-1)^p \cdot (-1)^p \cdot det(N'') = (-1)^{2p} \cdot det(N'') = det(N'')$.

Case 2, $n$ is odd.
If $n$ is odd, then $n = 2p-1$, $p \in \mathbb{Z}$,

Then $N$ has $p$ odd rows and $p-1$ even columns.

So,
$det(N)= \begin{vmatrix}
-a_{11} & a_{12} & \dots & -a_{1n}\\
a_{21} & -a_{22} & \dots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
-a_{n1} & a_{n2} & \dots & -a_{nn}
\end{vmatrix}$.

Multiplying the $p$ odd rows by -1 gives us
$det(N)=(-1)^p$$\begin{vmatrix}
a_{11} & -a_{12} & \dots & a_{1n}\\
a_{21} & -a_{22} & \dots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & -a_{n2} & \dots & a_{nn}
\end{vmatrix}$

Multiplying the $p-1$ even columns by -1 gives us
$det(N)=(-1)^n$$\begin{vmatrix}
a_{11} & a_{12} & \dots & a_{1n}\\
a_{21} & a_{22} & \dots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \dots & a_{nn}
\end{vmatrix}$

Since $p + (p-1) = 2p-1 = n$
\end{lemma}

Let $M:=(m_{ij})$ and $M^*=(m^*_{ij})$ be square matrices of size $n$, and let $m^*_{ij}= (-1)^{i+j+1}m_{ij}$.
Then $\det~M^* = (-1)^n \det M$.

\begin{proof}
First we multiply all odd rows of $M^*$ by -1, thus obtaining a new matrix $M'$ whose entries are given by

\begin{align*}
&m'_{2k-1,j}=(-1)m^*_{2k-1, j}= (-1)^{2k+j+1}m_{2k-1,j}
\quad j=1,\dots, n, \quad k=1,\dots,\lceil n/2\rceil,\\
&m'_{2k,j}=m^*_{2k,j},\quad j=1,\dots, n, \quad k=1,\dots,\lceil n/2\rceil.
\end{align*}

Note that

\begin{align}\label{mprime}
&m'_{2k-1,2l}=-m_{2k-1,2l},
\quad l=1,\dots,\lceil n/2\rceil, \quad k=1,\dots,\lceil n/2\rceil,\\
&m'_{2k-1,2l-1}=m_{2k-1,2l-1},
\quad l=1,\dots,\lceil n/2\rceil, \quad k=1,\dots,\lceil n/2\rceil,\\
&m'_{2k,2l}=m^*_{2k,2l}=-m_{2k,2l},
\quad l=1,\dots,\lceil n/2\rceil, \quad k=1,\dots,\lceil n/2\rceil,\\
&m'_{2k,2l-1}=m^*_{2k,2l-1}=m_{2k,2l-1},
\quad l=1,\dots,\lceil n/2\rceil, \quad k=1,\dots,\lceil n/2\rceil.\\
\end{align}
Then $\det M'=(-1)^{\lceil n/2\rceil}\det M^*$.
Clearly by multiplying all $n-\lceil n/2\rceil$ even columns of M' by -1 we reobtain matrix $M$, and, thus
$$\det M=(-1)^{n-\lceil n/2\rceil}\det M'=(-1)^{n-\lceil n/2\rceil}(-1)^{\lceil n/2\rceil}\det M^*=(-1)^n\det M^*$$
\end{proof}

\\

Let $p=\sum\limits_{i=0}^ka_i(x-1)^{2i}, x \in [0,2],$ and let $q=\sum\limits_{i=0}^kb_i(x+1)^{2i}, x \in [-2,0],$ be the B-symmetric polynomials forming a spline $s\in \BB_{2k}(\Delta_{[0,2]})$. Then $b_i=-a_i$ for all $i=1,\dots, k$.

\begin{proof}

We will prove this result using a reverse induction (probably a better wording).

We begin by taking the $(2k-1)$-order derivative of $p$ and $q$

\begin{align*}
&p^{(2k-1)}(x) = a_k(2k-1)!\binom{2k}{2k-1} (x-1),\\
&q^{(2k-1)}(x) = b_k(2k-1)!\binom{2k}{2k-1} (x+1).
\end{align*}
Evaluating both derivatives at $x=0$ shows that $b_k=-a_k$.

Next we take the $(2k-3)$-order derivative of $p$ and $q$

\begin{align*}
&p^{(2k-3)}(x) = (2k-3)!\Big[a_{k-1} \binom{2k-1}{2k-3} (x-1) + a_k \binom{2k}{2k-3} (x-1)^3\Big]
,\\
&q^{(2k-3)}(x)=(2k-3)!\Big[b_{k-1} \binom{2k-1}{2k-3}(x+1) + b_k \binom{2k}{2k-3} (x+1)^3\Big].
\end{align*}

Setting $p^{(2k-3)}(0)=q^{(2k-3)}(0)$ and taking into account that $b_k=-a_k$ gives us $b_{k-1}=-a_{k-1}$. Continuing this process yields the following: when we set $p^{(2j-1)}(0)=q^{(2j-1)}(0)$ and substitute $a_l=-b_l$, for $l>j$, we obtain that $b_j=-a_j$. This process terminates at the first order derivative, which completes the proof.
\end{proof}
This takes care of the smoothness conditions associated with odd-order derivatives. Our next result takes care of the ones associated with even order derivatives.

Let $p=\sum\limits_{i=0}^ka_i(x-1)^{2i}$ and let $q=b_0-\sum\limits_{i=1}^ka_i(x+1)^{2i}$ be the B-symmetric polynomials forming a spline $s\in \BB_{2k}(\Delta_{[0,2]})$.
Furthermore, let $p(1)$ and $q(-1)$ be given. Then $p\cup q$ forms a spline $s\in \BB_{2k}(\Delta_{[0,2]})$. Then the remaining coefficients $a_i$ are uniquely defined by the smoothness conditions.
\begin{proof}
Setting $p(0) = q(0)$ and using Lemma~\ref{smoothness} we obtain:
\begin{equation}\label{c0}
\sum_{i=1}^k a_i = \frac{b_0 - a_0}{2}.
\end{equation}

Using the differentiation formula and Lemma~\ref{smoothness}
\begin{alignat}{2}
&p^{(2m)}(x) = &(2m)!\sum_{i=m}^k a_i \binom{2i}{2m} (x-1)^{2(i-m)},\quad m=1,\dots,k-1,\\
&q^{(2m)}(x) = -&(2m)!\sum_{i=m}^k a_i \binom{2i}{2m} (x+1)^{2(i-m)},\quad m=1,\dots,k-1,
\end{alignat}
we evaluate both at $x=0$, and obtain
\begin{equation}\label{cm}
\sum_{i=m}^k a_i \binom{2i}{2m} = 0, \quad m = 1, \cdots, k-1.
\end{equation}
The system of $k$ linear equations~(\ref{c0}) and~(\ref{cm}) with $k$ unknowns gives rise to a $k\times k$ matrix $M=(m_{ij})$, whose entries are
$$m_{ij} = \binom{2j}{2(i-1)}, \quad 1 \le i,j \le k.$$
Note that it is a transpose of ..... + Lemma 4.1
\end{proof}

**Theorem**

For all $n\in \mathbb{N}$, the Euler number $E_{2n}=(-1)^n\det {\mathcal E}_{2n}$, where ${\mathcal E}_{2n}=(e_{ij})$ is a $(2n)\times(2n)$ matrix whose entries are given by
\begin{equation}
e_{ij} = \binom{i}{j-1}\cos\Big((i-j+1)\frac{\pi}{2}\Big),
\quad i=1,\dots, 2n,\quad j=1,\dots, 2n.
\end{equation}

**Theorem**
For all $n\in \mathbb{N}$, $\det {\mathcal E}_{2n}=\det A_n$, where $A_n$ is an $n\times n$ matrix obtained from ${\mathcal E}_{n}$ by removing all its odd numbered rows and even numbered columns.

**Theorem**
For all $n\in \mathbb{N}$, let $\widehat {\mathcal E}_n$ be an $n\times n$ matrix obtained from ${\mathcal E}_{2n}$ by removing all its odd numbered rows and even numbered columns. Let $\widetilde {\mathcal E}_n$ be an $n\times n$ matrix obtained from $\widehat {\mathcal E}_n$, where the entries of $\widetilde {\mathcal E}_n$ are equal to the absolute value of the entries in $\widetilde {\mathcal E}_n$

$$\det {\mathcal E}_{2n}=(-1)^n\det\widehat{\mathcal E}_{n}=\det\widetilde{\mathcal E}_{n}$$

\begin{proof}

We begin by writing $\mathcal{E}_{2n}$ explicitly:

$$\begin{bmatrix}
0 & 1 \choose 1 & 0 & 0 & 0 & ... & 0 & 0\\
-{2 \choose 0} & 0 & 2 \choose 2 & 0 & 0 & ... & 0 & 0\\
0 & -{3 \choose 1} & 0 & 3 \choose 3 & 0 & ... & 0 & 0\\
4 \choose 0 & 0 & -{4 \choose 2} & 0 & 4 \choose 4 & ... & 0 & 0\\
0 & 5 \choose 1 & 0 & -{5 \choose 3} & 0 & ... & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 2n-1 \choose 1 & 0 & -{2n-1 \choose 3} & 0 & ... & 0 & 2n-1 \choose 2n-1\\
-{2n \choose 0} & 0 & 2n \choose 2 & 0 & -{2n \choose 4} & ... & -{2n \choose 2n-2} & 0
\end{bmatrix}$$

Note that the matrix has a lot of zero entries. Indeed, if $i-j$ is even, then $\cos\left((i-j+1)\frac{\pi}{2}\right) = 0$. Thus, when $i$ and $j$ are both either even or odd, the corresponding $e_{ij}$ vanishes. Also, for all $i<j-1$, we have $\binom{i}{j-1} = 0$. Thus, all entries $e_{ij}$ located above the first superdiagonal of $\mathcal{E}_{2n}$ vanish as well.

We focus on the superdiagonal entry $e_{2r-1,2r}=\binom{2r-1}{2r-1}$, $r=1,\dots,n$.
It is located in an odd-numbered row $2r-1$, and the entries to the right of it are all equal to zero. Additionally, the entries in the odd-numbered columns to the left of $e_{2r-1,2r}$ are also all equal to zero, as observed above.

We compute the determinant of $\mathcal {E}_{2n}$ by the cofactor expansion across the first row. Then $\det \mathcal{E}_{2n} = -e_{12}\det \mathcal{E}^{(1)}_{2n-1}$, where $\mathcal{E}^{(1)}_{2n-1}=(e^{(1)})_{ij}$. Then

$$\mathcal {E}^{(1)}_{2n-1} = \begin{bmatrix}
-{2 \choose 0} & 2 \choose 2 & 0 & 0 & ... & 0 & 0\\
0 & 0 & 3 \choose 3 & 0 & ... & 0 & 0\\
4 \choose 0 & -{4 \choose 2} & 0 & 4 \choose 4 & ... & 0 & 0\\
0 & 0 & -{5 \choose 3} & 0 & ... & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & -{2n-1 \choose 3} & 0 & ... & 0 & 2n-1 \choose 2n-1\\
-{2n \choose 0} & 2n \choose 2 & 0 & -{2n \choose 4} & ... & -{2n \choose 2n-2} & 0
\end{bmatrix}.$$
Note that $\mathcal {E}^{(1)}_{2n-1}$ is obtained from $\mathcal {E}_{2n}$ by removing its first row and second column. Now we do the cofactor expansion across the second row of $\mathcal {E}^{(1)}_{2n-1}$.
The second row of $\mathcal {E}^{(1)}_{2n-1}$ is the third row of $\mathcal {E}_{2n}$ without $e_{32}$. The only non-zero entry in this row is $e^{(1)}_{23}=e_{34}$. Then $\det \mathcal{E}_{2n} = (-1)^2e_{34}\det \mathcal{E}^{(2)}_{2n-2}$, where $\mathcal{E}^{(2)}_{2n-2}=(e^{(2)})_{ij}$. Then
$$ \mathcal{E}^{(2)}_{2n-2}= \begin{bmatrix}
-{2 \choose 0} & 2 \choose 2 & 0 & ... & 0 & 0\\
4 \choose 0 & -{4 \choose 2} & 4 \choose 4 & ... & 0 & 0\\
0 & 0 & 0 & ... & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & ... & 0 & 2n-1 \choose 2n-1\\
-{2n \choose 0} & 2n \choose 2 & -{2n \choose 4} & ... & -{2n \choose 2n-2} & 0
\end{bmatrix}.$$
Note that $\mathcal {E}^{(2)}_{2n-2}$ is obtained from $\mathcal {E}_{2n}$ by removing its first and third row and and second and fourth column.

If we continue this pattern, we obtain $\mathcal {E}^{(l)}_{2n-l}$, $1\leq l\leq n$, from $\mathcal {E}_{2n}$ by removing all odd-numbered rows up to row $2l$ and even-numbered columns up to column $2l+1$ of the original matrix. Then
$$\det \mathcal{E}_{2n} = (-1)^le_{2l-1,2l}\det \mathcal{E}^{(l)}_{2n-l},$$
since all entries in the odd-numbered columns to the left of $e_{2l-1,2l}$ have been removed, and all the remaining entires are equal to zero.

All zeros, terms $e_{ij}$ where $i$ and $j$ have the same oddity, $i<2l-1$ and $j<2l-1$ vanish. That is all the zeros below the superdiagonal and every other zero above it.

The last row and column we can remove are row $2n-1$ and column $2n$. Then $\det {\mathcal E}_{2n}=(-1)^n\det \widehat {\mathcal E}_{2n}$, where

$$\widehat{\mathcal E}_{2}= \begin{bmatrix}
-{2 \choose 0} & 2 \choose 2 & 0 & ... & 0\\
4 \choose 0 & -{4 \choose 2} & 4 \choose 4 & ... & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-{2n \choose 0} & 2n \choose 2 & -{2n \choose 4} & ... & -{2n \choose 2n-2}
\end{bmatrix}.$$

By Lemma~\ref{absval}, we obtain
$$\det \widehat {\mathcal E}_{n}=(-1)^n\det \widetilde {\mathcal E}_{n},\quad \text{where}$$

$$\widetilde {\mathcal E}_{n} =\begin{bmatrix}
{2 \choose 0} & 2 \choose 2 & 0 & ... & 0\\
4 \choose 0 & {4 \choose 2} & 4 \choose 4 & ... & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
{2n \choose 0} & 2n \choose 2 & {2n \choose 4} & ... & {2n \choose 2n-2}
\end{bmatrix}.$$
This completes the proof.

\end{proof}