Let $n$ be a nonnegative integer, and let $B_n$ be the $n \times n$-matrix (over the rational numbers) whose $\left(i, j\right)$-th entry is $\dbinom{n+1}{2j-i}$ for all $i, j \in \left\{ 1, 2, \ldots, n \right\}$.

For example, \begin{equation} B_5 = \left(\begin{array}{rrrrr} 6 & 20 & 6 & 0 & 0 \\ 1 & 15 & 15 & 1 & 0 \\ 0 & 6 & 20 & 6 & 0 \\ 0 & 1 & 15 & 15 & 1 \\ 0 & 0 & 6 & 20 & 6 \end{array}\right) . \end{equation}

Question 1. Prove that the eigenvalues of $B_n$ are $2^1, 2^2, \ldots, 2^n$. (I know how to do this -- I'll write up the answer soon -- but there might be other approaches too.)

Question 2. Find a left eigenvector for each of these eigenvalues. What I know is that the row vector $v_n$ whose $i$-th entry is $\left(-1\right)^{i-1} \dbinom{n-1}{i-1}$ (for $i \in \left\{1,2,\ldots,n\right\}$) is a left eigenvector for eigenvalue $2^1$ (that is, $v_n B_n = 2 v_n$). But the other left eigenvectors are a mystery to me.

Question 3. Find a right eigenvector for each of these eigenvalues. For example, it appears to me that the row vector $w_n$ whose $i$-th entry is $\left(-1\right)^{i-1} / \dbinom{n-1}{i-1}$ (for $i \in \left\{1,2,\ldots,n\right\}$) is a right eigenvector for eigenvalue $2^1$ (that is, $B_n w_n = 2 w_n$). This (if correct) boils down to the identity \begin{equation} \sum_{k=1}^n \left(-1\right)^{k-1} \left(k-1\right)! \left(n-k\right)! \dbinom{n+1}{2k-i} = 2 \left(-1\right)^{i-1} \left(i-1\right)! \left(n-i\right)! \end{equation} for all $i \in \left\{1,2,\ldots,n\right\}$. Note that the entries of $w_n$ are the reciprocals to the corresponding entries of $v_n$ ! Needless to say, this pattern doesn't persist, but maybe there are subtler patterns.

I am going to put up an answer to Question 1 soon, as a stepping stone for the proof of https://math.stackexchange.com/questions/2886392 , but this shouldn't keep you from adding your ideas or answers.

Let $n$ be a nonnegative integer, and let $B$ be the $n \times n$-matrix (over the rational numbers) whose $\left(i, j\right)$-th entry is $\dbinom{n+1}{2j-i}$ for all $i, j \in \left\{ 1, 2, \ldots, n \right\}$.

For example, if $n = 5$, then \begin{equation} B = \left(\begin{array}{rrrrr} 6 & 20 & 6 & 0 & 0 \\ 1 & 15 & 15 & 1 & 0 \\ 0 & 6 & 20 & 6 & 0 \\ 0 & 1 & 15 & 15 & 1 \\ 0 & 0 & 6 & 20 & 6 \end{array}\right) . \end{equation}

Question 1. Prove that the eigenvalues of $B$ are $2^1, 2^2, \ldots, 2^n$. (I know how to do this -- I'll write up the answer soon -- but there might be other approaches too.)

Question 2. Find a left eigenvector for each of these eigenvalues. What I know is that the row vector $v$ whose $i$-th entry is $\left(-1\right)^{i-1} \dbinom{n-1}{i-1}$ (for $i \in \left\{1,2,\ldots,n\right\}$) is a left eigenvector for eigenvalue $2^1$ (that is, $v B = 2 v$). But the other left eigenvectors are a mystery to me.

Question 3. Find a right eigenvector for each of these eigenvalues. For example, it appears to me that the column vector $w$ whose $i$-th entry is $\left(-1\right)^{i-1} / \dbinom{n-1}{i-1}$ (for $i \in \left\{1,2,\ldots,n\right\}$) is a right eigenvector for eigenvalue $2^1$ (that is, $B w = 2 w$). This (if correct) boils down to the identity \begin{equation} \sum_{k=1}^n \left(-1\right)^{k-1} \left(k-1\right)! \left(n-k\right)! \dbinom{n+1}{2k-i} = 2 \left(-1\right)^{i-1} \left(i-1\right)! \left(n-i\right)! \end{equation} for all $i \in \left\{1,2,\ldots,n\right\}$. Note that the entries of $w$ are the reciprocals to the corresponding entries of $v$ ! Needless to say, this pattern doesn't persist, but maybe there are subtler patterns.

I am going to put up an answer to Question 1 soon, as a stepping stone for the proof of https://math.stackexchange.com/questions/2886392 , but this shouldn't keep you from adding your ideas or answers.

Let $n$ be a nonnegative integer, and let $B_n$ be the $n \times n$-matrix (over the rational numbers) whose $\left(i, j\right)$-th entry is $\dbinom{n+1}{2j-i}$ for all $i, j \in \left\{ 1, 2, \ldots, n \right\}$.

For example, \begin{equation} B_5 = \left(\begin{array}{rrrrr} 6 & 20 & 6 & 0 & 0 \\ 1 & 15 & 15 & 1 & 0 \\ 0 & 6 & 20 & 6 & 0 \\ 0 & 1 & 15 & 15 & 1 \\ 0 & 0 & 6 & 20 & 6 \end{array}\right) . \end{equation}

Question 1. Prove that the eigenvalues of $B_n$ are $2^1, 2^2, \ldots, 2^n$. (I know how to do this -- I'll write up the answer soon -- but there might be other approaches too.)

Question 2. Find a left eigenvector for each of these eigenvalues. What I know is that the row vector $v_n$ whose $i$-th entry is $\left(-1\right)^{i-1} \dbinom{n-1}{i-1}$ (for $i \in \left\{1,2,\ldots,n\right\}$) is a left eigenvector for eigenvalue $2^1$ (that is, $v_n B_n = 2 v_n$). But the other left eigenvectors are a mystery to me.

Question 3. Find a right eigenvector for each of these eigenvalues. For example, it appears to me that the row vector $w_n$ whose $i$-th entry is $\left(-1\right)^{i-1} / \dbinom{n-1}{i-1}$ (for $i \in \left\{1,2,\ldots,n\right\}$) is a right eigenvector for eigenvalue $2^1$ (that is, $B_n w_n = 2 w_n$). Note that the entries of $w_n$ are the reciprocals to the corresponding entries of $v_n$ ! Needless to say, this pattern doesn't persist, but maybe there are subtler patterns.

I am going to put up an answer to Question 1 soon, as a stepping stone for the proof of https://math.stackexchange.com/questions/2886392 , but this shouldn't keep you from adding your ideas or answers.

Let $n$ be a nonnegative integer, and let $B_n$ be the $n \times n$-matrix (over the rational numbers) whose $\left(i, j\right)$-th entry is $\dbinom{n+1}{2j-i}$ for all $i, j \in \left\{ 1, 2, \ldots, n \right\}$.

For example, \begin{equation} B_5 = \left(\begin{array}{rrrrr} 6 & 20 & 6 & 0 & 0 \\ 1 & 15 & 15 & 1 & 0 \\ 0 & 6 & 20 & 6 & 0 \\ 0 & 1 & 15 & 15 & 1 \\ 0 & 0 & 6 & 20 & 6 \end{array}\right) . \end{equation}

Question 1. Prove that the eigenvalues of $B_n$ are $2^1, 2^2, \ldots, 2^n$. (I know how to do this -- I'll write up the answer soon -- but there might be other approaches too.)

Question 2. Find a left eigenvector for each of these eigenvalues. What I know is that the row vector $v_n$ whose $i$-th entry is $\left(-1\right)^{i-1} \dbinom{n-1}{i-1}$ (for $i \in \left\{1,2,\ldots,n\right\}$) is a left eigenvector for eigenvalue $2^1$ (that is, $v_n B_n = 2 v_n$). But the other left eigenvectors are a mystery to me.

Question 3. Find a right eigenvector for each of these eigenvalues. For example, it appears to me that the row vector $w_n$ whose $i$-th entry is $\left(-1\right)^{i-1} / \dbinom{n-1}{i-1}$ (for $i \in \left\{1,2,\ldots,n\right\}$) is a right eigenvector for eigenvalue $2^1$ (that is, $B_n w_n = 2 w_n$). This (if correct) boils down to the identity \begin{equation} \sum_{k=1}^n \left(-1\right)^{k-1} \left(k-1\right)! \left(n-k\right)! \dbinom{n+1}{2k-i} = 2 \left(-1\right)^{i-1} \left(i-1\right)! \left(n-i\right)! \end{equation} for all $i \in \left\{1,2,\ldots,n\right\}$. Note that the entries of $w_n$ are the reciprocals to the corresponding entries of $v_n$ ! Needless to say, this pattern doesn't persist, but maybe there are subtler patterns.

I am going to put up an answer to Question 1 soon, as a stepping stone for the proof of https://math.stackexchange.com/questions/2886392 , but this shouldn't keep you from adding your ideas or answers.