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\begin{document}



\title{Linear Algebra Projects}

\author{}

\date{}


\maketitle


This project eventually answers the following question: {\em What is the
maximal number of linearly independent vectors in $\R^n$ so that the angles
between any two distinct vectors are obtuse?} Maybe you want to guess the
answer before you proceed, using $n = 2, 3$ as a guide. The following route
might not be the shortest, but it beautifully incorporates many different
aspects of linear algebra.

We write $\langle-, -\rangle$ for the standard inner product in the
following.

\begin{enumerate}
\item Let $\{e_1, \cdots, e_n\}$ be a basis of $\R^n$. Show that there is a
    basis $\{e_1^*, \cdots, e_n^*\}$ so that
        \[
        \langle e_i, e_j^* \rangle  =\left\{ \begin{aligned}
        & 1, && i = j \\ & 0, && i \not=j \end{aligned} \right., \quad
        1 \leq i, j \leq n.
        \]
    This basis is called the dual basis.

\item Show that we can find linearly independent vectors $v_1, \cdots,
    v_{n+1} \in \R^n$ so that
    \[
    \langle v_i, v_j \rangle < 0, \quad 1 \leq i \not= j \leq n+1.
    \]
    In other words, the angle between $v_i$ and $v_j$ is obtuse if $i
    \not=j$. (Hint: consider the standard simplex in $\R^n$ and vectors
    from the center of mass pointing to all vertices.)


\item Let $v_1, \cdots, v_{n+1}$ be as in the previous step. Then $\{v_1,
    \cdots, v_n\}$ is a basis. Let $\{v_1^*, \cdots, v_n^*\}$ be the dual
    basis. Assume that $v_{n+2} \in \R^n$ and $\langle v_i, v_{n+2} \rangle
    <0$ for all $i = 1, \cdots, n$. If we write
        \[
        v_{n+1} = a_1 v_1^* + \cdots + a_n v_n^*,
        \quad
        v_{n+2} = b_1 v_1^* + \cdots + b_n v_n^*,
        \]
    then $a_i$'s and $b_j$'s are all negative.

\item Let $A$ be an $n \times n$ matrix whose $(i, j)$ entry is $\langle
    v_i, v_j \rangle$ and $B$ be an $n \times n$ matrix whose $(i, j)$
    entry is $\langle v_i^*, v_j^* \rangle$. Show that $A$ and $B$ are
    symmetric positive definite and $A = B^{-1}$.

\item We will show below that all entries of $B$ are positive. Use this to
    conclude that
    \[
    \langle v_{n+1}, v_{n+2} \rangle = (a_1, \cdots, a_n) B \begin{pmatrix}
    b_1 \\ \vdots \\ b_n \end{pmatrix} > 0.
    \]
    Therefore the maximal number of linearly independent vectors in $\R^n$
    with pairwise obtuse angle is $n+1$.
\end{enumerate}

It remains to prove that all entries of $B$ are positive. Recall that $A$ and
$B$ being positive definite means that their principal minors are all
positive. In what follows, the fact that $A$ and $B$ are symmetric is
superfluous. Assume from now on that $A$ is any matrix satisfying the
following conditions.
\begin{itemize}
\item All principal minors are positive.

\item Diagonal entries are positive.

\item Off-diagonal entries are negative.
\end{itemize}
We will show that all entries of $A^{-1}$ are positive by induction on $n$.
The case $n = 1$ is trivial and we assume that the result is established for
all $(n-1) \times (n-1)$ matrices. Write
    \[
    A = \begin{pmatrix} A_1 & \alpha \\ \beta & a_{nn} \end{pmatrix}, \quad
    \alpha, \tp{\beta} \in \R^{n-1}, \quad a_{nn} \in \R.
    \]

\begin{enumerate}
\item Show that $A_1$ is invertible.

\item Prove that
    \[
    \begin{pmatrix} A_1 & \alpha \\ \beta & a_{nn} \end{pmatrix} =
    \begin{pmatrix} 1_{n-1} \\ \beta A_1^{-1} & 1 \end{pmatrix}
    \begin{pmatrix} A_1 \\ & a_{nn} - \beta A_1^{-1} \alpha \end{pmatrix}
    \begin{pmatrix} 1_{n-1} & A_1^{-1} \alpha \\ & 1 \end{pmatrix}.
    \]


\item Prove that
    \[
    A^{-1}  = \begin{pmatrix} A_1^{-1} +
    (a_{nn} - \beta A_1^{-1} \alpha)^{-1} A_1^{-1} \alpha \beta A_1^{-1} &
    -(a_{nn} - \beta A_1^{-1} \alpha)^{-1}A_1^{-1} \alpha\\
    -(a_{nn} - \beta A_1^{-1} \alpha)^{-1}\beta A_1^{-1} & (a_{nn} - \beta A_1^{-1} \alpha)^{-1}
    \end{pmatrix}.
    \]

\item Show that $a_{nn} - \beta A_1^{-1} \alpha > 0$.

\item Apply the induction hypothesis to $A_1^{-1}$ and get the conclusion
    for $A^{-1}$.
\end{enumerate}











\end{document}
