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\author{Thibaut Horel}
\title{CS 225 Problem Set 2 -- Solutions}

\begin{document}
\maketitle

Collaborator: Debmalya Mandal

\section*{Problem 2.9}

\paragraph{1.}
Let $\lambda$ be an eigenvalue of $M$ and $x$ be an associated eigenvector. Let
$i^*\in\{1,\ldots, n\}$ such that $|x_{i^*}| = \|x\|_{\infty}$. Since $x$ is an
eigenvector we have:
\begin{displaymath}
    \sum_{j=1}^n m_{i^*,j} x_j = \lambda x_{i^*}
\end{displaymath}
$x$ is non-zero, hence $x_{i^*}$ is non-zero. Dividing both sides by $x_{i^*}$,
we obtain:
\begin{displaymath}
    |\lambda| = 
    \left|\sum_{j=1}^n m_{i^*,j} \frac{x_j}{x_{i^*}}\right|
    \leq \sum_{j=1}^n m_{i^*,j} \left|\frac{x_j}{x_{i^*}}\right|
    \leq \sum_{j=1}^n m_{i^*, j} = 1
\end{displaymath}
where the first inequality used the triangle inequality and the second
inequality used the definition of $x_{i^*}$.

\paragraph{2.} Assume $G$ is disconnected. Hence it has at least two connected
components. We can write $V = C_1\cup C_2$ with $C_1\cap C_2 = \emptyset$ and
$ C_1\times C_2\cap E = \emptyset$. Then it is clear that $x^1
= \mathbf{1}_{C_1}$ (the indicator vector of $C_1$) and $x^2
= \mathbf{1}_{C_2}$ are eigenvectors of $M$ for the eigenvalue 1. For example
for $x^1$:
\begin{displaymath}
    \sum_{j=1}^n m_{i,j} x^1_j =
    \begin{cases}
        1&\text{if $i\in C_1$}\\
        0&\text{otherwise}
    \end{cases}
\end{displaymath}
where the first case follows from the fact that if $i\in C_1$, all the non-zero
terms $m_{i,j}$ will be preserved by the multiplication by $x^1_j$ since
$m_{i,j}\neq 0\Leftrightarrow j\in C_1$ and the fact that $\sum_{j=1}^n m_{i,j}
= 1$. Similarly if $i\in C_2$ $m_{i,j} = 0$ whenever $j\in C_1$.

Furthermore $x^1$ and $x^2$ are clearly orthogonal, so they span a vector space
of dimension 2. Hence the multiplicity of $1$ as an eigenvalue is at least 2.

\paragraph{} Let us now assume that $G$ is connected and let us consider an
eigenvector $x$ for the eigenvalue 1. We have $M^k x = x$ for all $k$.
Furthermore since $G$ is connected, we have that $M^n > 0$ (all the entries are
positive). Indeed, there always exists at least one path of length $n$ between
any two vertices. Writing $a_{i,j}$ the entries of $M^n$ and choosing $i^*\in
\{1,\ldots n\}$ such that $|x_{i^*}| = \|x\|_{\infty}$ we have:
\begin{displaymath}
    \sum_{j=1}^n a_{i^*,j}x_j = x_{i^*}
\end{displaymath}
Dividing by $x_{i^*}$ both sides (it is non-zero for the same reason as in
\textbf{1.}):
\begin{displaymath}
    1 = \left|\sum_{j=1}^n a_{i^*,j}\frac{x_j}{x_{i^*}}\right|
    \leq \sum_{j=1}^n a_{i^*,j}\left|\frac{x_j}{x_{i^*}}\right|
    \leq \sum_{j=1}^n a_{i^*,j} = 1
\end{displaymath}
where we again used the triangle inequality, the definition of $x_{i^*}$ and
the fact that the sum of the entries of the rows of $M^n$ is one (indeed we
have $M\textbf{1} = \textbf{1}$ where $\textbf{1}$ is the all-one vector, so
$M^n\textbf{1} = \textbf{1}$). Hence, the chain of inequalities is a chain of
equalities. Since $a_{i^*,j}\neq 0$ for all $j$, this implies that for all $j$:
$|\frac{x_j}{x_{i*}}|=1$ and all the $\frac{x_j}{x_{i^*}}$ have the same sign.
That is, for all $j$ $x_j = x_{i^*}$. Hence $x = \alpha \textbf{1}$ for some
$\alpha$.

We have just proved that $G$ connected implies that the multiplicity of $1$ is $1$
(the eigenspace is spanned by $\textbf{1}$). The contrapositive gives the
second part of the equivalence that we were asked to prove.

\paragraph{3.} Assuming that $G$ is connected, it is easy to see that $G$
bipartite is equivalent to $G^2$ being disconnected. Hence assuming $G$
connected and using \textbf{2.}:

$G$ bipartite $\Leftrightarrow$ $G^2$ disconnected $\Leftrightarrow$ $1$ is an eigenvalue of $M^2$
of multiplicity at least 2 $\Leftrightarrow$ $\exists x\perp \mathbf{1},\; M^2x
= x$ $\Leftrightarrow$ $\exists x\perp\mathbf{1},\; Mx
= \pm x$ $\Leftrightarrow$ $\exists x\perp\mathbf{1},\; Mx = -x$ $\Leftrightarrow$ $-1$ is an eigenvalue of
$M$.

The antepenultimate equivalence uses that the eigenvalues of $M^2$ are the
squares of the eigenvalues of $M$, and the penultimate equivalence uses that
$x$ cannot be an eigenvector for the eigenvalue $1$ because of \textbf{2}. The
second equivalence also uses \textbf{2.}.

\paragraph{4.} First observe that $\max_{\|x\|_2 = 1} \inprod{Mx, x}$
= $\lambda_{max}(M)$ for any symmetric matrix $M$. Indeed, using the spectral
theorem and writing $M = P^T DP$ with $D$ = $\text{diag}(\lambda_1,\dots,\lambda_n)$
and observing that $\|x\|_2 =1\Leftrightarrow \|Px\|_2 = 1$ we have:
\begin{displaymath}
    \max_{\|x\|_2 = 1} \inprod{Mx, x} = \max_{\|y\|_2=1} y^TDy = \max_{\|y\|_2
    = 1} \sum_{i=1}^n\lambda_i y_i^2
\end{displaymath}
the right-hand side maximum is clearly obtained when $y_i = 1$ for some
coordinated $i$ such that $\lambda_i = \lambda_{max}(M)$ and $y_j = 0$ for
$j\neq i$. For this $y$ the value is $\lambda_{max}(M)$.

Now observe that $\max_x\inprod{Mx, x}$ when $x$ is taken in the vector space
orthogonal to $\mathbf{1}$ is exactly the largest eigenvalue of $M$ when
restricted to the vector space orthogonal to $\mathbf{1}$ (restricting $M$ to
this subspace is allowed since this subspace is invariant by $M$). But since
the eigenspaces of $M$ are orthogonal, the largest eigenvalue of $M$ restricted
to the orthogonal of $\mathbf{1}$ is the second largest eigenvalue of $M$.

Hence:
\begin{displaymath}
    \lambda_2(M) = \max_{x} \inprod{Mx, x}
\end{displaymath}
where the maximum is taken over $x$ of length 1 in the orthogonal space of $\mathbf{1}$.

\paragrpha{} Let us now show the second identity given in the problem
statement.
\end{document}