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Let $G$ be a simple graph with incidence matrix $M$.

a) Show that the adjacency matrix of its line graph $L(G)$ is $M^t M − 2I$, where $I$ is the $m × m$ identity matrix.
b) Using the fact that $M^t M$ is positive-semidefinite, deduce that:
$\hspace{1cm}$i) each eigenvalue of $L(G)$ is at least $−2$,
$\hspace{1cm}$ii) if the rank of $M$ is less than $m$, then $−2$ is an eigenvalue of $L(G)$.

I could prove part a but could not prove part b. Please help me with it. Thank you.

rhkoulen
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Ramakrishna Nookala
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    I may not be of much help (this isn't my topic of choice), but I remember some factoid that the Graph Laplacian has eigenvalues greater than or equal to 0, and that matrix is partially derived from the adjacency matrix. Then we're subtracting 2I and it's asking to prove whether each eigenvalue is greater than to equal to -2? I'm really not sure, but it might be a starting point. – rhkoulen Nov 02 '21 at 02:03
  • My lin alg class caught up to some of these topics and I've laid out a proof that I think works! – rhkoulen Nov 06 '21 at 22:36

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First, let's observe that $M^tM$ is positive semidefinite. This means that it has only non-negative eigenvalues. So, let's find the eigenvalues for $M^tM-2I_m$. $$\vert (M^tM-2I_m)-\lambda I_m \vert=0$$ $$\vert M^tM-(\lambda+2)I_m \vert=0$$ Because we know that $M^tM$ has non-negative eigenvalues, $$(\lambda+2)\geq0$$ $$\lambda\geq-2$$ And there's point (i).

Point (ii) is a little more difficult.
First, it's very easy to see, by the IMT, that if $rank(M^tM)<m$, then $M^tM$ must have the eigenvalue $0$. Additionally, by the same argument above, we can show that this also implies $M^tM-2I_m$ must have the eigenvalue $-2$. Now, I'm not certain that notation $^t$ means transpose (I've always seen $^T$), but if it does, we can show that $rank(M)<m$ implies that $rank(M^tM)<m$.

This requires two very cool facts. First, column rank is equal to row rank: $rank(M)=rank(M^T)$, and rank cannot be increased by multiplication: $rank(AB)\leq min(rank(A),rank(B))$. $$rank(M^tM)\leq min(rank(M^t),rank(M))$$ $$rank(M^tM)\leq min(rank(M),rank(M))$$ $$rank(M^tM)\leq rank(M)$$ $$rank(M^tM)\leq rank(M)<m$$ $$rank(M^tM)<m$$ In summary, $rank(M)<m$ implies that $rank(M^tM)<m$ which implies that $M^tM$ has an eigenvalue $0$ which implies that $M^tM-2I_m$ has an eigenvalue $-2$.

rhkoulen
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