Chapter 31 The “Inverse” of \(R\)
Wednesday, April 14, 1993
Let \(\Gamma = (X, E)\) be any graph of diameter \(D\geq 2\). Fix a vertex \(x\in X\). Let \(E^*_i\equiv E^*_i(x)\), and \(T \equiv T(x)\).
Recall adjacency matrix \[\begin{align} A & = R + L + F\\ R & = \sum_{i=0}^D E^*_{i+1}AE^*_i,\\ L & = \sum_{i=0}^D E^*_{i-1}AE^*_i,\\ F & = \sum_{i=0}^D E^*_{i}AE^*_i. \end{align}\]
Observe \(R\) is not invertible (indeed \(RE^*_D = O\).) So, \(R^{-1}\) does not exist.
Below we find a matrix “\(R^{-1}\)”\(\in T(x)\) such that \(R^{-1}Rv = v\) for “almost all” \(v\in V\).
Lemma 31.1 Let \(\Gamma = (X, E)\) denote any graph, and the standard module \(V\) over \(\mathbb{C}\).
Fix a vertex \(x\in X\), write \[R\equiv R(x), \quad L \equiv L(x), \quad E^*_i\equiv E^*_i(x) \quad \text{for all }i.\] Then,
Proof.
\[V = (\mathrm{Ker}L) + (\mathrm{Ker}L)^\bot.\]
Claim 1. \(RL(\mathrm{Ker}L)^\bot \subseteq (\mathrm{Ker}L)^\bot\).
Proof of Claim 1. Pick \(v\in (\mathrm{Ker}L)^\bot\), and \(w\in \mathrm{Ker}L\). Show \[\langle RLv, w\rangle = 0.\] But \[\bar{R}^\top = R^\top = \left(\sum_{i=0}^D E^*_{i+1}AE^*_i\right)^\top = \sum_{i=0}^D E^*_iAE^*_{i+1} = L.\] So, \[\langle RLv, w\rangle = \langle Lv, \bar{R}^\top w\rangle = \langle Lv, Lw\rangle = 0.\]
Claim 2. \(RL: (\mathrm{Ker}L)^\bot \to (\mathrm{Ker}L)^\bot\) is an isomorphism of vector spaces.
Proof of Claim 2. It suffices to show above map is one-to-one.
Suppose there is a vector \(v\in (\mathrm{Ker}L)^\bot\) such that \(RLv = 0\).
Then, \[0 = \langle RLv , v\rangle = \langle Lv, \bar{R}^\top v\rangle = \|Lv\|^2.\] So \(Lv = 0\).
Hence \(v\in \mathrm{Ker}L \cap (\mathrm{Ker}L)^\bot = 0\).
This proves Claim 2.
Now “\(R^{-1}\) denote the unique matrix in \(\mathrm{Mat}(\mathbb{C})\) such that \[\begin{equation} R^{-1}v = \begin{cases} 0 & \text{if } v\in \mathrm{Ker}L \\ L(RL)^{-1}v & \text{if } v\in (\mathrm{Ker}L)^\bot. \end{cases}\tag{31.1} \end{equation}\] Observe that \((RL)^{-1}:(\mathrm{Ker}L)^\bot \to (\mathrm{Ker}L)^\bot\) exists by Claim 2.
Observe \(R^{-1}\) satisfies \((ia)\) by (31.1).
Claim 3. \(R^{-1}\) satisfies \((ib)\).
Proof of Claim 3. It suffices to check \[R^{-1}RLv = Lv\] for \(v\in \mathrm{Ker}L\) and \(v\in (\mathrm{Ker}L)^\bot\).
The case \(v\in \mathrm{Ker}L\) is clear. So assume \(v\in (\mathrm{Ker}L)^\bot\) by Claim 1. So, \[R^{-1}(RLv) = L(RL)^{-1}RLv = Lv\] as desired.
Uniqueness: Suppose a matrix \(\hat{R}^{-1}\in \mathrm{Mat}_X(\mathbb{C})\) satisfies \((ia), (ib)\). Then, \(\hat{R}^{-1}\) satisfies (31.1) above.
(Pf. The first part is clear. Let \(v\in (\mathrm{Ker}L)^\bot\). By Claim 2, there exists \(w\in (\mathrm{Ker}L)^\bot\) such that \(v\in RLw\). So \(\hat{R}^{-1}v = \hat{R}^{-1}RLw = Lw = L(RL)^{-1}v\).)
Therefore, \(\hat{R}^{-1}\) agrees with \(R^{-1}\) on a basis for \(V\), and \(\hat{R}^{-1} = R^{-1}\).
Without loss of generality we may assume that \(v\in \mathrm{Ker}L\) or \(v\in (\mathrm{Ker}L)^\bot\).
If \(v\in \mathrm{Ker}L\), then \(R^{-1}v = 0\in E^*_{i-1}V\).
If \(v\in (\mathrm{Ker}L)^\bot\), then \[R^{-1}v = L(RL)^{-1}v \in LE^*_iV \subseteq E^*_{i-1}V.\]
So \(V\), \(\mathrm{Ker}L\), each has basis consisting of vectors in \(\mathbb{Q}^{|X|}\).
Replacing the construction of \(R^{-1}\) with the base field replaced by \(\mathbb{Q}\), we find a matrix \(\tilde{R}^{-1}\in \mathrm{Mat}_X(\mathbb{Q})\) satisfying \((ia)\), \((ib)\).
Now \(R^{-1}\) and \(\tilde{R}^{-1}\) agree on a basis, and hence \(R^{-1} = \tilde{R}^{-1}\).
Let \(\theta\) be any eigenvalue of \(RL\). Let \(V_\theta\) denote the corresponding maximal eigenspace in \(V\). Then \[V = \sum_{\theta:\text{eigenvalue for }RL}V_\theta \quad (\text{orthogonal direct sum}).\] Let \(E_\theta: V\to V_\theta\) denote the orthogonal projection. Then \(E_\theta\) is a complex polynomial in \(RL\).
Thus \(E_\theta\in T(x)\).
HS MEMO
\(E_\theta\) is real. Since \(RL\) is an integral matrix, every eigenvalue of \(RL\) is an algebraic integer.
Claim 4. We have \[\begin{equation} R^{-1} = \sum_{\theta: \text{eigenvalue of }RL}\theta^{-1}LE_\theta. \tag{31.2} \end{equation}\] In particular, \(R^{-1}\in T(x)\).
Proof of Claim 4. Show two sides of (31.2) agree, when applied to arbitrary \(v\in V\).
Without loss of generality, we may assume that \(v\in V_\theta\) for some eigenvalue \(\theta\) of \(RL\).
Let \(\theta'\) denote any eigenvalue of \(RL\). \[E_{\theta'}v = \begin{cases} 0 & \text{if } \theta'\neq \theta,\\ v & \text{if } \theta' = \theta. \end{cases}\] \(\mathrm{RHS}\) of (31.2) applied to \(v\) equals \[\begin{cases} 0 & \text{if } \theta = 0,\\ \theta^{-1}Lv & \text{if } \theta \neq 0. \end{cases}\]
Show this equals \(R^{-1}v\).
Case \(\theta = 0\): Since \(RLv = 0\), \[0 = \langle v, RLv\rangle = \|Lv\|^2.\]
Hence \(Lv = 0\), or \(v\in \mathrm{Ker}L\). By \((ia)\), \(R^{-1}v = 0\).
Case \(\theta \neq 0\): Since \(RLv = \theta v\), \(v = \theta^{-1}RLv\). Hence, \[R^{-1}v = \theta^{-1}R^{-1}RLv = \theta^{-1}Lv\] by \((ib)\).