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Spatial and Network Econometrics · Tutorial 936 of 952

Invertibility and Stability of Spatial Multipliers

Learn when a spatial multiplier has a convergent series representation and how its terms describe repeated propagation through a spatial or network system.

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What You'll Learn

  • Derive the Neumann-series expansion of the spatial multiplier.
  • State the exact spectral-radius condition for convergence.
  • Distinguish convergence of feedback iteration from invertibility alone.
  • Interpret powers of a weights matrix as propagation over successive steps.
  • Apply stability checks to normalized and unnormalized weights matrices.
  • Recognize why a coefficient bound alone may be insufficient.

Why Stability Matters

In a spatial-lag model, an outcome at one location can depend on outcomes at connected locations, while those connected outcomes may themselves respond to the first location. The resulting feedback is not captured by a single pass across the weights matrix: it can circulate repeatedly. The spatial multiplier summarizes this repeated feedback, and its stability determines whether those successive rounds become negligible or instead fail to settle.

Let there be \(n\) locations or network nodes, let \(W\) be a fixed \(n\times n\) spatial weights matrix, and let \(\rho\) be a scalar spatial-lag coefficient. For a given vector \(g\) of non-spatial contributions, the spatial-lag equation is

$$ y=\rho Wy+g. $$

Here \(g\) may, for example, collect the covariate contribution and the disturbance in a spatial-lag model. Rearranging gives \((I-\rho W)y=g\). If \(I-\rho W\) is invertible, the algebraic solution is \(y=(I-\rho W)^{-1}g\). The question for this tutorial is when that inverse can also be represented as a convergent sum of successive rounds of feedback.

Definition: The spatial multiplier is \(S(\rho)=(I-\rho W)^{-1}\), when this inverse exists. The Neumann-series representation is \(S(\rho)=I+\rho W+\rho^2W^2+\cdots\), when the matrix series converges. In a finite-dimensional system, it converges exactly when the spectral radius of \(\rho W\) is less than one.

Deriving the Neumann Series

Write \(A=\rho W\). A finite geometric sum of matrices satisfies the same telescoping identity as a scalar geometric sum; multiplication is kept in the same order throughout.

$$ (I-A)\sum_{k=0}^{m}A^k =\sum_{k=0}^{m}A^k-\sum_{k=0}^{m}A^{k+1} =I-A^{m+1}. $$

If \(A^{m+1}\) tends to the zero matrix as \(m\) increases, the finite sum tends to a matrix \(S\) satisfying \((I-A)S=I\). For square matrices, a right inverse is an inverse, so \(S=(I-A)^{-1}\). Substituting \(A=\rho W\) gives the expansion.

$$ (I-\rho W)^{-1} =\sum_{k=0}^{\infty}(\rho W)^k =I+\rho W+\rho^2W^2+\rho^3W^3+\cdots. $$

The condition for this series to converge is not simply that \(I-\rho W\) be invertible. Let \(\lambda_1,\ldots,\lambda_n\) be the eigenvalues of \(W\), including complex eigenvalues if present, and define the spectral radius \(r(W)=\max_j|\lambda_j|\). The spectral radius of \(\rho W\) is \(|\rho|r(W)\). The exact convergence condition is therefore \(|\rho|r(W)<1\).

Theorem: For a finite square matrix \(W\), the Neumann series \(\sum_{k=0}^{\infty}(\rho W)^k\) converges if and only if \(r(\rho W)=|\rho|r(W)<1\). Under this condition its sum is \((I-\rho W)^{-1}\). Invertibility alone requires only that \(1-\rho\lambda_j\ne 0\) for every eigenvalue \(\lambda_j\) of \(W\); it does not guarantee convergence of the series.

To see why the spectral condition is exact, consider the powers of a matrix in its Jordan form. When every eigenvalue of \(A\) has modulus less than one, the entries in the powers of each Jordan block are sums of terms bounded by a polynomial in \(k\) times \(|\lambda|^k\). These terms tend to zero and are summable, so \(\sum A^k\) converges. In the other direction, if the matrix series converges, its terms \(A^k\) must tend to zero. For any eigenvalue \(\lambda\) and corresponding nonzero eigenvector \(v\), \(A^kv=\lambda^kv\). This can tend to zero only if \(|\lambda|<1\). Thus every eigenvalue must lie strictly inside the unit circle, which is precisely \(r(A)<1\).

What the Terms Represent

When \(W\) has nonnegative entries, its \(ij\) entry describes the one-step weight linking node \(j\) to node \(i\). The entry \((W^2)_{ij}\) sums products of weights over two-step paths from \(j\) to \(i\); more generally, \((W^k)_{ij}\) aggregates the weighted paths of length \(k\). Consequently, the multiplier adds the initial contribution \(g\), the first feedback round \(\rho Wg\), the second round \(\rho^2W^2g\), and all further rounds.

This path interpretation is clearest for nonnegative weights. If a matrix contains signed entries, its powers still describe repeated linear propagation, but individual path contributions can have either sign and need not be interpreted as positive exposure. In either case, the spectral condition determines whether the infinite sequence of rounds is summable.

Worked Example: Feedback Between Two Locations

Suppose two locations exchange feedback symmetrically, with \(W=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), \(\rho=0.4\), and an initial contribution \(g=(10,0)^\mathsf{T}\), measured in outcome points. The eigenvalues of \(W\) are \(1\) and \(-1\), so \(r(W)=1\) and \(|\rho|r(W)=0.4<1\). The series converges.

Since \(W^2=I\), the even powers contribute multiples of \(I\) and the odd powers multiples of \(W\). Thus

$$ (I-0.4W)^{-1} =\frac{1}{1-0.4^2}I+\frac{0.4}{1-0.4^2}W =\begin{pmatrix}1.1905&0.4762\\0.4762&1.1905\end{pmatrix}. $$

Multiplying by the stated initial contribution gives

$$ y=(I-0.4W)^{-1}g =\begin{pmatrix}1.1905&0.4762\\0.4762&1.1905\end{pmatrix} \begin{pmatrix}10\\0\end{pmatrix} =\begin{pmatrix}11.905\\4.762\end{pmatrix}. $$

The first location’s outcome includes the original 10 points and later feedback returning to it; the second receives feedback transmitted from the first. The entries are equilibrium outcomes in points for this hypothetical system, not a causal claim about any real locations.

Eigenvalues and the Meaning of Stability

The eigenvalues describe patterns, or modes, that are scaled by one application of \(W\). Along a mode associated with eigenvalue \(\lambda_j\), the spatial multiplier scales the corresponding component by \(1/(1-\rho\lambda_j)\), provided \(1-\rho\lambda_j\ne0\). If \(|\rho\lambda_j|\) approaches one, repeated feedback along that mode decays slowly and the multiplier can be large. If \(|\rho\lambda_j|>1\), repeated iteration along that mode grows rather than dies away.

This also connects the series condition to a feedback algorithm. Starting from any initial value \(y^{(0)}\), iterate \(y^{(m+1)}=\rho Wy^{(m)}+g\). If \(y^\ast\) is a fixed point, then the iteration error satisfies \(y^{(m+1)}-y^\ast=\rho W(y^{(m)}-y^\ast)\). Under \(|\rho|r(W)<1\), the error tends to zero. When that condition fails, the iteration need not converge, even if a fixed-point solution exists algebraically.

Worked Example: Stability With Row-Standardized Weights

Consider a three-node line with row-standardized weights \(W=\begin{pmatrix}0&1&0\\0.5&0&0.5\\0&1&0\end{pmatrix}\). Each row sums to one, and the eigenvalues are \(1,0,-1\); hence \(r(W)=1\). With \(\rho=0.8\), the exact condition is \(0.8<1\), so the series converges.

A uniform contribution \(g=c(1,1,1)^\mathsf{T}\) lies in the eigenmode with eigenvalue \(1\), since \(W(1,1,1)^\mathsf{T}=(1,1,1)^\mathsf{T}\). For this contribution the multiplier is \(1/(1-0.8)=5\), so \(y=5c(1,1,1)^\mathsf{T}\). For the alternating mode \((1,-1,1)^\mathsf{T}\), the eigenvalue is \(-1\), and the multiplier is \(1/(1+0.8)=0.5556\). A mode with eigenvalue zero is unchanged, with multiplier \(1\).

Thus the same \(\rho\) can amplify one spatial pattern and attenuate another. As \(\rho\) approaches one from below, the uniform mode’s multiplier grows without bound, while convergence still holds for every \(\rho<1\). This is a stability warning: a convergent series need not imply that the multiplier is numerically modest or insensitive to small changes in inputs.

For a row-standardized nonnegative matrix whose rows all sum to one, the constant vector is an eigenvector with eigenvalue one, and no eigenvalue has modulus greater than one. Therefore \(r(W)=1\), and the exact convergence condition reduces to \(|\rho|<1\). The row-standardization assumption matters: if rows do not sum to one, the spectral radius need not equal one.

Worked Example: Invertible but Not Neumann-Stable

Take the unnormalized weights matrix \(W=\begin{pmatrix}0&2\\1&0\end{pmatrix}\) and set \(\rho=0.8\). Its eigenvalues are \(\sqrt{2}\) and \(-\sqrt{2}\), so \(|\rho|r(W)=0.8\sqrt{2}\approx1.131>1\). The Neumann series does not converge.

Nevertheless, \(I-\rho W=\begin{pmatrix}1&-1.6\\-0.8&1\end{pmatrix}\) has determinant \(1-1.28=-0.28\), which is nonzero. It therefore has an algebraic inverse:

$$ (I-0.8W)^{-1} =\frac{1}{-0.28} \begin{pmatrix}1&1.6\\0.8&1\end{pmatrix} =\begin{pmatrix}-3.571&-5.714\\-2.857&-3.571\end{pmatrix}. $$

This inverse solves the simultaneous equations, but it is not the sum of \(I+0.8W+(0.8W)^2+\cdots\). Along the eigenmode with eigenvalue \(\sqrt{2}\), each iteration multiplies the deviation by \(0.8\sqrt{2}\approx1.131\), so that component grows rather than decays. Invertibility provides a unique algebraic solution; it does not make repeated feedback stable.

Practical Stability Checks

For an applied model, first identify the weights matrix actually used, including its normalization, and then assess its spectral radius. A convenient sufficient condition is \(|\rho|\lVert W\rVert<1\) for a compatible matrix norm, since \(r(W)\leq\lVert W\rVert\). This norm bound can be conservative: failure of the bound does not prove that the Neumann series diverges. The exact condition is based on \(r(W)\).

If \(W\) is symmetric, its eigenvalues are real and can be checked directly. For nonsymmetric weights matrices, eigenvalues may be complex, so use their moduli rather than comparing only their real parts. A coefficient bound such as \(|\rho|<1\) is sufficient for a row-standardized matrix with spectral radius one, but it is not a universal stability rule for arbitrary weights.

Common Mistake

  • Wrong reasoning: “\(I-\rho W\) has an inverse, so the geometric expansion must converge.” Invertibility only rules out an eigenvalue satisfying \(\rho\lambda_j=1\); it does not rule out eigenvalues with \(|\rho\lambda_j|>1\).
  • Correction: Check the spectral radius for the series and stability claim: require \(|\rho|r(W)<1\). Separately check \(1-\rho\lambda_j\ne0\) for all eigenvalues if the claim is only that the algebraic inverse exists.

The distinction matters because the series provides the interpretation of feedback as a sum of progressively longer propagation paths. When the series diverges, the inverse may still be computable, but it cannot be justified as the accumulated limit of those rounds. Large but convergent multipliers also call for care: near the stability boundary, small changes in \(\rho\), \(W\), or \(g\) can produce large changes in the equilibrium solution.

Key takeaway: The spatial multiplier has a convergent feedback expansion exactly when every eigenvalue of \(\rho W\) lies strictly inside the unit circle. Invertibility and stability are related but distinct conditions.

Check Your Understanding

Use the spectral condition and the feedback interpretation to answer each question.

  1. For \(W=\begin{pmatrix}0&1\\1&0\end{pmatrix}\) and \(\rho=-0.6\), does the Neumann series converge? State the spectral-radius calculation.
  2. Explain why \(W^k\) represents propagation over \(k\) steps when the entries of \(W\) are nonnegative weights.
  3. A row-standardized weights matrix has spectral radius one. What condition on \(\rho\) ensures Neumann-series convergence?
  4. Can \(I-\rho W\) be invertible even when the Neumann series diverges? Distinguish the two relevant conditions.
  5. What does a multiplier that becomes large as \(|\rho|r(W)\) approaches one indicate about feedback stability?
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