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Recipe

Apply the Neumann series to invert I-A or bound the resolvent

Use when: you need (I-A)^{-1} or (A-\lambda I)^{-1} written as an infinite matrix power series, or need to know exactly where such a series is valid

Goal: either the literal sum, when it terminates or has a recognizable form, or the precise region where the series is guaranteed to converge

Taught in The Neumann series and the resolvent

Quick recipe

  1. Compute \rho(A), the spectral radius
  2. Match the series you're building to its own convergence condition
  3. Check whether A is nilpotent
  4. Confirm the resulting sum really is the inverse
  1. 1

    Find A's eigenvalues and take the largest of their absolute values.

    Example

    For A=\begin{bmatrix}3&0\\0&1\end{bmatrix}, the eigenvalues are already 3 and 1, so \rho(A)=3.

    Why: Gelfand's formula shows this single number controls whether A^k\to0, the condition the whole series argument rests on.

  2. 2

    Decide whether the question asks for (I-A)^{-1} itself or the resolvent (A-\lambda I)^{-1} at a specific \lambda.

    Example

    For the resolvent at A=\begin{bmatrix}3&0\\0&1\end{bmatrix} with \rho(A)=3, the valid region is every \lambda with |\lambda|>3.

    • solving for (I-A)^{-1} directly — the series converges exactly when \rho(A)<1.
    • solving for the resolvent (A-\lambda I)^{-1} — the series converges exactly when |\lambda|>\rho(A).

    Why: rewriting A-\lambda I as -\lambda(I-A/\lambda) turns its convergence condition into \rho(A/\lambda)<1, which is exactly |\lambda|>\rho(A).

  3. 3

    Compute increasing powers of A; a power that comes out to the zero matrix truncates the whole series there.

    Example

    For A=\begin{bmatrix}0&0.5\\0&0\end{bmatrix}, A^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}, so every power past A^1 vanishes.

    • A is nilpotent — sum only the finitely many nonzero terms directly, since \rho(A)=0<1 automatically.
    • A isn't nilpotent — the series genuinely has infinitely many nonzero terms, so Step 2's region is the honest final answer unless a closed form is already known.

    Why: a nilpotent matrix always has \rho(A)=0, which trivially satisfies the convergence bound and turns the infinite sum finite in practice.

  4. 4

    Multiply I-A by the sum found in Step 3 and check the product comes out to I.

    Example

    (I-A)(I+A)=I-A^2=I-\begin{bmatrix}0&0\\0&0\end{bmatrix}=I, confirming (I-A)^{-1}=I+A.

    Why: this checks the definition of matrix inverse directly, catching a dropped term or sign error in the series before trusting the answer.

Common mistakes

  • Reporting the Neumann series' own convergence condition, \rho(A)<1, when the resolvent condition |\lambda|>\rho(A) was actually asked for.
  • Forgetting that a nilpotent matrix already satisfies the convergence bound automatically, and needlessly checking \rho(A)<1 by computing eigenvalues.
  • Stopping the sum at the wrong power, one term before or after the power that first hits the zero matrix.
  • Skipping the Step 4 check and trusting a hand-summed series without verifying (I-A) times the sum equals I.