Recipe

Bound or confirm the extreme eigenvalues using the Rayleigh quotient

Use when: you want to estimate, bound, or confirm \lambda_{\max} or \lambda_{\min} of a symmetric matrix without first solving the full characteristic equation

Goal: a certified lower bound on \lambda_{\max} (or upper bound on \lambda_{\min}), tightened to an exact value once a genuine eigenvector turns up

Taught in Rayleigh-Ritz and the extreme eigenvalues

1. 1

   Pick several nonzero test vectors, starting with the standard basis vectors, and compute R_A(x) at each one.

   Example

   For A=\begin{bmatrix}5&0\\0&1\end{bmatrix}, test (1,0), (0,1), and (1,1): R_A(1,0)=5, R_A(0,1)=1, R_A(1,1)=3.

   Why: Rayleigh-Ritz guarantees every one of these numbers already lands inside [\lambda_{\min},\lambda_{\max}], so each sampled value is itself a genuine bound.

2. 2

   The biggest number found so far is a certified lower bound; it can never exceed the true \lambda_{\max}.

   Example

   The largest sampled value above is R_A(1,0)=5, so \lambda_{\max}\ge5 is already certain.

   • the direction giving the largest value looks like a simple vector, such as a coordinate axis — test it directly for being an eigenvector next.
   • no candidate looks like an eigenvector yet — sample more directions, especially ones between the best two found so far, and repeat this step.

   Why: a weighted average of eigenvalues can never exceed \lambda_{\max}, so the largest quotient found by any sampling is a lower bound that is never an overshoot.

3. 3

   Multiply A by the leading candidate x; if the result is exactly a scalar multiple of x, that scalar is the exact eigenvalue.

   Example

   A(1,0)=(5,0)=5(1,0), confirming (1,0) is a genuine eigenvector, so \lambda_{\max}=5 exactly, not merely a lower bound.

   Why: the theorem guarantees equality only at an actual eigenvector, so this check is what upgrades "at least this large" to "exactly this."

4. 4

   Once \lambda_{\min} and \lambda_{\max} are known, compute R_A at a vector known not to be an eigenvector and confirm it lands strictly between them.

   Example

   For A=\begin{bmatrix}3&1\\1&3\end{bmatrix} with \lambda_{\min}=2,\lambda_{\max}=4, R_A(1,0)=3, strictly between the two.

   Why: landing exactly on an endpoint without the vector being an eigenvector would contradict the theorem, so this check catches an arithmetic slip in an earlier step.