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For a Hermitian family, the first derivative of a simple eigenvalue is the corresponding Rayleigh quotient direction

Example

Let

$A (t) = (\begin{array}{cc} 1 & 0 \\ 0 & 3 \end{array}) + t (\begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array}) .$

At $t = 0$ , the simple eigenvalue $λ = 1$ has unit eigenvector $x = e_{1}$ , so

$λ^{'} (0) = x^{*} A^{'} (0) x = e_{1}^{*} (\begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array}) e_{1} = 2.$

Facts & Assumptions

Given: The Hermitian family $A (t)$ above.

[L1]

For a Hermitian simple eigenvalue, the derivative simplifies to $x^{*} A^{'} x$ (For a Hermitian simple eigenvalue, one may take $y = x$ and the first-order formulas simplify accordingly).

Verification

technique · direct

1.1algebra

The base matrix $A (0) = diag (1, 3)$ is Hermitian, and its simple eigenvalue $1$ has unit eigenvector $e_{1}$ .

2.1L1step 1.1algebra∎

Applying [L1] gives $λ^{'} (0) = e_{1}^{*} A^{'} (0) e_{1} = 2$ . This is exactly the Rayleigh quotient of the perturbation in the eigenvector direction.

Depends on

For a Hermitian simple eigenvalue, one may take $y=x$ and the first-order formulas simplify accordingly

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Alan Edelman and Steven G. Johnson, Matrix Calculus for Machine Learning and Beyond (standard reference, not scraped)