ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-31

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For a nonnormal matrix, the simple eigenvalue derivative uses distinct left and right eigenvectors

Example

Let

$A = (\begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array}), H = (\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}) .$

For the simple eigenvalue $λ = 1$ , a right eigenvector is $x = e_{1}$ and a left eigenvector is $y = (1, - 1)^{T}$ , with $y^{*} x = 1$ . Therefore

$D λ (A) [H] = y^{*} H x = - 1,$

whereas the Hermitian-style expression $x^{*} H x$ would give $0$ .

Facts & Assumptions

Given: The matrices $A$ and $H$ above.

[L1]

For a simple eigenvalue normalized by $y^{*} x = 1$ , the derivative is $D λ (A) [H] = y^{*} H x$ (Along a differentiable matrix path, a simple eigenvalue satisfies $λ^{'} = y^{*} A^{'} x$ under the normalization $y^{*} x = 1$ ).

Verification

technique · direct

1.1

One checks directly that $A e_{1} = e_{1}$ , so $x = e_{1}$ is a right eigenvector for $λ = 1$ . Also $(1, - 1) A = (1, - 1)$ , so $y = (1, - 1)^{T}$ is a compatible left eigenvector and $y^{*} x = 1$ .

algebra

2.1

Therefore [L1] gives $D λ (A) [H] = y^{*} H x = (1, - 1) (\begin{array}{c} 0 \\ 1 \end{array}) = - 1$ . But $x^{*} H x = e_{1}^{*} H e_{1} = 0$ , so the nonnormal case genuinely uses distinct left and right eigenvectors.

L1step 1.1algebra∎

Depends on

Along a differentiable matrix path, a simple eigenvalue satisfies $\lambda'=y^*A'x$ under the normalization $y^*x=1$

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

David Bindel, CS 6210: Matrix Computations - Perturbation theory (standard reference, not scraped)