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February 5

Hessian Matrix Meaning

Let $f:{ℝ}^n\to ℝ$ be a smooth function. Let $x\in {ℝ}^n$, such that the gradient of f at x is zero. Let H be the Hessian matrix of f at the point x. Let V be the vector space spanned by the eigenvectors corresponding to negative eigenvalues of H. Let $y\in V$. Then, f(x)>f(y)? or maybe f(x)>f(x+y)?

In other words, does negative eigenvalue imply maximum point at the direction of the corresponding eigenvector, or maybe this is a maximum in another direction, and not in the direction of the eigenvector? עברית (talk) 06:45, 5 February 2016 (UTC)

See Morse lemma. Sławomir
Biały 12:23, 5 February 2016 (UTC)

Template:EcYou're kind of circling around the second derivative test for functions of several variables. The Taylor expansion for f at x is

$f(𝐱+𝐲)\approx f(𝐱)+{𝐲}^{\mathrm{T}}\mathrm{D}f(𝐱)+\frac{1}{2!}{𝐲}^{\mathrm{T}}{\mathrm{D}}^{2}f(𝐱)𝐲+\cdots$

where Df is the gradient and D²f is the Hessian. In this case the gradient is 0 at x so this reduces to

$f(𝐱+𝐲)\approx f(𝐱)+\frac{1}{2!}{𝐲}^{\mathrm{T}}{\mathrm{D}}^{2}f(𝐱)𝐲+\cdots .$

Let e be an eigenvector with eigenvalue λ, and wlog take e to be length 1. If y = te, then

$f(𝐱+𝐲)\approx f(𝐱)+\frac{1}{2!}\lambda t^2+\cdots$

so f has a local minimum or maximum along the line parallel to e though x, depending on whether λ is positive or negative. If e, f ... are several linearly independent eigenvectors, with eigenvalues λ, μ, ... , and y = te + uf + ... , then

$f(𝐱+𝐲)\approx f(𝐱)+\frac{1}{2!}(\lambda t^2+\mu u^2+\cdots )+\cdots$

so f has a local minimum or maximum in the relevant space though x provided λ, μ, ... have the same sign. (The eigenvectors may be taken to be orthogonal since D²f is symmetric.) Note, this is only valid for y sufficiently small, otherwise the higher order terms in the Taylor series become significant and the approximation is no longer valid. --RDBury (talk) 12:46, 5 February 2016 (UTC)

Oh, great! Thank you! :) עברית (talk) 08:38, 6 February 2016 (UTC)