Small Matrix Perturbation

• categories: Linear algebra, Observation

Change of Inverse Matrix

Let $A(t)$ be a differentiable matrix-valued function, where $A(t)$ is invertible. The time derivative of the inverse $A^{-1}(t)$ is:

$\frac{d}{dt}{A}^{-1}(t)=-A^{-1}(t)\frac{dA(t)}{dt}{A}^{-1}(t)$

Derivation:

1. Start from $A(t)A^{-1}(t)=I$, where $I$ is the identity matrix.
2. Differentiate both sides with respect to $t$: $\frac{dA(t)}{dt}{A}^{-1}(t)+A(t)\frac{d{A}^{-1}(t)}{dt}=0.$
3. Solve for $\frac{d{A}^{-1}(t)}{dt}$: $\frac{d{A}^{-1}(t)}{dt}=-A^{-1}(t)\frac{dA(t)}{dt}{A}^{-1}(t).$

Change of Eigenvalues

Let ${\lambda }_i(t)$ be the eigenvalues of $A(t)$ and $v_i(t)$ the corresponding eigenvectors. For a differentiable $A(t)$:

First-Order Change in Eigenvalues:

If $A(t)$ is diagonalizable, the rate of change of the eigenvalue ${\lambda }_i(t)$ is given by:

$\frac{d{\lambda }_i(t)}{dt}=v_i(t{)}^T\frac{dA(t)}{dt}{v}_i(t)$

Conditions:

1. The eigenvectors $v_i(t)$ are normalized: $v_i(t{)}^{\ast }{v}_i(t)=1$.
2. This result assumes that ${\lambda }_i(t)$ is a simple eigenvalue (no degeneracy).

Change of Singular Values

Let ${\sigma }_i(t)$ be the singular values of $A(t)$, defined as the square roots of the eigenvalues of $A(t{)}^{T}A(t)$ or $A(t)A(t{)}^T$.

First-Order Change in Singular Values:

The rate of change of ${\sigma }_i(t)$ is: $\frac{d{\sigma }_i(t)}{dt}=\frac{u_i(t{)}^T\frac{dA(t)}{dt}{v}_i(t)}{{\sigma }_i(t)}$ $u_i(t)$ and $v_i(t)$ are the left and right singular vectors corresponding to ${\sigma }_i(t)$. This assumes ${\sigma }_i(t)\ne 0$.

Conditions:

1. Singular value decomposition: $A(t)=U(t)\Sigma (t)V(t{)}^T$, where $U(t)$ and $V(t)$ are orthogonal matrices, and $\Sigma (t)$ is diagonal.
2. The singular values are real and non-negative: ${\sigma }_i(t)\ge 0$.

For small perturbations, these formulas can be used to approximate how the inverse matrix, eigenvalues, and singular values evolve over time or with respect to a parameter.