Matrix Differentiation and First-order Spectral Perturbation: Examples and Counterexamples

Since $A+tH=\mathrm{diag}(2+t,1-t)$, one has $(A+tH{)}^{-1}=\mathrm{diag}\text{\hspace{0.17em}⁣}(\frac{1}{2+t},\frac{1}{1-t})=\left(\begin{array}{cc}\frac{1}{2}-\frac{1}{4}t& 0\\ 0& 1+t\end{array}\right)+O(t^2)\mathrm{.}$ Also direct multiplication gives $-A^{-1}H{A}^{-1}=\left(\begin{array}{cc}-\frac{1}{4}& 0\\ 0& 1\end{array}\right)$.

Step 1.1 shows that the coefficient of $t$ in the exact inverse expansion is $\left(\begin{array}{cc}-\frac{1}{4}& 0\\ 0& 1\end{array}\right)$, which is exactly $-A^{-1}H{A}^{-1}$. This matches [L1] entry by entry.

At $A=I_2$, one has $\mathrm{adj}(I_2)=I_2$, so [L1] gives $D\det (I_2)[H]=\mathrm{tr}(H)=h_{11}+h_{22}$.

At $S=\mathrm{diag}(1,0)$, $\det (S+tH)=\det \left(\begin{array}{cc}1+t{h}_{11}& t{h}_{12}\\ t{h}_{21}& t{h}_{22}\end{array}\right)=t{h}_{22}+t^2(h_{11}{h}_{22}-h_{12}{h}_{21}),$ so $D\det (S)[H]=h_{22}$. Since $\mathrm{adj}(S)=\mathrm{diag}(0,1)$, this is $\mathrm{tr}(\mathrm{adj}(S)H)$, exactly as [L1] predicts. But $S^{-1}$ does not exist, so Jacobi's inverse-locus formula is unavailable.

Expanding the square gives the displayed scalar function. Differentiating entrywise yields ${\mathrm{\partial }}_{x_1}f=x_1+x_2-1$ and ${\mathrm{\partial }}_{x_2}f=x_1+2x_2-1$, so the gradient is as shown.

Also $A^{\ast }(Ax-b)=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)\left(\begin{array}{c}{x}_1+x_2-1\\ x_2\end{array}\right)=\left(\begin{array}{c}{x}_1+x_2-1\\ x_1+2x_2-1\end{array}\right)$, and $A^{\ast }A=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$. This matches [L1].

One checks directly that $A{e}_1=e_1$, so $x=e_1$ is a right eigenvector for $\lambda =1$. Also $(1,-1)A=(1,-1)$, so $y=(1,-1{)}^{\mathsf{T}}$ is a compatible left eigenvector and $y^{\ast }x=1$.

Therefore [L1] gives $D\lambda (A)[H]=y^{\ast }Hx=(1,-1)\left(\begin{array}{c}0\\ 1\end{array}\right)=-1$. But $x^{\ast }Hx=e_1^{\ast }H{e}_1=0$, so the nonnormal case genuinely uses distinct left and right eigenvectors.

The base matrix $A(0)=\mathrm{diag}(1,3)$ is Hermitian, and its simple eigenvalue $1$ has unit eigenvector $e_1$.

Applying [L1] gives ${\lambda }^{\mathrm{\prime }}(0)=e_1^{\ast }{A}^{\mathrm{\prime }}(0)e_1=2$. This is exactly the Rayleigh quotient of the perturbation in the eigenvector direction.

Direct multiplication shows $A(t)(t,1{)}^{\mathsf{T}}=2(t,1{)}^{\mathsf{T}}$, so both displayed branches are valid right eigenvectors for $\lambda =2$. Differentiating gives the displayed derivatives at $t=0$.

The difference is $x_2^{\mathrm{\prime }}(0)-x_1^{\mathrm{\prime }}(0)=-x_1(0)$, a multiple of the eigenvector itself. Thus the eigenvector derivative depends on the normalization, exactly as [L1] warns. Both branches span the same eigendirection, so they define the same spectral projector.

Computing the determinant of $zI-A_{\epsilon }$ gives $z^2-\epsilon$, so the roots are $\pm \sqrt{\epsilon }$.

The separation between the two eigenvalues is therefore $2\sqrt{\epsilon }$, not a quantity linear in $\epsilon$. This is exactly the square-root splitting described in [L1].

Because $A$ is diagonal with entries $2$ and $1$, its singular values are $2$ and $1$. The largest one is simple, with $u=v=e_1$.

Applying [L1] gives $D\sigma (A)[H]=\mathrm{Re}(e_1^{\ast }H{e}_1)=1$.

For $t<0$, the larger eigenvalue is $-t$ with eigendirection $\mathrm{span}\{e_2\}$, while for $t>0$ the larger eigenvalue is $t$ with eigendirection $\mathrm{span}\{e_1\}$.

Any normalization still has to represent those two different eigendirections on the two sides of the crossing, so the ordered branch cannot be continuous or differentiable through $t=0$. This is exactly the phenomenon recorded in [L1].