Jordan, contour, similarity, and normal-functional-calculus formulas agree

Let $C$ be any finite complex square matrix whose eigenvalues lie in $\mathrm{\Omega }$, let ${\lambda }_1,\dots ,{\lambda }_s$ be its distinct eigenvalues, and let $m_i$ be the largest Jordan-block size at ${\lambda }_i$. By [F2], there is a unique polynomial $p$ of degree less than ${\sum }_i{m}_i$ with $p^{(k)}({\lambda }_i)=f^{(k)}({\lambda }_i)$ for $k<m_i$.

On a block $J_d(\lambda )=\lambda I+N$, the finite geometric identity $(zI-J_d(\lambda ){)}^{-1}={\sum }_{k=0}^{d-1}\frac{N^k}{(z-\lambda {)}^{k+1}}$ holds whenever $z\mathrm{\ne }\lambda$, since multiplication by $(z-\lambda )I-N$ cancels adjacent terms and $N^d=0$.

For every integer $k\ge 0$, the higher product rule is $(fg{)}^{(k)}={\sum }_{j=0}^k\left(\genfrac{}{}{0px}{}{k}{j}\right)f^{(j)}{g}^{(k-j)}$: the case $k=0$ is the product itself; differentiate the finite sum using [F6] and combine adjacent terms by $\left(\genfrac{}{}{0px}{}{k}{j-1}\right)+\left(\genfrac{}{}{0px}{}{k}{j}\right)=\left(\genfrac{}{}{0px}{}{k+1}{j}\right)$ to obtain the successor case. Thus on each Jordan block, multiplying the finite Taylor jets has degree-$k$ coefficient ${\sum }_{j=0}^k{f}^{(j)}(\lambda )g^{(k-j)}(\lambda )\mathrm{/}(j!(k-j)!)=(fg{)}^{(k)}(\lambda )\mathrm{/}k!$. Since $N_d^d=0$, terms of degree at least $d$ vanish, and the resulting block is exactly the jet of $fg$. Conjugating block by block proves $(fg)(A)=f(A)g(A)$.

For each Jordan block $J_d({\lambda }_i)={\lambda }_{i}I+N_d$, the exact finite Taylor expansion of $p$ at ${\lambda }_i$ gives $p(J_d({\lambda }_i))={\sum }_{k<d}{p}^{(k)}({\lambda }_i)N_d^k\mathrm{/}k!$ because $N_d^d=0$. Since $d\le m_i$, these derivatives equal those of $f$; hence $p(C)$ equals the Jordan expression in [F1].

Apply [F4] to each eigenvalue $\lambda$ with index one. Integrating the finite sum in step 1.2 term by term therefore gives $\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}f(z)(zI-J_d(\lambda ){)}^{-1}dz={\sum }_{k=0}^{d-1}\frac{f^{(k)}(\lambda )}{k!}{N}^k\mathrm{.}$ The integral is a finite sum of scalar integrals, so entrywise linearity justifies the operation.

By [F3], any other Jordan representation has the same block multiset, so the same nodes and maximal sizes determine the same unique polynomial $p$. Thus its Jordan expression is again $p(C)$. For invertible $X$, $(XC{X}^{-1}{)}^k=X{C}^k{X}^{-1}$ for every $k$, so $p(XC{X}^{-1})=Xp(C)X^{-1}$. This proves independence and similarity invariance for every $C$.

Let $T$ be block upper triangular and let $p$ be the Hermite polynomial obtained from steps 1.1–2.1 with $C=T$. By [F3], if $m_{\lambda }$ is the largest Jordan-block size of $T$ at $\lambda$, then $a(x)={\prod }_{\lambda \in \sigma (T)}(x-\lambda {)}^{m_{\lambda }}$ annihilates $T$. Every power of $T$ is block upper triangular with $i$th diagonal block $T_{ii}^k$, so $a(T_{ii})=0$ for each $i$. Any eigenvalue $\mu$ of $T_{ii}$ therefore satisfies $a(\mu )=0$, hence $\mu \in \sigma (T)$. If $J_d(\lambda )$ is a Jordan block of $T_{ii}$, then $a(J_d(\lambda ))=0$ forces $d\le m_{\lambda }$: the factor multiplying $(J_d(\lambda )-\lambda I{)}^{m_{\lambda }}$ is invertible, so that nilpotent power must vanish. Thus the Hermite conditions defining $p$ give $p(T_{ii})=f(T_{ii})$. Every power of $T$ also shows that $p(T)$ is block upper triangular with these diagonal blocks; by step 2.1, $f(T)=p(T)$. This proves the claimed block structure and diagonal blocks.

Conjugating the blockwise identity in step 2.2 by the Jordan basis matrix $Z$ and using the definition in [F1] proves the contour formula. If $A$ is normal, [F5] supplies an orthonormal eigenbasis and all Jordan blocks have size one; the same formula is ${\sum }_{\lambda }f(\lambda )P_{\lambda }$, which is the stated spectral functional calculus.