An important consequence of the Jordan form in $C^n$ is that given a polynomial $P$ with complex coefficients and a matrix $A$, the value of $P(A)$ depends on $P$ and its derivatives only at the eigenvalues of $A$. For example,

$$P\left( \left[ \begin{array}{cccc} \lambda & 1 & 0 & 0 \\ & \lambda & 1 & 0 \\ & & \lambda & 1 \\ & & & \lambda \end{array}% \right] \right) =\left[ \begin{array}{cccc} P\left( \lambda \right) & P^{\prime }\left( \lambda \right) & P^{\prime \prime }\left( \lambda \right) /2 & P^{\prime \prime \prime }\left( \lambda \right) /6 \\ & P\left( \lambda \right) & P^{\prime }\left( \lambda \right) & P^{\prime \prime }\left( \lambda \right) /2 \\ & & P\left( \lambda \right) & P^{\prime }\left( \lambda \right) \\ & & & P\left( \lambda \right) \end{array}% \right] $$

In particular, in the finite-dimensional case if $f:D \rightarrow C$ is an analytic function on a domain containing the spectrum of $A$ then one may uniquely define $$f(A)=P(A),$$ where $P$ is any polynomial which interpolates $f$ and its derivatives at the eigenvalues of $A$ $$P^{(n)}(\lambda_k)=f^{(n)}(\lambda_k),$$ where for each eigenvalue $\lambda_k$ the order of the derivative $n$ is less than or equal to the longest chain of 1's in the Jordan block.

This definition has nice properties, such as $f(g(A))=(f\circ g)(A)$, and it agrees with the power series definition for the matrix exponential, ect.

Indeed, this is just the finite-dimensional version of the usual definition of $f(A)$ under the so-called "Dunford calculus," https://en.wikipedia.org/wiki/Holomorphic_functional_calculus which uses the Cauchy integral formula and applies in infinite-dimensions.

If one views the purpose of the Jordan form in finite-dimensions as the creation of an analytic calculus, then the Dunford calculus extends this purpose to infinite dimensions.

Now the spectral theorem allows one to define the Borel functional calculus, computing $f(A)$ for Borel measureable $f:\sigma(A)\rightarrow R$ and self-adjoint $A$. It's not obvious to me how could obtain the Borel calculus from a holomorphic calculus in infinite dimensions, in part because the spectrum of $A$ need no longer be discrete.

An important consequence of the Jordan form in $C^n$ is that given a polynomial $P$ with complex coefficients and a matrix $A$, the value of $P(A)$ depends on $P$ and its derivatives only at the eigenvalues of $A$. For example,

$$P\left( \left[ \begin{array}{cccc} \lambda & 1 & 0 & 0 \\ & \lambda & 1 & 0 \\ & & \lambda & 1 \\ & & & \lambda \end{array}% \right] \right) =\left[ \begin{array}{cccc} P\left( \lambda \right) & P^{\prime }\left( \lambda \right) & P^{\prime \prime }\left( \lambda \right) /2 & P^{\prime \prime \prime }\left( \lambda \right) /6 \\ & P\left( \lambda \right) & P^{\prime }\left( \lambda \right) & P^{\prime \prime }\left( \lambda \right) /2 \\ & & P\left( \lambda \right) & P^{\prime }\left( \lambda \right) \\ & & & P\left( \lambda \right) \end{array}% \right] $$

In particular, in the finite-dimensional case if $f:D \rightarrow C$ is an analytic function on a domain containing the spectrum of $A$ then one may uniquely define $$f(A)=P(A),$$ where $P$ is any polynomial which interpolates $f$ and its derivatives at the eigenvalues of $A$ $$P^{(n)}(\lambda_k)=f^{(n)}(\lambda_k),$$ where for each eigenvalue $\lambda_k$ the order of the derivative $n$ is less than or equal to the longest chain of 1's in the Jordan block.

This definition has nice properties, such as $f(g(A))=(f\circ g)(A)$, and it agrees with the power series definition for the matrix exponential, ect.

Indeed, this is just the finite-dimensional version of the usual definition of $f(A)$ under the so-called "Dunford calculus," https://en.wikipedia.org/wiki/Holomorphic_functional_calculus which uses the Cauchy integral formula and applies in infinite-dimensions.

If one views the purpose of the Jordan form in finite-dimensions as the creation of an analytic calculus, then the Dunford calculus extends this purpose to infinite dimensions. So one might argue that the Dunford calculus essentially is the generalization of the Jordan form.

Now the spectral theorem allows one to define the Borel functional calculus, computing $f(A)$ for Borel measureable $f:\sigma(A)\rightarrow R$ and self-adjoint $A$. It's not obvious to me how could obtain the Borel calculus from a holomorphic calculus in infinite dimensions, in part because the spectrum of $A$ need no longer be discrete.