Some 🇺🇸 State Space Models have partial 🕹️ Controllability: there are some components that are controllable, and others that are not. Kalman decomposition is a way to transform such a system (via a change of basis) to explicitly separate these components.

Let our system be with transfer function , and controllability matrix with rank . Let be a basis for the range space of , and add to form a full basis for . Then, define our similarity transformation as

The transformed system has the structure

Writing out the matrix multiplication, we have

where are the controllable states and are the uncontrollable states.

Intuitively, the first equation describes how controllable states can be affected by the input , and thus forms the controllable system (also called reduced system)

which also realizes . The second equation describes the uncontrollable system , which ignore the input. Note that the uncontrollable states can affect the controllable ones via , but this is fine since the overall can be affected by ; conversely, the controllable states cannot affect the uncontrollable ones.

Zero Blocks#

The key to the decomposition is the zero blocks, which ensure that (1) controllable states do not affect uncontrollable states () and (2) inputs do not affect uncontrollable states (). We'll show below how these zero blocks arise.

First, observe that , and from the Cayley-Hamilton theorem we know that can be expressed in terms of . Then, the columns of are linear combinations of those in , and so for each basis vector in range space of , can be written in terms of ,

Collect these into matrix to get

Next, the remaining basis vectors span the entire and can be expressed in terms of all , so

Putting this together, we have

and thus .

Second, observe that the columns of are among the columns of (by construction of ), so they can be written in terms of ,

and hence .

Controllable Component#

We can formally show that the reduced system is controllable by finding the rank of its controllability matrix. First, we know that has rank since similarity transform does not affect rank. Then,

and thus is also rank . By Cayley-Hamilton, the range space of this matrix is same as that of , the controllability matrix for .

Transfer Function#

Finally, we'll show that Kalman decomposition also preserves the transfer function . First, we know that a change of basis via does not change the transfer function,

Then, plugging in the matrices and simplifying the zero blocks, we arrive at

thus showing that this transfer function is the same as that of .