Product between a power series and polynomial of finite degree

Question

Consider the matrix $A_{n \times n}$ with its characteristic polynomial, being $a(z)=det(zI-A)$ of degree $n$:

$ a(z)= a_nz^{-n} + ... + a_1z^{-1} +a_0 $

Consider now the product

$P(z)=a(z)(zI-A)^{-1}$.

My goal is to show that $P(z)$ is a polynomial in $z$ of degree $n$ $P(z)= P_nz^{-n}+ \dots P_1z^{-1}$ with coefficients

$[P_n \ \dots P_1]= [A^{n-1} \ A^{n-2} \ ... I]\begin{bmatrix} a_0 & 0 & 0 & ... & 0\\ a_1 & a_0 & 0 & ... & 0\\ \vdots & & \ddots & & \vdots\\ \vdots & & & \ddots & \vdots \\ a_{n-1} & \dots & & & a_0\\ \end{bmatrix}$.

It is not clear to me how the coefficients $A_i$ with $i\geq n$ are canceled out.

Here few steps to prove it: first I noticed that $(zI-A)^{-1}$ is the power series expansion

$\frac{1}{zI-A} = z^{-1}\frac{1}{I-Az^{-1}} = z^{-1}\sum_{n=0}^{\infty}\Big(Az^{-1}\Big)^n = z^{-1}[I+Az^{-1}+A^2z^{-2}+...]$

Then, the product is

$P(z)= a(z)z^{-1}[I+Az^{-1}+A^2z^{-2}+...]= [a_nz^{-n} + ... +a_0][z^{-1}+Az^{-2}+A^2z^{-3}+...]$

that can be rewritten as

$[P_n \ \dots \ P_1]\begin{bmatrix} z^{-n}\\ \vdots \\ z^{-1}\end{bmatrix}= [a_n \ \dots \ a_0]\begin{bmatrix} z^{-n}\\ \vdots \\z^{0}\end{bmatrix}[\dots \ A^{n-1} \ \dots \ I]\begin{bmatrix} \vdots \\ z^{-n} \\ \vdots \\ z^{-1}\end{bmatrix}$

And now I am stuck in this step and i don't know how to proceed to get the desired representation. Should I simply discard the coefficients with index $i\geq n$ just because the $P(z)$ is of degree $n$?

Thanks for any help and suggestions

Answer 1

For an adjoint(or adjugate) matrix $\text{adj}(zI-A)$, we have that $$ (zI-A)\text{adj}(zI-A)=\det(zI-A)I=a(z)I, $$ thus $\text{adj}(zI-A)=a(z)(zI-A)^{-1}$ for $z$ not in the spectrum of $A$, i.e. $a(z)\ne 0$. On the other hand, we can see from the definition of adjoint matrix that every entry of $\text{adj}(zI-A)$ is a polynomial in $z$ of degree at most $n-1$. Hence there is a sequence of matrices such that $$ a(z)(zI-A)^{-1}=A_0+A_1z+\cdots +z^{n-1}A_{n-1}. $$ Now, we can match each coefficient of $z^j$ in $$\begin{align*} a(z)I&=a_0I+za_1I+\cdots +z^n a_nI \\&=(zI-A)(A_0+A_1z+\cdots + z^{n-1}A_{n-1}) \\&=-AA_0 +z(A_0-AA_1)+z^2(A_1-AA_2)+\cdots+z^nA_{n-1}. \end{align*}$$ to obtain $$ A_0=-a_0A^{-1}\\ A_1=-a_0A^{-2}-a_1A^{-1}\\ \vdots \\ A_{n-2}=-a_0A^{-n+1}-\cdots -a_{n-2}A^{-1}\\A_{n-1}=-a_0A^{-n}-\cdots -a_{n-2}A^{-2}-a_{n-1}A^{-1}=a_nI. $$

Answer 2

all the row and column transformation we used to diagonal a matrix is elementary and invertible! so the condition of scalar matrix is its character vector be unique, then multiple your analytic function with $A,A^2,A^3,……$ it is sufficient to get the unique condition, such that every coefficient equals zero. therefore you can represent your diagonal procedure as a multiple between power matrix with one column and a power of scalar.

is it helpful? thank you!