This book contains an algorithm which claims that a matrix $sI - A$, where $A$ is some $n \times n $ square matrix and $s$ a variable can be expanded into

$$adj(sI - A) = K_0 s^{n-1} + K_1 s^{n-2} +...+K_{n-1}$$

where the $K_i$s are matrices.

Of course when I first read it I thought it was completely crazy. For example, given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, the adjoint is $adj(A) = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$ obviously not a polynomial...

But then I realized the $K_is$ are matrices, so if one can pull out the $s$ somehow, then the adjoint matrix $adj(sI-A)$ might just turn into a polynomial with matrix coefficients.

Does anyone know how to achieve this? Is there a well known method?


Well, there is explanation in the book at the same page:

Note that since each element of $\Gamma(s)$ (the adjoint matrix of $sI-A$) is obtained from the determinant of $(n-1)\times(n-1)$ matrix, each element $\Gamma_{ij}(s)$ is a polynomial in $s$ of degree no greater than $s^{n-1}$

And since we can group terms with the same degree and put the common multiplier before the grouped terms, the decomposition of ${\rm adj}\: (sI-A)$ as $K_{n-1} s^{n-1} + K_{n-2} s^{n-2} + \dots + K_1 s + K_0, \; K_i \in \mathbb{R}^{n \times n}$ is quite straightforward.



We can use the Cayley Hamilton Theorem to do this.

$M^{-1}=\frac{adj(M)}{det(M)}$. So,$adj(M)=M^{-1}det(M)$.

If $M=sI-A$, then $adj(sI-A)=(sI-A)^{-1}det(sI-A)$.

But $det(sI-A)$ is the characteristic polynomial of $A$.

  • $\begingroup$ Isn't $(sI-A)^{-1}$ a matrix of polynomials too? $\endgroup$ – Evgeny Aug 26 '15 at 5:35
  • $\begingroup$ Yes. But it is finally a matrix. det(sI-A) is a polynomial with variable 's' and matrix coefficients. so, we have a matrix (whose entries are polynomials in 's') multiplied by a polynomial in 's'. $\endgroup$ – Srinivas K Aug 26 '15 at 6:47
  • $\begingroup$ The only thing that confuses me is that ${\rm det}\: (sI-A)$ is a polynomial of degree $n$ and we multiply it by another polynomial, so the final degree will be greater than $n$. It seems that I miss something in my reasoning... $\endgroup$ – Evgeny Aug 26 '15 at 6:56
  • $\begingroup$ even i'm doubtful about that. need to think further. thanks. $\endgroup$ – Srinivas K Aug 26 '15 at 7:14

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