Using a piece from my last question I want to show how to find $A^{-1}$ as a polynomial expression in $A$ of degree < $\deg m_A$ where the leading coefficient of the polynomial is $\dfrac{-1}{m_A(0)}$ where $m_A$ is the minimal polynomial of $A$.
Method of finding inverse of a Matrix using minimal polynomials
0
$\begingroup$
linear-algebra
matrices
polynomials
inverse
-
0Yes. It is the lowest degree polynomial that does that. And it appears that you are calling the coefficient of $I$ in the minimal polynomial $m_A(0).$ Write it out, solve for $I,$ and see what happens. – 2012-12-07
1 Answers
6
If $m_A(x)=x^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0$ then since $p(A)=0\Rightarrow $ $A^m+a_{m-1}A^{m-1}+\cdots+a_1A+a_0I=0\Rightarrow a_0I=-(A^m+a_{m-1}A^{m-1}+\cdots+a_1A)\Rightarrow\\ a_0I=A(-A^{m-1}-a_{m-1}A^{m-2}-\cdots-a_1)\Rightarrow \\I=A\left(-\dfrac{1}{a_0}A^{m-1}-\dfrac{a_{m-1}}{a_0}A^{m-2}-\cdots-\dfrac{a_1}{a_0}\right)\Rightarrow\\ A^{-1}=-\dfrac{1}{a_0}A^{m-1}-\dfrac{a_{m-1}}{a_0}A^{m-2}-\cdots-\dfrac{a_1}{a_0}.$