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I came a cross a question that gives a matrix $A$, and asks to write $A^{-1}$ as a polynomial in $A$ with real coefficients.

I don't know what this means, and googling didn't clarify very much. I found the inverse $A$, but I do not know what it means to write it as a polynomial in $A$.

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    What's the characteristic equation of $A$? – Angina Seng Dec 17 '19 at 05:16
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    For example, you might write $A^{-1} = 8 I + (3/2) A + \pi A^2$. The right side is a polynomial in $A$, i.e. a sum of finitely many coefficients times nonnegative powers of $A$ (with $A^0 = I$ the identity matrix), where the coefficients are real numbers. – Robert Israel Dec 17 '19 at 05:27
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    Given a polynomial $$f(x) = a_0 + a_1x + \cdots + a_nx^n$$ and a square matrix $A$, $f(A)$ is the matrix given by $$f(A) = a_0I + a_1A + \cdots + a_nA^n.$$ – azif00 Dec 17 '19 at 05:37
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    Let $\chi_A(\lambda) = \det(\lambda I_n -A)$ be the characteristic polynomial of $A$, Cayley Hamilton theorem says that $\chi_A(A) = 0$. When $\det A \ne 0$, you can rearrange $\chi_A(A) = 0$ to the form $A g(A) = I_n$ for some polynomial $g(\lambda)$, that means... – achille hui Dec 17 '19 at 05:46

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This can be done for any invertible $A$ using the Cayley–Hamilton theorem that any matrix satisfies its own characteristic equation. If $a_i$ are the coefficients of the characteristic polynomial then $a_0I + a_1A + \cdots + a_nA^n=0$ with $a_0=\det A$. Hence $$A^{-1}=-\frac1{a_0}(a_1I + \cdots + a_nA^{n-1}).$$

You can find the characteristic polynomial by computing the determinant $\det(A-\lambda I)=a_0 + a_1\lambda + \cdots + a_n\lambda^n$.

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