Write $m_{I-A}$ (the minimal polynomial of $I-A$) in terms of $m_A$ and conclude: $\deg(m_A)=\deg(m_{I-A})$

Question

Given a matrix $A$ and its minimal polynomial $m_A$.

Write $m_{I-A}$ (the minimal polynomial of $I-A$) in terms of $m_A$ and conclude: $\deg(m_A)=\deg(m_{I-A})$

Any hint or direction would be appreciated.

Answer 1

If $\deg m_A = k$ then $$m_{I-A}(x) = (-1)^k m_A(1-x)$$

Indeed, we have that $(-1)^k m_A(1-x)$ annihilates $I - A$ so $m_{I-A}(x) \,| \,(-1)^k m_A(1-x)$.

Also note that $m_{I-A}(1-x)$ annihilates $A$ so $m_A(x) \,| \,m_{I-A}(1-x)$.

Therefore $\deg m_A = \deg_{I-A}$ so $(-1)^k m_A(1-x)$ is monic polynomial of the same degree as $m_{I-A}$ which annihilates $I-A$. By minimality, we conclude that $m_{I-A}(x) = (-1)^k m_A(1-x)$.