i'm stuck and unsure regarding this question. i am requested to check if any of those three claims (same question) are true:
a) $A$ is an invertible matrix that follow $A^{-1}=-A$, then the minimal polynom of $A$ (in $\mathbb R$) is $t^2+1$
regarding the two following claims, the following condition occurs:V is a linear space of dimension $5$, $T: V \to V$ is an invertible linear transformation
b) the free element of the minimal polynom of $T$ is different than $0$.
c) $T^{-1}$ can be presented by a polynom in $T$ whose degree is less or equal to $4$
what i think:
a) if $A$ is invertible then it follows $AA^{-1}=I$, and also $A^2+1=0$, but $A(-A)=I$, so it's a false claim, because the eigen values cannot be $0$ in an invertible matrix.
b) an invertible transformation is one iff its eigenvalues(roots) are not zero. so this claim is true, because otherwise it would be a contradiction to the detail and cannot be invertible.
c) i don't know and if someone can, i would appreciate an explanation regarding this one.
thank you very much for your help
a) It is true. Since $A^2=\operatorname{Id}$, the minimal polynomial of $A$ in $\mathbb R$ must divide $t^2+1$ and the only non-constant monic polynomial with real coefficients that does that is $t^2+1$.
b) It is true. The constant term is the product of the eigenvalues and, since the transformation is invertible, none of them can be $0$.
c) It is also true. Since the space has dimension $5$, it follows from the Hamilton-Cayley theorem that the are numbers $a_0,a_1,\ldots,a_4\in\mathbb R$ such that $T^5+a_4T^4+a_3T^3+a_2T^2+a_1T+a_0\operatorname{Id}=0$. And you know that $a_0\neq0$ (from the previous item). Therefore$$\frac1{a_0}T^4+\frac{a_3}{a_0}T^2+\frac{a_2}{a_0}T+\frac{a_1}{a_0}\operatorname{Id}+T^{-1}=0.$$