Let $A\in M_{n\times n}(\Bbb R)$ so that $I\notin span(A,A^2,...,A^n)$. Prove that $\det(A)=0$.

Question

Let $A\in M_{n\times n}(\Bbb R)$ so that $I\notin span(A,A^2,...,A^n)$. Prove that $\det(A)=0$.

I was thinking of showing $A$ is not invertible, meaning it has an eigenvalue of $0$. Since no matter the power you give to $A$, it is still not the identity, you can deduce that at least one eigenvalue is indeed $0$. However this doesn't work if you have different eigenvalues in the Matrix, so I got stuck.

Help will be most appreciated.

Answer 1

If $\det A \ne 0$, then the constant term of $p(t) = t^n + c_{n-1} t^{n-1} + \cdots + c_1 t + c_0 = \det(tI-A)$, the characteristic polynomial, is nonzero.

By Cayley-Hamilton, $p(A) = A^n + c_{n-1} A^{n-1} + \cdots + c_1 A + c_0 I_n = 0$.

Therefore, $I_n = -\frac{c_1}{c_0} A - \cdots - \frac{c_{n-1}}{c_0} A^{n-1} - \frac1{c_0} A^n \in \operatorname{span}(A, \cdots, A^n)$.

Answer 2

By the Cayley-Hamilton theorem, you can express $A^n$ as $c_{n-1}A^{n-1}+\cdots+c_0\operatorname{Id}$. But then $c_0=0$, since otherwise $\operatorname{Id}$ would be a linear combination of $A,A^2,\ldots,A^{n-1}$. So,$$A.(A^{n-1}-c_{n-1}A^{n-2}-\cdots-c_1\operatorname{Id})=0.$$If $\det(A)\neq0$, $A$ would be invertible and it would follow from the previous equality that $A^{n-1}=c_{n-1}A^{n-1}+\cdots+c_1\operatorname{Id}$. But then, by the same reason as before, $c_1=0$ and so on. So, each $c_k$ is $0$, and so $A^n=0$. So, $\det A=0$