If $1,-1,0$ are eigen values of $A$ then $\det(I+A^{100})=$?

Question

As the question states, if $1,-1,0$ are eigen values of a matrix $A$ then I need to find what $\det(I+A^{100})$ is.

Now I know that $\det A=0$, $\det (I+A)=0$ and $\det(I-A)=0$. But I don't know what to do further.

I tried expanding $(I+A)^{100}$ but that did not simplify things in any way.

Thank you.

Answer 1

try to think on these lines since the eigenvalues of $A^{100}$ are $1,1,0$ and the eigenvalues of I are $1,1,1$ hence the characteristic equation of $A^{100}+I$ becomes $(x-2)(x-2)(x-1)$ assuming it to be matrix of $3*3$ order

Answer 2

It is non-zero. The exact value depends on the multiplicities of the eigenvalues of $A$.

Answer 3

It will have the same eigenvectors, and for those eigenvectors $A^{100}x = \lambda^{100}x$.