Why is $\det(I + A^{50}) = 4$ here? I’m trying to understand the reasoning behind the following:

Let $A \in \mathbb{R}^{3\times3}$ with eigenvalues $1$, $-1$, $0$. What is $\det \left(I + A^{50} \right)$?

The given answer is 4.
A is obviously singular due to eigenvalue 0. Moreover, spectral decomposition of $A$ exists:
$$
A = T diag(1, -1, 0) T^{-1}\\
A^{50} = T diag(1, -1, 0)^{50} T^{-1}
$$
But I can’t see why $\det(I + T diag(1, -1, 0)^{50}T^{-1})$ would be 4 either, due to the limited knowledge about $T$. Can somebody shed light on this for me?

If $v_1$, $v_{-1}$, $v_0$ are respective eigenvectirs to eigenvalues $1,-1,0$, then they are necessarily linearly independent and, by dimension, form a basis. We have $(I+A^{50})v_\lambda=v_\lambda+\lambda^{50}v_\lambda$, so $v_1\mapsto 2v_1$, $v_{-1}\mapsto 2v_{-1}$, $v_0\mapsto v_0$. From this, clearly $\det(I+A^{50})=2\cdot 2\cdot 1=4$.
