Why is $\det(I + A^{50}) = 4$ here?

Question

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?

Answer 1

Let $D$ be the diagonal matrix in question.

Then, $I+A^{50}=I+TD^{50}T^{-1}=T(I+D^{50})T^{-1}$ and hence,

$$ \det(I+A^{50})=\det(T)\det(I+D^{50})\det(T^{-1})=\det(I+D^{50}) $$ Can you take it from here?

Answer 2

According Cayley-Hamilton, we have $A(A-I)(A+I) =0_{3\times 3}$.

Hence, $A^3 = A \Rightarrow A^{50} = A^2$

$\Rightarrow \det (I+A^{50}) = \det diag(2,2,1) = 4$

Answer 3

Remember that $I = TT^{-1}$ and that the determinant has the property $\det (ABC) = \det(A)\det(B)\det(C)$. So the determinant simplifies to the determinant of the sum of two diagonal matrices.

Answer 4

As you said, you can diagonalize $A=PDP^{-1}$ with $D=\operatorname{diag}(1,-1,0)$. Then $A^{50}=PD^{50}P^{-1}$. $D^{50}=\operatorname{diag}(1,1,0)$. So, $$\det(I+A^{50})=\det(P^{-1}(I+A^{50})P)=\det(I+D^{50})=\det(\operatorname{diag}(2,2,1))=4.$$ The first equality follows from the fact that determinants are invariant under change of basis. The second follows from distributing.

Answer 5

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 $\det(I+A^{50})=2\cdot 2\cdot 1=4$.

Answer 6

Eigenvalues of $A^{50}$ are the the eigenvalues of $A$ to the power of $50$ that is $\{0,1,1\}$

Thus eigenvalues of $I+A^{50}$ are $1+\{0,1,1\}=\{1,2,2\}$

Thus $$\det (I+A^{50})=4$$ which is product of the eigenvalues.