- Let $A= \pmatrix{10&25\\ 0&-10}$. Find $A^{100}$.
- Let $B \in M_2(\mathbb{R})$. For which $ x > 0$ can we have $B^{100}= \pmatrix{-1&0\\ 0&-1-x}$?
For case one I find $A^2={10}^2 I$ and $A^4={10}^4 I$ ... then $A^{100}={10}^{100} I$.
For case two $\det(B^{100})=1+x$ and $B $ is invertible matrix.
Hints:
Diagonalize $A$: $$A = \pmatrix{1 & -5 \\ 0 & 4}\pmatrix{10 & 0 \\ 0 & -10}\pmatrix{1 & -5 \\ 0 & 4}^{-1}$$
so $$A^{100} = \pmatrix{1 & -5 \\ 0 & 4}\pmatrix{10^{100} & 0 \\ 0 & (-10)^{100}}\pmatrix{1 & -5 \\ 0 & 4}^{-1}$$
I'm reading the second question like this.
$B \in M_2(\mathbb{R})$ is given. For which $x > 0$ is $B^{100} = \pmatrix{-1&0\\ 0&-1-x}$ possible?
We know that $$\{-1, -1-x\} = \sigma(B^{100}) = \sigma(B)^{100}$$ so there exists $\lambda \in \sigma(B)$ such that $\lambda^{100} = -1$. It has to be $\lambda \in \mathbb{C} \setminus \mathbb{R}$, but then since $B$ is a real matrix, we know that also $\overline{\lambda} \in \sigma(B)$. Hence, $\sigma(B) = \{\lambda, \overline{\lambda}\}$. But then $\sigma(B)^{100} = \{-1\}$ so $x = 0$.
Hence, no such $x > 0$ exists.