I am new to linear algebra, and was asked the following question:
$A$ is a matrix of order n x n, $n \geq 2$, with characteristic polynomial $p(λ)=λ^n-1$. Is $A$ diagonalizable over R? over C? If $A$ is diagonalizable then draw a diagonal matrix similar to A.
If $A$ is diagonalizable, then $p(λ)=λ^n-1=0$, in other words $λ^n=1$
Over C: $λ^n=1$ will have n solutions, so there will be n eigenvalues, with algebraic multiplicity 1.
$1 \leq$ geometric multiplicity $\leq $ algebraic multiplicity
so the geometric multiplicity is 1. Geometric multiplicity = algebraic multiplicity, so $A$ is diagonalizable.
Over R: $λ^n=1$ has a different number of solutions depending whether n is odd or even.
For example $λ^2=1$ gives λ=1, λ=-1; while $λ^3=1$ gives λ=1. Therefore we cannot check the geometric multiplicity, and we cannot know if A is diagonalizable.
I am lost as to the last part of the question, i.e. how to draw a diagonal matrix similar to A, over C. Any suggestions would be great!
Thank you!
A similar matrix to $A$ over $\mathbb C$ is
$$\bar{A} = \mbox{diag}(1, e^{2i \pi /n}, \dots , e^{2i (n-1) \pi/n})$$ as the $n$ roots of $p_n(\lambda) = \lambda^n-1$ are $1, e^{2i \pi /n}, \dots , e^{2i (n-1) \pi/n}$.