# Diagonalizability of matrix over C and R respectively

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!

• If $A$ is diagonalizable over $\Bbb R$ then all the complex eigenvalues must be real... – David C. Ullrich Jan 19 at 17:19
• You should also search the site before posting for similar questions. This can be helpful, e.g., see here. – Dietrich Burde Jan 19 at 17:22

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 $$1, e^{2i \pi /n}, \dots , e^{2i (n-1) \pi/n}$$.