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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!

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    $\begingroup$ If $A$ is diagonalizable over $\Bbb R$ then all the complex eigenvalues must be real... $\endgroup$ – David C. Ullrich Jan 19 at 17:19
  • $\begingroup$ You should also search the site before posting for similar questions. This can be helpful, e.g., see here. $\endgroup$ – Dietrich Burde Jan 19 at 17:22
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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 $1, e^{2i \pi /n}, \dots , e^{2i (n-1) \pi/n}$.

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  • $\begingroup$ Thank you! and the rest of the answer is correct? $\endgroup$ – dalta Jan 19 at 17:26
  • $\begingroup$ The rest seems indeed correct. $\endgroup$ – mathcounterexamples.net Jan 19 at 17:29

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