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!

  • 1
    $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.