# Existence of diagonalizable matrix

Let $$A \in \mathbb{C}^{4 \times4}$$ and $$A^5 = I$$.

1. Is $$A$$ always diagonalizable?

2. How to show $$|\mbox{tr} (A)| \leq 4$$?

3. If $$\mbox{tr} (A) = 4$$, can we determine matrix $$A$$?

Some ideas:

1. A matrix is diagonalizable if and only for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue.

2. A matrix or linear map is diagonalizable over the field $$F$$ if and only if its minimal polynomial is a product of distinct linear factors over $$F$$.

But how can we obtain the eigenvalue from the first method? And how can we calculate minimal polynomial of $$A$$? By the way, given $$A \in \mathbb{C}^{4\times4}$$, can we say we have a linear operator $$T \in \mathcal{L}(\mathbb{C}^{4\times4})$$?

And how can we calculate minimal polynomial of $$A$$

We can't - but we can say enough about it. We have a polynomial that annihilates $$A$$, so that polynomial $$x^5-1$$ must be some multiple of the minimal polynomial. Factoring $$x^5-1=(x-1)(x-\omega)(x-\omega^2)(x-\omega^3)(x-\omega^4)$$ where $$\omega$$ is a primitive $$5$$th root of unity, the minimal polynomial must be a product of some subset of those factors. That can be any subset of one to four factors; zero is impossible because $$I\neq 0$$, and five is impossible because it's a $$4\times 4$$ matrix and its minimal polynomial can't have degree greater than $$4$$. In any case, that minimal polynomial is definitely a product of distinct linear factors.

Whichever factors we choose, we'll get the corresponding number as an eigenvalue. If we chose fewer than four factors, some of those eigenvalues will be doubled up with higher multiplicity. So then, how can we show that the geometric multiplicity is the same as the algebraic multiplicity for each of them? Try putting it into Jordan form, and see what that polynomial $$A^5-1$$ does to a Jordan block.

By the way, given $$A\in \mathbb{C}^{4\times 4}$$, can we say we have a linear operator $$T\in\mathcal{L}(\mathbb{C}^{4\times 4})$$

We really shouldn't. We need both a domain and a codomain for our linear operator - that $$\mathcal{L}$$ should have two arguments. Make both of those arguments $$\mathbb{C}^4$$, and we're talking about the operator given by multiplying our matrix by a vector (represented with respect to the standard basis). I'll guess this is the operation you meant - in which case the notation you used is just wrong.

We could also make both of the arguments $$\mathbb{C}^{4\times 4}$$, for multiplication of a matrix by $$A$$. That is a linear operator, but it's not something we're likely to use.

Yes it is diagonalisable. Since $$A^5-1 = 0$$ and the polynomial $$X^5-1$$ has simple roots over $$\mathbb{C}$$.

Thus the eigenvalues of $$A$$ are included in the root of $$X^5-1$$. The root of this polynomial are $$5-$$ root of unity thus they have module $$1$$. Hence $$\mid tr(A) \mid \leq 4$$.

Moreover if $$\mid tr(A) \mid =4$$ it means that a subset of length $$4$$ of $$\{ e^{2\pi i k /5} \mid k \in [0,4] \}$$, has a sum with module equal to $$4$$, I am sure you can continue from here.

And yes given a matrix we have a linear operator but not from $$L(\mathbb{ C}^{4 \times 4})$$ but from $$L(\mathbb{C}^4)$$. It’s the linear operator which sends the canonical basis on the column of $$A$$.