Let $A \in \mathbb{C}^{4 \times4}$ and $A^5 = I$.
Is $A$ always diagonalizable?
How to show $|\mbox{tr} (A)| \leq 4$?
If $\mbox{tr} (A) = 4$, can we determine matrix $A$?
Some ideas:
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A matrix is diagonalizable if and only for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue.
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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 $T \in \mathcal{L}(\mathbb{C}^{4\times4})$?
Best Answer
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 $A$.