Complex square matrix with distinct eigenvalues

Question

Is there a simple way to show that if $A$ is a complex square matrix with distinct eigenvalues ​​then $A$ is similar to a matrix whose all entries are nonzero.

Answer 1

I'm not sure if this answer is simple enough. Anyway, we can prove a slightly stronger result:

Lemma. Let $n\ge2$. If $A\in M_n(\mathbb{C})$ is not a scalar multiple of the identity matrix, then $A$ is similar to some matrix with all entries nonzero.

Proof. For any matrix that is not a scalar multiple of the identity matrix, its rational canonical form is not diagonal. So, we may assume that $A$ is not a diagonal matrix. Now, for any $x\ne0$, all nonzero off-diagonal entries of $$ B=\operatorname{diag}\left(x^2,x^{2^2},x^{2^3},\ldots,x^{2^n}\right)\,A\,\operatorname{diag}\left(x^{-2},x^{-2^2},x^{-2^3},\ldots,x^{-2^n}\right) $$ are multiples of distinct powers of $x$. Let $J$ be the all-one matrix. Then \begin{align*} C&=(I+tJ)B(I+tJ)^{-1}\\ &=(I+tJ)B\left(I-\frac{t}{nt+1}J\right)\\ &=\frac{1}{nt+1}(I+tJ)\,B\,\left[I+t(nI-J)\right]\\ &=\frac{1}{nt+1}\left[t^2(nJB-JBJ) + t(JB+nB-BJ) + B\right]. \end{align*} The expression inside the pair of square brackets is quadratic in $t$. Its leading coefficient is $P=nJB-JBJ$, whose $(i,j)$-th entry is $$ p_{ij}=n\times(j\text{-th column sum of } B)-(\text{sum of all entries in } B). $$ Since $B$ has at least one nonzero off-diagonal entry and those nonzero off-diagonal entries of $B$ are multiples of distinct powers of $x$, each $p_{ij}$ is a nonzero polynomial in $x$. Consequently, for appropriate values of $x$ and $t$, all entries of $C$ are nonzero.