Complex matrix similar to a matrix with identical diagonal entries Let $A$ be a complex matrix. Show that it is similar to a matrix with identical diagonal entries.
I do have some sense, but could not prove it.
 A: It is equivalent to show that if $A$ is a matrix with trace $0$, then there is a similar matrix whose diagonal entries are all zero.
We prove that this is the case inductively, using Lemma 2 from this paper, which states that if $A$ is not a non-zero scalar multiple of the identity, then there is a similar matrix with a $0$ as its first diagonal element (the paper contains a proof).
Base case: $A$ is $1 \times 1$
Trivially, $A = 0$ if its trace is zero.
Inductive step: if $A$ is $n \times n$, then it is similar to a matrix of the form
$$
\pmatrix{0&r^T\\c&A'}
$$
We note that $A'$ has a trace of zero.  By our inductive hypothesis, there is some $B = SA'S^{-1}$ with zeros on the diagonal.  Define 
$$
S_0 = \pmatrix{1&0\\0&S}
$$
we have
$$
S_0\pmatrix{0&r^T\\c&A'} S_0^{-1} = 
\pmatrix{0&r^T S^{-1}\\Sc & SA'S^{-1}} = 
\pmatrix{0&r^T S^{-1}\\Sc & B}
$$
By the transitivity of similarity, we conclude that $A$ is similar to a matrix with $0$s along the diagonal, as desired.

Here's how we extend this to the general case: Given a matrix $A$, let $a = \text{trace}(A)/n$.  Let $B = A - a I$.  Note that $B$ has a trace of zero, so we may state that $B$ is similar to a matrix $B' = SBS^{-1}$ with $0$s on the diagonal. 
Define $A' = SAS^{-1}$.  We have
$$
A' = SAS^{-1} = S(B + aI)S^{-1} = SBS^{-1} + aI = B' + aI
$$
Thus, all the diagonal entries of $A'$ are $a$.  So, $A$ is similar to a matrix with identical diagonal entries.
