According to the answer given here, the diagonal generalized Gell-Mann matrices are not unique. But what exactly is meant by this?
Are they just saying that we can multiply the diagonal matrices by a constant and still have a valid generator of the group since the matrices will still be orthogonal? If so, then why are the other generalized Gell-Mann matrices not also being called non-unique (you can multiply them by a constant and still have an orthogonal set of matrices).
Or are they saying that we could have completely different diagonal matrices as long as they are traceless and satisfy orthogonality under the Hilbert-Schmidt norm: $(A,B) =Tr(AB)$ ? For example we could have $\textrm{diag}(1 ,0 , -1)$ and $\textrm{diag}( 1/\sqrt{3} , -2/\sqrt{3} , 1/\sqrt{3})$, or even something more radically different such as $\textrm{diag}(a,b,c)$ and $\textrm{diag}(d,e,f)$ where none of $a,b,c$ are $0$?
If deriving the n×n matrices in the same way Gell-Mann did, there is only one choice for the diagonal matrices.
Gell-Mann generated the two diagonal 3×3 matrices by starting with the Pauli $z$ matrix and padding it with zeros to make it 3×3, and letting the other diagonal matrix be arbitrary:
$$\tag{1} \lambda_3 = \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 0 \\ \end{pmatrix}, \lambda_8=\begin{pmatrix} a & 0 & 0\\ 0 & b & 0\\ 0 & 0 & c \\ \end{pmatrix}. $$
Then he used the following constraints:
\begin{eqnarray} \textrm{Tr}(\lambda_3\lambda_8) = 0 \tag{2}\\ \textrm{Tr}(\lambda_8) = 0 \tag{3}\\ \textrm{Tr}(\lambda_8^2) = 2 .\tag{4} \end{eqnarray}
The first constraint forces us to choose $b = a$,
the second constraint forces us to choose $c = -2a$,
and the third constraint forces us to choose $a = \frac{1}{\sqrt{3}}$.
Let's now derive the three diagonal 4×4 matrices in the same, way, by starting with the 3×3 matrices padded with zeros, and adding an arbitrary diagonal matrix:
$$\tag{5} \Lambda_3 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}, \Lambda_8=\frac{1}{\sqrt{3}}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}, \Lambda_{15} = \begin{pmatrix} a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & 0 & 0 & d \\ \end{pmatrix}. $$
Then we have the following constraints:
\begin{eqnarray} \textrm{Tr}(\Lambda_3\Lambda_{15}) = 0 \tag{6}\\ \textrm{Tr}(\Lambda_8\Lambda_{15}) = 0 \tag{7}\\ \textrm{Tr}(\Lambda_{15}) = 0 \tag{8}\\ \textrm{Tr}(\Lambda_{15}^2) = 2 .\tag{9} \end{eqnarray}
The first constraint forces us to choose $b = a$
The second constraint forces us to choose $c = a$
The third constraint forces us to choose $d = -3a$
The fourth constraint forces us to choose $a = \frac{1}{\sqrt{6}}$.
This matches when you see on Pg 368 of this book (PDF), Eq. 11 of this paper (PDF), Pg 7 of this talk (PDF), Eq 3 of this paper (PDF), and the definition of "generalized Gell-Mann matrices" given here.
In general we carry over padded versions of the $n-2$ diagonal matrices from the $(n-1)$×$(n-1)$ case, and the orthogonality conditions force us to make all entries of the last diagonal matrix to be equal except for the last entry. The last entry $x$ has to be the negative of the sum of the rest of the entries (which are all the same, call them $a$ so $x = -(n-1)a$), due to the condition that the matrices are all traceless. The condition that the trace of each matrix multiplied by itself is 2, gives us only one choice for $a$.