What is meant here, when it is said that "diagonal generalized Gell-Mann matrices are not unique"? 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 $0$?
 A: 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 $a$.
