Vtyjkyuk

I’m looking for a formula which given a column-major linear index to an element in a symmetric $n times n$ matrix returns an index pointing to the corresponding location on the opposite side of the diagonal of the same matrix.
Say for example I have a matrix:
$$
A =
beginbmatrix
0 & 1 &2 \
1 & 0 & 2 \
2 & 2 & 0
endbmatrix
$$
And I enter $0$ into the formula, I would need it to return $0$ (since index $0$ is on the diagonal).

If I enter $1$ to the formula, I would need it to return $3$, since index $3$ points to the same location but mirrored to the opposite side of the diagonal.

If I enter $2$, it should return $6$. And so on...

This exact question has been asked before, but with row-major indexing instead of column-major. However, as far as I can tell, the accepted and upvoted answer on that tread is completely wrong so it offers little help.

Does anyone know how to solve this?

I take two steps to solve the problem:

1) Find the matrix index in the form of $(i, j)$ corresponding to the number entered

2) swap $(i, j)$ to $(j, i)$, which is for "mirroring". Find the index of the vector form of the column based matrix.

The solution works for $1$ based, but can be converted easily to $0$ based since their difference for the same location is $1$.

notation

$leftlfloor fracln rightrfloor = l // n$

Let $(i_l, j_l)$ be a mapping:
beginequation
labelvecmap
beginsplit
& (i_l, j_l): 1, ldots, n^2 to 1, ldots, ntimes 1, ldots, n\
& l to (i_l, j_l)\
& i_l = l - n(j_l-1) \
& j_l = leftlfloor fracln rightrfloor \
endsplit
endequation

For exmaple:
$$
beginsplit
&l = 1, (i_1, j_1) = (1, 1)\
&l = n, (i_l, j_l) = (n, 1)\
&l = n+1, (i_l, j_l) = (1, 2)\
&l = n^2, (i_l, j_l) = (n, n)\
endsplit
$$

We want the "$l_*$" of $(j_l, i_l)$, which is $l_* = n(i_l - 1) + j_l$.

Explanation
Enter $i = 2$, the corresponding $l=3$ because I assume the starting index of series is 1 rather than 0. In your example, n = 3. Hence $i_l = 3, j_l = 1$, this gives $l_* = 3(3- 1) + 1 = 7$. Again because starting index is 1, converting by minus 1 you get 6.