Permutation of matrix elements Let each row and each column of a $n\times n$ matrix $A$ be a permutation of $\{1,2,\ldots,n\}$ and
let $A$ be symmetric.
(a) If $n$ is odd, prove that each of $1,2,\ldots,n$ occurs on the principle diagonal of $A$.
(b) For every even number $n$, show that there exists an $A$ in which not all of $1,2,\ldots,n$
appear on the diagonal.
My knowledge of matrices is pretty basic. I have noticed similar types of questions before. What topics do I need to learn to solve these types of problems?
 A: Part (a)
If $n$ is odd, then an odd number of each element must occur.
If an element occurs in a symmetric table at an odd number of locations, then that element must occur along a diagonal an odd number of times.
If one element occurs $3$ or more times along a diagonal then another must occur zero times.
Zero is not odd.
Part (b)
Pair up the numbers $1 .. n$ any partition you like, such as $c_1 = \{1,2\}, c_2 = \{3, 4\}, \dots,  c_{n/2} = \{n-1, n\}$.
Create ${n\over 2}$ 2x2 tables $B_k = \begin{bmatrix} c_{k,1} & c_{k,2} \\ c_{k,2} & c_{k,1}\end{bmatrix}$.
Create a table $M_{i,j} = B_{(i + j - 1) \text{mod} (n/2)}$:
$$M = \begin{bmatrix}
B_1 & B_2 & B_3 & ... & B_{n/2}\\
B_2 & B_3 & B_4 & ... & B_{1}\\
B_3 & B_4 & B_5 & ... & B_{2}\\
    &     & \vdots & & \\
B_{n/2} & B_1 & B_2 & ... & B_{n/2 - 1}\\
\end{bmatrix}$$
Continuing the example for 6x6, that would be
$$\left[\begin{array} {cc|cc|cc}
1 &2 &3 &4 &5 &6 \\
2 &1 &4 &3 &6 &5 \\ \hline
3 &4 &5 &6 &1 &2 \\
4 &3 &6 &5 &1 &2 \\ \hline
5 &6 &1 &2 &3 &4 \\
6 &5 &2 &1 &4 &3 \\
\end{array}\right]$$
Uniqueness of the elements of each row and column follows from construction.

Personal pet peeve, I'm against calling all 2D tables matrices.  A table only becomes a matrix when linear algebra is involved; the same way an integer only becomes a numerator when a fraction is involved.
