Miscellaneous · Q1: $AB-BA$ is skew symmetric

Basics you need

Solution

1. Check that every matrix in the problem exists

   A symmetric matrix must be square (Definition 4 is stated for square matrices). Let A and B both be $n\times n$⁠. Then:

   • $AB$⁠: $(n\times n)(n\times n)$⁠, columns of A $=n=$ rows of B, so it is defined and is $n\times n$⁠.
   • $BA$⁠: defined in the same way, also $n\times n$⁠.
   • $AB - BA$⁠: a difference of two $n\times n$ matrices, so it is defined and is $n\times n$⁠.

   So $C = AB - BA$ is a square matrix, which is what a skew symmetric matrix has to be.

   The question does not say "of the same order", but it has to be meant. If A were $2\times2$ and B were $3\times3$⁠, the product $AB$ would not exist at all and there would be nothing to prove. Checking orders first also confirms that C is square, which Definition 5 requires before we may call it skew symmetric.
   Product rule for orders: $(m\times n)(n\times p)$ gives $m\times p$⁠. Sums and differences need equal orders. Basics: pair up, multiply, add

2. Transpose the difference term by term

   Let $C = AB - BA$⁠. The transpose of a difference is the difference of the transposes:

   $$C' = (AB - BA)' = (AB)' - (BA)'.$$
   This comes from two listed rules: $(X+Y)' = X'+Y'$ and $(kX)' = kX'$⁠. Writing $AB - BA = AB + (-1)(BA)$ and applying both gives $(AB)' + (-1)(BA)' = (AB)' - (BA)'$⁠. Transposing just moves every entry to its mirror position, and subtracting entry by entry commutes with that move.
   Properties (iii) and (ii) of Section 3.5.1. Lesson: transpose

3. Reverse each product

   Use $(XY)' = Y'X'$ on each term: with $X = A, Y = B$ in the first and $X = B, Y = A$ in the second.

   $$C' = B'A' - A'B'.$$
   The order reverses because the $(i,j)$ entry of $XY$ is row $i$ of X times column $j$ of Y. After transposing, that number sits at position $(j,i)$⁠, and it equals row $j$ of $Y'$ (which is column $j$ of Y) times column $i$ of $X'$ (which is row $i$ of X). That is the $(j,i)$ entry of $Y'X'$⁠, with Y first.
   Property (iv) of Section 3.5.1: $(AB)' = B'A'$⁠. Lesson: transpose

4. Use the symmetry of A and B

   Since $A' = A$ and $B' = B$⁠, replace each primed letter:

   $$\begin{aligned} C' &= B'A' - A'B' \\ &= BA - AB \\ &= -(AB - BA) \\ &= -C. \end{aligned}$$

   So $C' = -C$⁠, and C is square (Step 1). By Definition 5, $AB - BA$ is skew symmetric.

   The third line just takes out the factor $-1$⁠: $BA - AB = (-1)(AB - BA)$⁠, because matrix addition is commutative and $-(X - Y) = Y - X$ entry by entry. Notice what the proof did not use: we never assumed $AB = BA$⁠. In fact if $AB = BA$ then $C = O$⁠, which is skew symmetric for a trivial reason. The interesting case is $AB \ne BA$⁠, and the proof covers it.
   Definitions 4 and 5 (Section 3.6): $A' = A$ means symmetric, $C' = -C$ means skew symmetric. Lesson: symmetric and skew symmetric matrices

5. What this says about the entries

   Every skew symmetric matrix has zeros on its diagonal: putting $i = j$ in $c_{ji} = -c_{ii}$ gives $c_{ii} = -c_{ii}$⁠, so $2c_{ii} = 0$ and $c_{ii} = 0$⁠. So for any symmetric A and B of the same order, every diagonal entry of $AB - BA$ is 0, and the entries across the diagonal are negatives of each other.

   This gives a quick way to spot an arithmetic slip when you compute $AB - BA$ for actual matrices: if a diagonal entry is not 0, something has gone wrong. We use exactly this in the check below.
   The remark after Definition 5 (Section 3.6): the diagonal of a skew symmetric matrix is zero. Basics: mirror patterns

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