The Transpose

Definition and the four laws

Definition. The transpose $A'$ (also written $A^T$) of $A = [a_{ij}]_{m \times n}$ is the $n \times m$ matrix obtained by interchanging rows and columns: $\ (A')_{ij} = a_{ji}$.

Row 1 of $A$ becomes column 1 of $A'$, and the order flips from $m \times n$ to $n \times m$.

The four laws (memorise):

$$\textbf{(1)}\ (A')' = A \qquad \textbf{(2)}\ (A + B)' = A' + B' \qquad \textbf{(3)}\ (kA)' = kA' \qquad \textbf{(4)}\ (AB)' = B'A'$$

The fourth is the reversal law — the transpose of a product reverses the order (like the inverse's socks-shoes rule). Orders force it: if $A$ is $m \times n$ and $B$ is $n \times p$, then $(AB)'$ is $p \times m$, and only $B'A' = (p \times n)(n \times m)$ has that shape — $A'B'$ generally is not even defined.

Numerical check of the reversal law. For $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$, $C = \begin{pmatrix} 2 & 1 \\ 1 & 4 \end{pmatrix}$: $$AC = \begin{pmatrix} 4 & 9 \\ 3 & 12 \end{pmatrix} \implies (AC)' = \begin{pmatrix} 4 & 3 \\ 9 & 12 \end{pmatrix}, \qquad C'A' = \begin{pmatrix} 2 & 1 \\ 1 & 4 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 3 \\ 9 & 12 \end{pmatrix}$$ both agree ✓.

Common mistakes to avoid

Mistake 1 — $(AB)' = A'B'$. The order must reverse; the wrong version usually is not even conformable.

Mistake 2 — forgetting the order flip. The transpose of a $2 \times 3$ matrix is $3 \times 2$; entries do not just "reflect in place" unless the matrix is square.

Mistake 3 — transposing only part of an expression: $(2A + 3B)' = 2A' + 3B'$ — the transpose distributes over the whole linear combination.

Symmetric and Skew-Symmetric Matrices

The two families

Definitions (square matrices only). $A$ is symmetric if $A' = A$ (mirror symmetry across the diagonal: $a_{ij} = a_{ji}$). $\ A$ is skew-symmetric if $A' = -A$ (i.e. $a_{ij} = -a_{ji}$).

Diagonal entries of a skew-symmetric matrix vanish: putting $j = i$ in $a_{ij} = -a_{ji}$ gives $a_{ii} = -a_{ii}$, so $a_{ii} = 0$ — every skew-symmetric matrix has a zero diagonal.

Two building lemmas

For any square matrix $A$:

$$A + A' \text{ is always symmetric}, \qquad A - A' \text{ is always skew-symmetric}$$

Proof of the first: $(A + A')' = A' + (A')' = A' + A = A + A'$ ✓ — it equals its own transpose. The second is identical with a sign: $(A - A')' = A' - A = -(A - A')$.

The decomposition theorem

Theorem. Every square matrix is expressible, uniquely, as the sum of a symmetric and a skew-symmetric matrix: $$A = \underbrace{\frac{1}{2}(A + A')}_{P \ \text{symmetric}} + \underbrace{\frac{1}{2}(A - A')}_{Q \ \text{skew-symmetric}}$$

Worked decomposition. $B = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$:

Step 1 — compute $B + B'$ and halve: $$P = \frac{1}{2}\begin{pmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{pmatrix} = \begin{pmatrix} 2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{pmatrix} \ (\text{symmetric})$$

Step 2 — compute $B - B'$ and halve: $$Q = \frac{1}{2}\begin{pmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -\frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{pmatrix} \ (\text{skew, zero diagonal})$$

Step 3 — check: $P + Q = B$ ✓.

Products of symmetric matrices (the subtle facts)

For symmetric $A$ and $B$ (same order):

$$AB \text{ is symmetric} \iff AB = BA$$ $$AB + BA \text{ is always symmetric}, \qquad AB - BA \text{ is always skew-symmetric}$$

Why the first: $(AB)' = B'A' = BA$, so $(AB)' = AB$ exactly when $BA = AB$. The other two follow the same one-line transpose computation. These three statements are the section's favourite MCQs.

Common mistakes to avoid

Mistake 1 — applying 'symmetric' to non-square matrices. $A' = A$ forces the order to be square before anything else.

Mistake 2 — forgetting the $\frac{1}{2}$'s in the decomposition — $A + A'$ alone is symmetric but does not sum with $A - A'$ to $A$.

Mistake 3 — assuming a product of symmetric matrices is symmetric. Only when they commute.

Mistake 4 — nonzero diagonal in a claimed skew-symmetric matrix — instant contradiction.

Solved Examples

Example 1 — Transposing and combining

For $A = \begin{pmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{pmatrix}$ (order $3 \times 2$), write $A'$ and state its order.

Step 1 — rows become columns: $$A' = \begin{pmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{pmatrix}$$

Answer: $A'$ has order $2 \times 3$ — the order always flips.

Example 2 — Verifying the reversal law

For $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ and $C = \begin{pmatrix} 2 & 1 \\ 1 & 4 \end{pmatrix}$, verify that $(AC)' = C'A'$.

Step 1 — the product: $AC = \begin{pmatrix} 4 & 9 \\ 3 & 12 \end{pmatrix}$, so $(AC)' = \begin{pmatrix} 4 & 3 \\ 9 & 12 \end{pmatrix}$.

Step 2 — the reversed product of transposes: $$C'A' = \begin{pmatrix} 2 & 1 \\ 1 & 4 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 3 \\ 9 & 12 \end{pmatrix}$$

Answer: the two agree ✓ — and note $A'C'$ would give a different matrix: the reversal is essential.

Example 3 — Why skew diagonals vanish

Show that every diagonal entry of a skew-symmetric matrix is zero.

Step 1 — apply the definition at a diagonal address: $a_{ij} = -a_{ji}$ with $j = i$ reads $a_{ii} = -a_{ii}$.

Step 2 — solve: $2a_{ii} = 0$, so $a_{ii} = 0$.

Answer: all diagonal entries vanish — the fastest way to disqualify a matrix from being skew-symmetric is one nonzero diagonal entry.

Example 4 — $A + A'$ and $A - A'$

For any square matrix $A$, prove that $A + A'$ is symmetric and $A - A'$ is skew-symmetric.

Step 1 — transpose the sum: $$(A + A')' = A' + (A')' = A' + A = A + A' \implies \text{symmetric}$$

Step 2 — transpose the difference: $$(A - A')' = A' - A = -(A - A') \implies \text{skew-symmetric}$$

Answer: proved in two lines each — the building blocks of the decomposition theorem.

Example 5 — The full decomposition

Express $B = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.

Step 1 — symmetric part: $$P = \frac{1}{2}(B + B') = \begin{pmatrix} 2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{pmatrix}$$

Step 2 — skew part: $$Q = \frac{1}{2}(B - B') = \begin{pmatrix} 0 & -\frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{pmatrix}$$

Step 3 — verify: $P' = P$ ✓, $\ Q' = -Q$ with zero diagonal ✓, $\ P + Q = B$ ✓.

Answer: $B = P + Q$ as above — the standard 5-mark presentation, checks included.

Example 6 — Products of symmetric matrices

If $A$ and $B$ are symmetric matrices of the same order, show that $AB$ is symmetric if and only if $A$ and $B$ commute.

Step 1 — transpose the product with the reversal law: $$(AB)' = B'A' = BA \quad (\text{using } A' = A, \ B' = B)$$

Step 2 — read off both directions: $(AB)' = AB \iff BA = AB$.

Answer: symmetric times symmetric is symmetric exactly when the factors commute — and since matrices usually do not commute, $AB$ is usually not symmetric.

Example 7 — Always-symmetric and always-skew combinations

For symmetric $A$ and $B$, classify $AB + BA$ and $AB - BA$.

Step 1 — transpose the first: $$(AB + BA)' = (AB)' + (BA)' = BA + AB = AB + BA \implies \text{symmetric}$$

Step 2 — transpose the second: $$(AB - BA)' = BA - AB = -(AB - BA) \implies \text{skew-symmetric}$$

Answer: $AB + BA$ is always symmetric; $AB - BA$ is always skew-symmetric — the symmetrised and antisymmetrised products, a favourite one-line MCQ pair.