Transpose; Symmetric & Skew-Symmetric

The big idea. The transpose of a matrix $A$, written $A^T$, is the matrix you get by flipping $A$ across its main diagonal — every row becomes a column and every column becomes a row.

Concretely, suppose $A$ is an $m\times n$ matrix (that is, $m$ rows and $n$ columns) with entries $a_{ij}$ (the number sitting in row $i$, column $j$). Then $A^T$ is the $n\times m$ matrix whose entries are $(A^T)_{ij} = A_{ji}.$ Read that carefully: the entry in row $i$, column $j$ of $A^T$ is the entry in row $j$, column $i$ of $A$. The two indices simply swap places.

A worked flip. $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \quad\Longrightarrow\quad A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}.$ Row 1 of $A$, namely $(1,2,3)$, became column 1 of $A^T$. The shape changed from $2\times 3$ to $3\times 2$ — the transpose always swaps the two dimensions.

The diagonal stays put. The main diagonal consists of the entries $a_{ii}$ (equal row and column index). Since $(A^T)_{ii} = A_{ii}$, those entries never move — they are the hinge the matrix flips around.

Transposition interacts very cleanly with the other matrix operations. Let $A, B$ be matrices of compatible sizes and let $c$ be a scalar (an ordinary number). Then:

• Involution: $(A^T)^T = A$. Flipping twice returns the original matrix.
• Additivity: $(A+B)^T = A^T + B^T$ (here $A$ and $B$ must be the same size).
• Scalars pass through: $(cA)^T = c\,A^T$.
• Reversal rule for products: $(AB)^T = B^T A^T$ — the order of the factors reverses.

Why the order reverses. If $A$ is $m\times n$ and $B$ is $n\times p$, then $AB$ is $m\times p$, so $(AB)^T$ is $p\times m$. For a product of the transposed pieces to even be defined and have that shape, you need $B^T$ (size $p\times n$) times $A^T$ (size $n\times m$), giving $p\times m$. Trying $A^T B^T$ would be $(n\times m)(p\times n)$ — generally not even a legal product. Entry-by-entry the identity reads $\big[(AB)^T\big]_{ij} = (AB)_{ji} = \sum_k a_{jk}b_{ki} = \sum_k (B^T)_{ik}(A^T)_{kj} = (B^T A^T)_{ij}.$ The rule extends to more factors: $(ABC)^T = C^T B^T A^T$, and so on.

A square matrix $A$ (same number of rows and columns) is symmetric when it equals its own transpose: $A = A^T, \qquad\text{equivalently}\qquad a_{ij} = a_{ji} \ \text{ for all } i,j.$ Geometrically the entries are mirror images across the main diagonal: whatever sits in row $i$, column $j$ is copied into row $j$, column $i$. For example $A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 5 & 7 \\ 0 & 7 & 3 \end{bmatrix}, \qquad A = A^T.$ Check the mirror: $a_{12} = -1 = a_{21}$, and $a_{23} = 7 = a_{32}$, and so on. Only square matrices can be symmetric, because $A$ and $A^T$ must share the same shape before you can even compare them entry-for-entry.

A square matrix $A$ is skew-symmetric (also called antisymmetric) when $A = -A^T, \qquad\text{equivalently}\qquad a_{ij} = -a_{ji} \ \text{ for all } i,j.$ Now each entry is the negative of its mirror image across the diagonal.

The diagonal is forced to zero. Put $i=j$ in the entry condition: $a_{ii} = -a_{ii}$. Adding $a_{ii}$ to both sides gives $2a_{ii} = 0$, hence $a_{ii} = 0$. So every diagonal entry of a skew-symmetric matrix must be $0$. Example: $A = \begin{bmatrix} 0 & 3 & -2 \\ -3 & 0 & 4 \\ 2 & -4 & 0 \end{bmatrix}, \qquad A^T = \begin{bmatrix} 0 & -3 & 2 \\ 3 & 0 & -4 \\ -2 & 4 & 0 \end{bmatrix} = -A.$

Bonus — splitting any square matrix. Every square matrix decomposes into a symmetric part plus a skew-symmetric part: $A = \underbrace{\tfrac12\big(A + A^T\big)}_{\text{symmetric}} \; + \; \underbrace{\tfrac12\big(A - A^T\big)}_{\text{skew-symmetric}}.$ This is a handy sanity check that the two notions together account for all of $A$.