3.3 Transpose, symmetric, and skew-symmetric matrices

Motivation

Once matrix multiplication is available, the shape of a matrix matters as much as its numerical entries. A row of coefficients and a column of coordinates play different roles in a product, yet many arguments need to pass from one role to the other without changing the underlying data. Transpose performs exactly that conversion: rows become columns, columns become rows, and the dimensions change accordingly.

For square matrices, this index swap can be pictured as reflection across the main diagonal. Matrices fixed by that reflection are symmetric; matrices sent to their negatives are skew-symmetric. These conditions are more than visual patterns. They determine how products behave, explain why multiplication order matters, and lead to a canonical way of separating any square matrix into two structurally simpler parts.

Throughout this note, matrix entries are real numbers. All transpose identities remain valid over an arbitrary field, but the statements that divide by $2$ or deduce $a_{ii}=0$ from $a_{ii}=-a_{ii}$ require that $2$ be nonzero and invertible. Thus they also hold over any field of characteristic not equal to $2$. In characteristic $2$, symmetric and skew-symmetric conditions are no longer distinct, so those conclusions must be reformulated.

Transpose swaps rows and columns

Definition

Transpose

If $A = [a_{ij}]$ is an $m \times n$ matrix, then its transpose $A^T$ is the $n \times m$ matrix whose $(j,i)$ entry is $a_{ij}$.

Equivalently,

$$(A^T)_{ji} = a_{ij}.$$

So every row of $A$ becomes a column of $A^T$, and every column of $A$ becomes a row.

This operation does not alter the entries themselves; it changes where each entry is read. If $A$ is $m \times n$, then the row index of $A$ ranges from $1$ to $m$ and its column index ranges from $1$ to $n$. After transposition, those index ranges exchange places. This dimension check should be made before using any transpose identity, because an equality of matrices is meaningful only when both sides have the same size.

The sequence below sets out the geometric picture before the examples: transpose is an index swap, and for square matrices that index swap becomes reflection across the main diagonal.

Follow transpose as index-swapping and diagonal reflection, then connect that picture to symmetric matrices, skew-symmetric matrices, product order reversal, and the symmetric/skew decomposition.

1. Index swap

   The entry a_ij in A appears at position (j,i) in A^T, so an m x n matrix becomes an n x m matrix.

2. Diagonal reflection

   For a square matrix, transpose keeps the main diagonal fixed and swaps entries across that diagonal.

3. Symmetric rule

   A matrix is symmetric when A^T = A, which means each off-diagonal pair satisfies a_ij = a_ji.

4. Skew rule

   A matrix is skew-symmetric when A^T = -A, so paired entries have opposite signs and diagonal entries must be zero.

5. Product order

   Transpose reverses products: (AB)^T = B^T A^T. The order changes because row-column pairings turn around.

6. Two-part split

   For square A, the matrix 1/2(A+A^T) is symmetric, 1/2(A-A^T) is skew-symmetric, and the two parts add back to A.

Transpose swaps the two indices. For square matrices, that is diagonal reflection: symmetric matrices keep paired entries equal, skew-symmetric matrices make paired entries opposite, and every square matrix splits into symmetric and skew-symmetric parts.

Guided visual comparison

The live comparison below lets you switch among several examples. Compare $A$ with $A^T$, then look at what happens to entries on opposite sides of the main diagonal.

Worked example

Compute a transpose

Let

$$A = \begin{bmatrix} 1 & 4 & -2 \\ 0 & 3 & 5 \end{bmatrix}.$$

Then

$$A^T = \begin{bmatrix} 1 & 0 \\ 4 & 3 \\ -2 & 5 \end{bmatrix}.$$

The $2 \times 3$ matrix becomes a $3 \times 2$ matrix because rows and columns swap roles.

Three basic identities should be treated as part of the definition-level vocabulary of the subject:

$$(A^T)^T = A, \qquad (A + B)^T = A^T + B^T, \qquad (cA)^T = cA^T.$$

When products are defined, transpose reverses the order:

$$(AB)^T = B^T A^T.$$

That reversal is not a cosmetic detail. It reflects the fact that matrix multiplication is built from row-column pairings, and transposition switches the role of rows and columns.

Worked example

Transpose identities in action

Take

$$A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 3 & 0 \\ 4 & 5 \end{bmatrix}.$$

Then

$$A + B = \begin{bmatrix} 4 & 2 \\ 4 & 6 \end{bmatrix}, \qquad (A + B)^T = \begin{bmatrix} 4 & 4 \\ 2 & 6 \end{bmatrix}.$$

On the other hand,

$$A^T + B^T = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 2 & 6 \end{bmatrix}.$$

The product rule is similar, but the order matters:

$$AB = \begin{bmatrix} 11 & 10 \\ 4 & 5 \end{bmatrix}, \qquad (AB)^T = \begin{bmatrix} 11 & 4 \\ 10 & 5 \end{bmatrix},$$

while

$$B^T A^T = \begin{bmatrix} 3 & 4 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 4 \\ 10 & 5 \end{bmatrix}.$$

Theorem

Basic properties of transpose

Let $A$ and $B$ have the same size when they are added, let $c$ be a scalar, and suppose $A$ is $m \times n$ and $B$ is $n \times p$ when they are multiplied. Then:

1. $(A^T)^T = A$
2. $(A + B)^T = A^T + B^T$
3. $(cA)^T = cA^T$
4. $(AB)^T = B^T A^T$

The first three rules follow immediately by comparing entries. The product rule deserves a complete calculation because it explains why the order reverses.

Proof

Entrywise proof of the product rule

Let $A=[a_{ij}]$ be $m \times n$ and $B=[b_{ij}]$ be $n \times p$. Then $AB$ is $m \times p$, so both $(AB)^T$ and $B^T A^T$ are $p \times m$. For $1 \le i \le p$ and $1 \le j \le m$, the $(i,j)$ entry of the first matrix is

$$\bigl((AB)^T\bigr)_{ij} =(AB)_{ji} =\sum_{k=1}^{n} a_{jk}b_{ki}.$$

On the other hand, the $(i,j)$ entry of $B^T A^T$ is

$$\begin{aligned} (B^T A^T)_{ij} &=\sum_{k=1}^{n}(B^T)_{ik}(A^T)_{kj} \\ &=\sum_{k=1}^{n}b_{ki}a_{jk} \\ &=\sum_{k=1}^{n}a_{jk}b_{ki}. \end{aligned}$$

The last equality uses commutativity of scalar multiplication, not commutativity of matrix multiplication. Corresponding entries therefore agree, so $(AB)^T=B^T A^T$. The dimensions also explain the reversed order: $B^T$ is $p \times n$ and $A^T$ is $n \times m$, whereas $A^T B^T$ would generally not even be defined.

Why transpose matters beyond rearranging entries

Suppose $A$ is $m \times n$, $x$ is a column in $\mathbb{R}^n$, and $y$ is a column in $\mathbb{R}^m$. The product $Ax$ lies in $\mathbb{R}^m$, so its dot product with $y$ is defined. Written as matrix multiplication, the same scalar can be regrouped as

$$(Ax)^T y=x^T A^T y=x^T(A^T y).$$

The first equality uses the product-transpose rule; the second only changes parentheses, using associativity. Thus an application of $A$ to $x$ can be transferred to an application of $A^T$ to $y$ when the two sides occur inside a dot product. Later courses describe $A^T$ as the matrix that moves a linear map from one side of the Euclidean inner product to the other.

The dimensions clarify the roles. The matrix $A$ sends an $n$-component input to an $m$-component output, whereas $A^T$ sends an $m$-component coefficient vector back to $n$ components. This does not say that $A^T$ undoes $A$; that would be a statement about an inverse. It says instead that transpose records how the same coefficients are read after rows and columns exchange roles. This distinction becomes essential in orthogonality, projections, and least-squares problems, where a rectangular matrix cannot have an ordinary two-sided inverse but always has a transpose.

This identity also provides a reliable dimension audit: if either side is not a scalar, then at least one matrix or vector has been placed in the wrong order.

Symmetric and skew-symmetric matrices

Definition

Symmetric and skew-symmetric matrices

Let $A$ be a square matrix.

• $A$ is symmetric if $A^T = A$.
• $A$ is skew-symmetric if $A^T = -A$.

Notice the word square. If $A$ is not square, then $A$ and $A^T$ do not even have the same size, so the equations $A^T = A$ and $A^T = -A$ cannot be true.

Symmetry says the matrix matches its reflection across the main diagonal. Skew-symmetry says the reflected matrix is the negative of the original. Entrywise, these statements are

$$A\text{ symmetric} \iff a_{ij}=a_{ji}, \qquad A\text{ skew-symmetric} \iff a_{ij}=-a_{ji}$$

for every pair of indices $i,j$. The diagonal deserves special attention. If $A$ is skew-symmetric, then $a_{ii}=-a_{ii}$, so $2a_{ii}=0$. Over the real numbers this forces $a_{ii}=0$. This short argument is exactly where the characteristic-not-$2$ assumption enters.

Worked example

Recognize symmetry

The matrix

$$\begin{bmatrix} 2 & -1 & 4 \\ -1 & 3 & 0 \\ 4 & 0 & 5 \end{bmatrix}$$

is symmetric, because the $(i,j)$ and $(j,i)$ entries agree.

The matrix

$$\begin{bmatrix} 0 & 2 & -1 \\ -2 & 0 & 4 \\ 1 & -4 & 0 \end{bmatrix}$$

is skew-symmetric, because the transpose changes every off-diagonal entry's sign and leaves the diagonal as $0$.

Three quick consequences are especially useful:

• every diagonal entry of a real skew-symmetric matrix is $0$
• the zero matrix is both symmetric and skew-symmetric
• the identity matrix is symmetric

The third item is a special case of the fact that diagonal matrices are fixed by transpose.

Theorem

Useful transpose-based identities

For every square matrix $A$,

$$(A + A^T)^T = A + A^T, \qquad (A - A^T)^T = -(A - A^T).$$

So $A + A^T$ is symmetric and $A - A^T$ is skew-symmetric.

Proof

Why the identities hold

Apply the basic transpose rules:

$$(A + A^T)^T = A^T + (A^T)^T = A^T + A = A + A^T.$$

For the difference,

$$(A - A^T)^T = A^T - (A^T)^T = A^T - A = -(A - A^T).$$

The previous result is the starting point for the most important structural fact in this section.

Theorem

Decomposition into symmetric and skew-symmetric parts

Every real square matrix $A$ can be written uniquely as

$$A = S + K,$$

where $S$ is symmetric and $K$ is skew-symmetric. In fact,

$$S = \frac{1}{2}(A + A^T), \qquad K = \frac{1}{2}(A - A^T).$$

The conclusion is both an existence statement and a uniqueness statement. A formula alone is not enough unless both parts are checked.

Proof

Existence and uniqueness of the decomposition

For existence, define

$$S=\frac12(A+A^T), \qquad K=\frac12(A-A^T).$$

The preceding transpose identities give $S^T=S$ and $K^T=-K$, so $S$ is symmetric and $K$ is skew-symmetric. Moreover,

$$S+K =\frac12(A+A^T)+\frac12(A-A^T) =A.$$

Thus a decomposition exists. To prove uniqueness, suppose instead that some symmetric matrix $S$ and some skew-symmetric matrix $K$ satisfy $A=S+K$. Taking transposes and using the defining properties gives a second equation,

$$A^T=S^T+K^T=S-K.$$

Adding and subtracting the two equations yields

$$A+A^T=2S, \qquad A-A^T=2K.$$

Because $2$ is invertible over the real numbers,

$$S=\frac12(A+A^T), \qquad K=\frac12(A-A^T).$$

Any proposed decomposition must therefore use exactly these two matrices. This proves uniqueness as well as existence. The same proof works over every field of characteristic not equal to $2$.

Worked example

Decompose a matrix

Let

$$A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & -1 & 3 \\ 4 & 2 & 0 \end{bmatrix}.$$

Then

$$A^T = \begin{bmatrix} 1 & 2 & 4 \\ 0 & -1 & 2 \\ 1 & 3 & 0 \end{bmatrix}.$$

So

$$\frac{1}{2}(A + A^T) = \begin{bmatrix} 1 & 1 & \frac{5}{2} \\ 1 & -1 & \frac{5}{2} \\ \frac{5}{2} & \frac{5}{2} & 0 \end{bmatrix},$$

and

$$\frac{1}{2}(A - A^T) = \begin{bmatrix} 0 & -1 & -\frac{3}{2} \\ 1 & 0 & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} & 0 \end{bmatrix}.$$

The first matrix is symmetric, the second is skew-symmetric, and their sum is $A$.

There is also a useful symmetric pattern whenever a matrix is multiplied by its transpose. It converts a possibly rectangular matrix into square matrices that encode row or column interactions.

Theorem

Products with a transpose are symmetric

If the product is defined, then both $A^T A$ and $A A^T$ are symmetric.

Proof

Proof by transposition

If $A$ is $m \times n$, then $A^T A$ is $n \times n$ and $AA^T$ is $m \times m$, so symmetry is dimensionally possible in both cases. For the first product,

$$(A^T A)^T = A^T (A^T)^T = A^T A.$$

For the second,

$$(AA^T)^T=(A^T)^T A^T=AA^T.$$

Each product equals its transpose, which is precisely the definition of a symmetric matrix.

This identity is one of the main reasons transpose shows up again in later topics such as orthogonality, projections, and least-squares problems. For example, when $x$ is a compatible real column vector,

$$x^T A^T A x=(Ax)^T(Ax)=\lVert Ax\rVert^2\ge 0.$$

Thus $A^T A$ carries information about the lengths of vectors after applying $A$. This observation will later connect transpose to geometry, not merely to entry rearrangement.

Commuting and non-commuting matrices

Definition

Commuting matrices

Two square matrices $A$ and $B$ of the same order commute if

$$AB = BA.$$

For addition, commutativity is automatic. For multiplication, it is exceptional. This is one of the most important differences between scalar algebra and matrix algebra.

The zero matrix and the identity matrix commute with every square matrix of the same size. Diagonal matrices of the same size also commute with one another, because their products remain diagonal and the diagonal entries multiply in the ordinary commutative way.

Transpose gives a precise test for when products of structured matrices retain structure. If $A$ and $B$ are symmetric, then $(AB)^T=BA$; consequently $AB$ is symmetric exactly when $AB=BA$. For two skew-symmetric matrices the same formula $(AB)^T=BA$ appears, because the two minus signs cancel. By contrast, if $S$ is symmetric and $K$ is skew-symmetric, then $(SK)^T=-KS$, so a commuting product $SK=KS$ is skew-symmetric rather than symmetric.

Worked example

A non-commuting pair

Take

$$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}.$$

Then

$$AB = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}, \qquad BA = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}.$$

So $AB \ne BA$.

That kind of example is not a curiosity. It is the reason matrix identities must always preserve the order of factors.

Theorem

A useful skew-symmetric commuting test

If $A$ and $B$ are skew-symmetric square matrices of the same order, then

$$AB \text{ is symmetric } \iff AB = BA.$$

Proof

Why this is true

Because $A^T = -A$ and $B^T = -B$,

$$(AB)^T = B^T A^T = (-B)(-A) = BA.$$

So $AB$ is symmetric exactly when $(AB)^T = AB$, and that happens exactly when $BA = AB$.

Worked example

Symmetric times skew-symmetric

Let

$$S = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}, \qquad K = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.$$

Here $S$ is symmetric and $K$ is skew-symmetric, but they do not commute:

$$SK = \begin{bmatrix} 0 & 2 \\ -3 & 0 \end{bmatrix}, \qquad KS = \begin{bmatrix} 0 & 3 \\ -2 & 0 \end{bmatrix}.$$

If two matrices of this kind do commute, then their product is skew-symmetric, because

$$(SK)^T = K^T S^T = (-K)S = -KS = -SK.$$

Common mistakes

Common mistake

Transpose reverses products

The correct identity is $(AB)^T = B^T A^T$, not $A^T B^T$.

Common mistake

Symmetric does not mean commuting

Symmetric is a property of one matrix. Commuting is a property of two matrices. They are different statements.

Common mistake

A zero diagonal does not guarantee skew-symmetry

If $A^T=-A$, then every diagonal entry is $0$, but the converse is false. For example, $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ has a zero diagonal and is symmetric, not skew-symmetric. The off-diagonal conditions $a_{ij}=-a_{ji}$ must also be checked.

Common mistake

Transpose is not the same as inverse

Transpose always exists and only exchanges indices. An inverse exists only for certain square matrices and must satisfy $AA^{-1}=A^{-1}A=I$. The symbols $A^T$ and $A^{-1}$ therefore describe different operations, even though they coincide for special matrices studied later.

Summary

Transpose exchanges the two matrix indices, changes an $m \times n$ matrix into an $n \times m$ matrix, and reverses multiplication order. Symmetric and skew-symmetric matrices are square matrices characterized by $A^T=A$ and $A^T=-A$; over the real numbers, the latter condition forces a zero diagonal.

The formulas

$$\frac12(A+A^T) \quad\text{and}\quad \frac12(A-A^T)$$

are not arbitrary tricks. They are the uniquely determined symmetric and skew-symmetric parts of $A$, obtained by solving the two equations $A=S+K$ and $A^T=S-K$. Products such as $A^T A$ are automatically symmetric, while products of two structured square matrices may require commutativity to retain symmetry or skew-symmetry. These ideas prepare the algebra needed for orthogonality, quadratic forms, projections, and least-squares problems.

Quick checks

Checkpoint

If $A$ is $3 \times 2$, what is the size of $A^T$?

Swap rows and columns.

Solution · Answer

$A^T$ is $2 \times 3$.

Checkpoint

What must every diagonal entry of a real skew-symmetric matrix be?

Use $A^T = -A$ on the diagonal.

Solution · Answer

Every diagonal entry must be $0$.

Checkpoint

Which identity is correct: $(AB)^T = A^T B^T$ or $(AB)^T = B^T A^T$?

Compare an arbitrary entry on the two sides, or test a noncommuting pair of square matrices.

Solution · Answer

$(AB)^T = B^T A^T$.

Checkpoint

If $S$ is symmetric and $K$ is skew-symmetric, what is $S^T$ and what is $K^T$?

Read the definitions directly.

Solution · Answer

$S^T = S$ and $K^T = -K$.

Guided exercises

Checkpoint

Find the symmetric and skew-symmetric parts of $A = \begin{bmatrix}2&1\\4&3\end{bmatrix}$.

Use $S = 1/2 (A + A^T)$ and $K = 1/2 (A - A^T)$.

Solution · Guided solution

First compute

$$A^T = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}.$$

Then

$$S = \frac{1}{2} \begin{bmatrix} 4 & 5 \\ 5 & 6 \end{bmatrix} = \begin{bmatrix} 2 & \frac{5}{2} \\ \frac{5}{2} & 3 \end{bmatrix},$$

and

$$K = \frac{1}{2} \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{3}{2} \\ \frac{3}{2} & 0 \end{bmatrix}.$$

Check that $S$ is symmetric, $K$ is skew-symmetric, and $A = S + K$.

Checkpoint

Let $A$ and $B$ be symmetric matrices. If $AB = BA$, why is $AB$ symmetric?

Use the transpose identity and the commuting assumption.

Solution · Guided solution

Because

$$(AB)^T = B^T A^T = BA = AB.$$

So $AB$ equals its transpose, which is exactly the definition of symmetric.

Checkpoint

For a square matrix $A$, explain in words why $A + A^T$ is always symmetric.

You do not need a full formal proof, but you must mention transpose.

Solution · Guided solution

If you transpose $A + A^T$, you get $(A + A^T)^T = A^T + (A^T)^T = A^T + A = A + A^T$, so the matrix equals its own transpose and is therefore symmetric.

Checkpoint

For a square matrix $A$, explain in words why $A - A^T$ is always skew-symmetric.

Again, focus on what happens after taking the transpose.

Solution · Guided solution

If you transpose $A - A^T$, you get $(A - A^T)^T = A^T - A = -(A - A^T)$, so the transpose is the negative of the original matrix.

Related notes

This note builds on 3.2 Matrix multiplication, identity matrices, and linear systems. Continue to 3.4 Special matrices to study diagonal, triangular, identity, and elementary matrix families.