Skew Symmetric Matrix - Properties, Theorems, and Examples

What Is a Skew Symmetric Matrix?

A skew symmetric matrix (also called antisymmetric) is a square matrix $A$ that satisfies

$$A^T = -A$$

where $A^T$ is the transpose, the matrix formed by swapping rows and columns. Entry by entry, the condition reads $a_{ij} = -a_{ji}$: every entry is the negative of its mirror across the main diagonal.

A 3x3 example:

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

Check one pair: $a_{12} = 2$ and $a_{21} = -2$, which are negatives, as required.

Variable glossary. $a_{ij}$ is the entry in row $i$, column $j$. $A^T$ is the transpose. The trace is the sum of the diagonal entries. A singular matrix has determinant zero (no inverse).

Why Are the Diagonal Entries Always Zero?

This is the first thing readers ask, and it falls straight out of the definition. For a diagonal entry, $i = j$, so the condition $a_{ij} = -a_{ji}$ becomes:

$a_{ii} = -a_{ii}$

Add $a_{ii}$ to both sides:

$2a_{ii} = 0$

$a_{ii} = 0$

Every diagonal entry must equal its own negative, and the only number that does is zero. So the diagonal is forced to be all zeros, and you never get to choose those entries. An immediate consequence: the trace (sum of the diagonal) is zero for every skew symmetric matrix.

What Are the Properties of a Skew Symmetric Matrix?

The defining condition $A^T = -A$ propagates into a tidy set of properties.

• Zero diagonal, zero trace. As derived above, every $a_{ii} = 0$, so the trace is 0.

• Sums stay skew symmetric. If $A$ and $B$ are both skew symmetric, so is $A + B$, and so is any scalar multiple $kA$.

• Odd-order determinant is zero. Any skew symmetric matrix of odd order is singular, so its determinant is 0 (proof below).

• $A^2$ is symmetric. Squaring a skew symmetric matrix produces a symmetric one.

• Eigenvalues are zero or purely imaginary. A real skew symmetric matrix has no nonzero real eigenvalues.

The contrast with its partner, the symmetric matrix (where $A^T = A$), is the cleanest way to keep skew symmetric straight: symmetric mirrors equally, skew symmetric mirrors with a sign flip.

The Decomposition Theorem: Every Square Matrix as Symmetric Plus Skew

Here is the result that gives skew symmetric matrices their importance. Any square matrix can be written as the sum of a symmetric matrix and a skew symmetric matrix, in exactly one way:

$$A = \underbrace{\tfrac{1}{2}(A + A^T)}{\text{symmetric}} + \underbrace{\tfrac{1}{2}(A - A^T)}{\text{skew symmetric}}$$

The first piece, $\tfrac{1}{2}(A + A^T)$, is always symmetric. The second piece, $\tfrac{1}{2}(A - A^T)$, is always skew symmetric. Add them and the transposes cancel back to $A$. This decomposition is why the skew symmetric part of a matrix appears throughout physics and engineering — it captures the "rotational" content of a transformation, while the symmetric part captures the "stretching" content.

Why Is the Determinant of an Odd-Order Skew Symmetric Matrix Zero?

A short, clean proof. Start from $A^T = -A$ and take determinants of both sides:

$|A^T| = |-A|$

Use two facts: $|A^T| = |A|$ (transposing does not change the determinant), and $|-A| = (-1)^n |A|$ for an $n \times n$ matrix (factoring $-1$ out of each of the $n$ rows):

$|A| = (-1)^n |A|$

When $n$ is odd, $(-1)^n = -1$, so:

$|A| = -|A|$

$2|A| = 0$

$|A| = 0$

So every odd-order skew symmetric matrix is singular. (For even $n$, $(-1)^n = 1$ and the equation gives $|A| = |A|$, no information — even-order determinants need not be zero.)

Examples of a Skew Symmetric Matrix

Example 1

Is $\begin{bmatrix} 0 & 6 \ -6 & 0 \end{bmatrix}$ skew symmetric?

Check $a_{ij} = -a_{ji}$. The diagonal is all zeros. Off-diagonal: $a_{12} = 6$, $a_{21} = -6$, which are negatives.

Final answer: Yes, it is skew symmetric.

Example 2

Is $\begin{bmatrix} 0 & 4 \ 4 & 0 \end{bmatrix}$ skew symmetric?

This is the wrong-path-first example.

Wrong attempt. A student sees zeros on the diagonal and declares it skew symmetric, reasoning that the zero diagonal is the defining feature.

Why it is wrong. The zero diagonal is necessary but not sufficient. The off-diagonal entries must also be negatives of each other. Here $a_{12} = 4$ and $a_{21} = 4$ — equal, not opposite.

Correct. Test $a_{12} = -a_{21}$: is $4 = -4$? No. The mirror entries are equal, which makes this matrix symmetric, not skew symmetric.

Final answer: No. It is symmetric, not skew symmetric — the zero diagonal alone is not enough.

Example 3

Find $x$ so that $\begin{bmatrix} 0 & x \ -7 & 0 \end{bmatrix}$ is skew symmetric.

The condition $a_{12} = -a_{21}$ requires:

$x = -(-7)$

$x = 7$

Final answer: $x = 7$.

Example 4

Find the determinant of the $3 \times 3$ skew symmetric matrix $\begin{bmatrix} 0 & 2 & -5 \ -2 & 0 & 3 \ 5 & -3 & 0 \end{bmatrix}$.

The matrix is skew symmetric and has odd order (3), so the theorem says the determinant is zero. Confirm by cofactor expansion along the first row:

$|A| = 0 \cdot M_{11} - 2\begin{vmatrix} -2 & 3 \ 5 & 0 \end{vmatrix} + (-5)\begin{vmatrix} -2 & 0 \ 5 & -3 \end{vmatrix}$

$|A| = 0 - 2(0 - 15) - 5(6 - 0)$

$|A| = -2(-15) - 5(6)$

$|A| = 30 - 30$

$|A| = 0$

Final answer: $|A| = 0$, matching the odd-order rule.

Example 5

Decompose $A = \begin{bmatrix} 2 & 5 \ 1 & 4 \end{bmatrix}$ into symmetric and skew symmetric parts.

First find $A^T = \begin{bmatrix} 2 & 1 \ 5 & 4 \end{bmatrix}$.

Symmetric part $\tfrac{1}{2}(A + A^T)$:

$$\tfrac{1}{2}\begin{bmatrix} 4 & 6 \ 6 & 8 \end{bmatrix} = \begin{bmatrix} 2 & 3 \ 3 & 4 \end{bmatrix}$$

Skew symmetric part $\tfrac{1}{2}(A - A^T)$:

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

Final answer: $A = \begin{bmatrix} 2 & 3 \ 3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix}$, a symmetric matrix plus a skew symmetric one.

Example 6

Verify that $A^2$ is symmetric for $A = \begin{bmatrix} 0 & 3 \ -3 & 0 \end{bmatrix}$.

Compute $A^2$:

$$\begin{bmatrix} 0 & 3 \ -3 & 0 \end{bmatrix}\begin{bmatrix} 0 & 3 \ -3 & 0 \end{bmatrix} = \begin{bmatrix} -9 & 0 \ 0 & -9 \end{bmatrix}$$

The result is diagonal, hence symmetric (it equals its own transpose).

Final answer: $A^2 = \begin{bmatrix} -9 & 0 \ 0 & -9 \end{bmatrix}$, which is symmetric, as the property predicts.

Why Skew Symmetric Matrices Matter: "The math of pure rotation"

Skew symmetric matrices exist because rotation and the cross product needed an algebraic home. When a transformation rotates without stretching, a rigid spin, its generator is skew symmetric, because rotation has no "diagonal" stretching content, only the off-diagonal mixing of axes. The forced-zero diagonal is the algebra noticing that pure rotation cannot scale.

The destination is concrete:

• 3D rotations and robotics. The angular velocity of a spinning rigid body is encoded as a $3 \times 3$ skew symmetric matrix; multiplying by it produces the cross product $\boldsymbol{\omega} \times \mathbf{v}$. Robot arms and spacecraft attitude control run on exactly this.

• The cross product itself. Any cross product $\mathbf{a} \times \mathbf{b}$ can be written as a skew symmetric matrix built from $\mathbf{a}$ times the vector $\mathbf{b}$, the so-called "cross-product matrix."

• Physics field theory. The electromagnetic field tensor in special relativity is antisymmetric, packaging the electric and magnetic fields into one skew symmetric object.

Anywhere a system rotates or circulates rather than expands, a skew symmetric matrix is doing the bookkeeping.

Skew Symmetric Matrix - Where Intuition Breaks

Mistake 1: Believing the zero diagonal alone makes a matrix skew symmetric

Where it slips in: Spotting the zero diagonal and stopping the check there.

Don't do this: Call $\begin{bmatrix} 0 & 4 \ 4 & 0 \end{bmatrix}$ skew symmetric because the diagonal is zero.

The correct way: Test both conditions, the zero diagonal and $a_{ij} = -a_{ji}$ off the diagonal. The mirror entries must be negatives, not equals. The point of confusion between symmetric and skew symmetric is the single most common source of wrong answers here: a zero diagonal is shared by some symmetric matrices too, so it can never decide the question on its own.

Mistake 2: Mixing up the sign in the decomposition theorem

Where it slips in: Splitting a matrix into symmetric and skew parts.

Don't do this: Use $\tfrac{1}{2}(A + A^T)$ for the skew part.

The correct way: The sum $\tfrac{1}{2}(A + A^T)$ is symmetric; the difference $\tfrac{1}{2}(A - A^T)$ is skew symmetric. The plus sign keeps things equal (symmetric); the minus sign creates the negation (skew). The rusher who races through the formula swaps the two halves and gets parts that do not have the claimed symmetry, and a quick transpose check catches it.

Mistake 3: Assuming every skew symmetric determinant is zero

Where it slips in: Applying the odd-order rule to an even-order matrix.

Don't do this: Declare a $4 \times 4$ skew symmetric matrix singular by reflex.

The correct way: The zero-determinant rule holds only for odd order. Even-order skew symmetric matrices can have nonzero determinants (in fact, a perfect-square value via the Pfaffian). The step where $(-1)^n$ decides the outcome looks skippable, but skipping it is exactly where the even case goes wrong.

The Mathematicians Behind the Skew Symmetric Matrix

Arthur Cayley (1821-1895, England) developed the algebra of matrices, including the symmetric and skew symmetric classifications, in his foundational memoir on matrices of 1858. Ferdinand Georg Frobenius (1849-1917, Germany) extended the theory of bilinear and quadratic forms that underpins the symmetric-plus-skew decomposition, work surveyed at MacTutor.

Skew Symmetric Matrix - In Five Bullets

• A skew symmetric matrix satisfies $A^T = -A$: its diagonal is all zeros and each entry is the negative of its mirror across the diagonal.

• The zero diagonal follows from $a_{ii} = -a_{ii}$, which also forces the trace to zero.

• Every square matrix splits uniquely into a symmetric part $\tfrac{1}{2}(A + A^T)$ plus a skew symmetric part $\tfrac{1}{2}(A - A^T)$.

• The determinant of an odd-order skew symmetric matrix is always zero; even-order ones need not be.

• Skew symmetric matrices encode rotation, the cross product, and circulating physical fields, wherever a system turns rather than stretches.

Practice Questions on Skew Symmetric Matrices

Work through these, then check the answers below.

1. Is $\begin{bmatrix} 0 & -3 \ 3 & 0 \end{bmatrix}$ skew symmetric?

2. Is $\begin{bmatrix} 0 & 5 \ 5 & 0 \end{bmatrix}$ skew symmetric?

3. Find $y$ so that $\begin{bmatrix} 0 & y \ -8 & 0 \end{bmatrix}$ is skew symmetric.

4. Find the determinant of $\begin{bmatrix} 0 & 1 & -4 \ -1 & 0 & 6 \ 4 & -6 & 0 \end{bmatrix}$.

5. Decompose $A = \begin{bmatrix} 1 & 7 \ 3 & 2 \end{bmatrix}$ into a symmetric and a skew symmetric part.

Answer to Question 1: Yes. The diagonal is zero and $a_{12} = -3 = -a_{21}$.

Answer to Question 2: No. The mirror entries are equal ($5 = 5$), not negatives, so it is symmetric, not skew symmetric.

Answer to Question 3: The condition $a_{12} = -a_{21}$ gives $y = -(-8) = 8$.

Answer to Question 4: The matrix is skew symmetric and of odd order (3), so the determinant is 0.

Answer to Question 5: $A = \begin{bmatrix} 1 & 5 \ 5 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix}$, using $\tfrac{1}{2}(A + A^T)$ and $\tfrac{1}{2}(A - A^T)$.

Sharpen Your Skew Symmetric Matrix Skills - Three Practice Problems

Practice these to solidify your understanding: test three matrices for skew symmetry (include one symmetric trap with a zero diagonal), decompose a 2x2 matrix into symmetric and skew parts, and confirm the determinant of a 3x3 skew symmetric matrix is zero by direct expansion. If the symmetric-versus-skew distinction blurs, come back to the $A^T = -A$ versus $A^T = A$ contrast. Next, the partner concept symmetric matrix and the broader triangular matrix deepen the picture.

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