Mathematics · Ch 3 — Matrices

Symmetric and Skew Symmetric Matrices

3.6

Symmetric and Skew Symmetric Matrices

A square matrix can have special symmetry properties that simplify calculations and reveal structure. The two fundamental types are symmetric and skew symmetric matrices.

Symmetric Matrices

A square matrix $A = [a_{ij}]$ is symmetric if it equals its own transpose:

A' = A, \quad \text{i.e.} \quad a_{ij} = a_{ji} \text{ for all } i, j

Every entry above the main diagonal is mirrored by an equal entry below it; the diagonal entries are unrestricted.

Example:

A = \begin{bmatrix} 3 & \sqrt{2} & 3 \\ \sqrt{2} & -1.5 & -1 \\ 3 & -1 & 1 \end{bmatrix}

is symmetric: $(1,2)$ and $(2,1)$ are both $\sqrt{2}$ , $(1,3)$ and $(3,1)$ are both $3$ , $(2,3)$ and $(3,2)$ are both $-1$ .

Note

Only square matrices can be symmetric or skew symmetric, since the equality $A' = A$ (or $A' = -A$ ) requires $A$ and $A'$ to have the same order.

Skew Symmetric Matrices

A square matrix $A = [a_{ij}]$ is skew symmetric if its transpose equals its negative:

A' = -A, \quad \text{i.e.} \quad a_{ji} = -a_{ij} \text{ for all } i, j

Setting $i = j$ gives $a_{ii} = -a_{ii}$ , so $2a_{ii} = 0$ :

Important

All diagonal elements of a skew symmetric matrix are zero.

Example:

B = \begin{bmatrix} 0 & e & f \\ -e & 0 & g \\ -f & -g & 0 \end{bmatrix}

is skew symmetric ( $B' = -B$ ): zeros on the main diagonal, and each off-diagonal entry is the negative of its mirror image across the diagonal.

Theorem 1: Building Symmetric and Skew Symmetric Matrices from Any Square Matrix

For any square matrix $A$ with real entries, $A + A'$ is symmetric and $A - A'$ is skew symmetric.

›Proof

$A + A'$ is symmetric. Let $B = A + A'$ . Then $B' = (A + A')' = A' + (A')' = A' + A = A + A' = B$ . So $B$ is symmetric.

$A - A'$ is skew symmetric. Let $C = A - A'$ . Then $C' = (A - A')' = A' - (A')' = A' - A = -(A - A') = -C$ . So $C$ is skew symmetric.

(using $(X \pm Y)' = X' \pm Y'$ and $(A')' = A$ .)

Theorem 2: Expressing Any Square Matrix as a Sum

Any square matrix can be expressed as the sum of a symmetric matrix and a skew symmetric matrix.

›Proof

For any square matrix $A$ , write

$A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A')$

By Theorem 1, $A + A'$ is symmetric and $A - A'$ is skew symmetric; scaling by $\frac{1}{2}$ preserves each property (since $(kX)' = kX'$ ). Thus $A = P + Q$ with $P = \frac{1}{2}(A + A')$ symmetric and $Q = \frac{1}{2}(A - A')$ skew symmetric.

Tip

To decompose any square matrix $A$ : compute $A'$ ; form the symmetric part $P = \frac{1}{2}(A + A')$ and the skew symmetric part $Q = \frac{1}{2}(A - A')$ ; then $A = P + Q$ .

A

Definition 4Symmetric

Definition

A square matrix $A = [a_{ij}]$ is called symmetric if it equals its own transpose. That is:

A' = A

Equivalently, for every possible row index $i$ and column index $j$ :

a_{ij} = a_{ji}

This means the entry at the $i$ -th row and $j$ -th column is exactly the same as the entry at the $j$ -th row and $i$ -th column — the matrix is mirrored along its main diagonal.

Intuition

Think of folding the matrix along the main diagonal (top-left to bottom-right). If the two halves match perfectly, the matrix is symmetric. It's like a reflection: what you see above the diagonal is exactly what you see below it.

Example

The matrix …

Definition 5Skew Symmetric Matrix

Definition

A skew symmetric matrix is a square matrix $A = [a_{ij}]$ such that its transpose equals its negative:

A' = -A

Equivalently, for every entry:

a_{ji} = -a_{ij} \quad \text{for all possible } i, j

Key Consequence (Diagonal Entries)

If we set $i = j$ , the condition becomes:

a_{ii} = -a_{ii} \quad \Rightarrow \quad 2a_{ii} = 0 \quad \Rightarrow \quad a_{ii} = 0

So all diagonal elements of a skew symmetric matrix are zero.

Intuition

Think of a skew symmetric matrix as a "mirror image with a sign flip": the entry at $(i, j)$ is the negative of the entry at $(j, i)$ . This forces the diagonal to be zero because a number cannot be its own negative unless it is zero.

Example …

Theorem 1

Theorem 2: Every Square Matrix is the Sum of a Symmetric and a Skew-Symmetric Matrix

Statement:

Let $A$ be any square matrix with real entries. Then $A$ can be uniquely expressed as

A = P + Q,

where $P$ is a symmetric matrix ( $P' = P$ ) and $Q$ is a skew-symmetric matrix ( $Q' = -Q$ ).

Why This Theorem Matters

This decomposition is a standard tool in linear algebra. It is used to separate a matrix into its "even" and "odd" parts under transposition, much like splitting a function into even and odd components. You will encounter it in problems that ask you to write a given matrix as a sum of a symmetric and a skew-symmetric matrix (see Example 22 in the textbook).

The Complete Proof

›Proof

Let $A$ be any square matrix. Consider the following two matrices:

$P = \frac{1}{2}(A + A') \quad \text{and} \quad Q = \frac{1}{2}(A - A').$

Step 1: Show $P$ is symmetric.

Compute the transpose of $P$ :

$P' = \left[\frac{1}{2}(A + A')\right]' = \frac{1}{2}(A + A')' = \frac{1}{2}\bigl(A' + (A')'\bigr).$

Since $(A')' = A$ , we get

$P' = \frac{1}{2}(A' + A) = \frac{1}{2}(A + A') = P.$

Hence $P' = P$ , so $P$ is symmetric.

Step 2: Show $Q$ is skew-symmetric.

Compute the transpose of $Q$ :

$Q' = \left[\frac{1}{2}(A - A')\right]' = \frac{1}{2}(A - A')' = \frac{1}{2}\bigl(A' - (A')'\bigr).$

Again using $(A')' = A$ , we obtain

$Q' = \frac{1}{2}(A' - A) = -\frac{1}{2}(A - A') = -Q.$

Thus $Q' = -Q$ , so $Q$ is skew-symmetric.

Step 3: Verify that $A = P + Q$ .

Add $P$ and $Q$ :

$P + Q = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') = \frac{1}{2}\bigl[(A + A') + (A - A')\bigr] = \frac{1}{2}(2A) = A.$

Therefore $A$ is expressed as the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$ .

Step 4: Uniqueness (optional but often asked).

Suppose $A = P_1 + Q_1$ where $P_1$ is symmetric and $Q_1$ is skew-symmetric. Then $A' = P_1' + Q_1' = P_1 - Q_1$ . Solving the system

$P_1 + Q_1 = A, \quad P_1 - Q_1 = A'$

gives $P_1 = \frac{1}{2}(A + A')$ and $Q_1 = \frac{1}{2}(A - A')$ , which are exactly $P$ and $Q$ . Hence the decomposition is unique.

Key Points to Remember

The theorem works for any square matrix with real entries. The textbook assumes real numbers, but the proof holds for matrices over any field where $2$ is invertible.
The two building blocks are:
- Symmetric part: $\displaystyle P = \frac{1}{2}(A + A')$
- Skew-symmetric part: $\displaystyle Q = \frac{1}{2}(A - A')$
The proof uses only two properties of transpose: $(A + B)' = A' + B'$ and $(A')' = A$ .

Tip

To quickly check your work: after finding $P$ and $Q$ , verify that $P' = P$ and $Q' = -Q$ . Then add them to confirm you get back $A$ . …

Theorem 2

Theorem 2: Every Square Matrix is the Sum of a Symmetric and a Skew-Symmetric Matrix

Statement:

Let $A$ be any square matrix with real entries. Then $A$ can be uniquely expressed as

A = P + Q,

where $P$ is a symmetric matrix ( $P' = P$ ) and $Q$ is a skew-symmetric matrix ( $Q' = -Q$ ).

Why This Theorem Matters

The Complete Proof

›Proof

Let $A$ be any square matrix. Consider the following two matrices:

$P = \frac{1}{2}(A + A') \quad \text{and} \quad Q = \frac{1}{2}(A - A').$

Step 1: Show $P$ is symmetric.

Compute the transpose of $P$ :

$P' = \left[\frac{1}{2}(A + A')\right]' = \frac{1}{2}(A + A')' = \frac{1}{2}\bigl(A' + (A')'\bigr).$

Since $(A')' = A$ , we get

$P' = \frac{1}{2}(A' + A) = \frac{1}{2}(A + A') = P.$

Hence $P' = P$ , so $P$ is symmetric.

Step 2: Show $Q$ is skew-symmetric.

Compute the transpose of $Q$ :

$Q' = \left[\frac{1}{2}(A - A')\right]' = \frac{1}{2}(A - A')' = \frac{1}{2}\bigl(A' - (A')'\bigr).$

Again using $(A')' = A$ , we obtain

$Q' = \frac{1}{2}(A' - A) = -\frac{1}{2}(A - A') = -Q.$

Thus $Q' = -Q$ , so $Q$ is skew-symmetric.

Step 3: Verify that $A = P + Q$ .

Add $P$ and $Q$ :

$P + Q = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') = \frac{1}{2}\bigl[(A + A') + (A - A')\bigr] = \frac{1}{2}(2A) = A.$

Therefore $A$ is expressed as the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$ .

Step 4: Uniqueness (optional but often asked).

Suppose $A = P_1 + Q_1$ where $P_1$ is symmetric and $Q_1$ is skew-symmetric. Then $A' = P_1' + Q_1' = P_1 - Q_1$ . Solving the system

$P_1 + Q_1 = A, \quad P_1 - Q_1 = A'$

gives $P_1 = \frac{1}{2}(A + A')$ and $Q_1 = \frac{1}{2}(A - A')$ , which are exactly $P$ and $Q$ . Hence the decomposition is unique.

Key Points to Remember

The theorem works for any square matrix with real entries. The textbook assumes real numbers, but the proof holds for matrices over any field where $2$ is invertible.
The two building blocks are:
- Symmetric part: $\displaystyle P = \frac{1}{2}(A + A')$
- Skew-symmetric part: $\displaystyle Q = \frac{1}{2}(A - A')$
The proof uses only two properties of transpose: $(A + B)' = A' + B'$ and $(A')' = A$ .

Tip

To quickly check your work: after finding $P$ and $Q$ , verify that $P' = P$ and $Q' = -Q$ . Then add them to confirm you get back $A$ . …

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