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# Symmetric and Skew Symmetric Matrices

## Symmetric and Skew Symmetric Matrices

The process of taking transpose of a matrix generally produce a new matrix but some times this new matrix is same as the original matrix or some times it may gives the negative of the original matrix. See the following examples.

### Examples of Transpose of Matrix:

Transpose of a matrix is it self :

 \begin{aligned} & D=\left[\begin{matrix}1&0&3\\0&1&2\\3&2&9\end{matrix}\right]_{3\times 3} \\ \end{aligned}

Transpose of D is

 \begin{aligned} D^T&=\left[\begin{matrix}1&0&3\\0&1&2\\3&2&9\end{matrix}\right]_{3\times 3}\\ &=D \end{aligned}

Transpose of a matrix is the negative of the original matrix :

 \begin{aligned} E&=\left[\begin{matrix}0&2&-5\\-2&0&3\\5&-3&0\end{matrix}\right] \end{aligned}

Transpose of E is

 \begin{aligned} E^T&=\left[\begin{matrix}0&-2&5\\2&0&-3\\-5&3&0\end{matrix}\right]\\ &=-\left[\begin{matrix}0&2&-5\\-2&0&3\\5&-3&0\end{matrix}\right]\\ &=-E \end{aligned}

In example 1 we get transpose of the matrix is it self and in Example 2 we get the transpose of the matrix is negative of the original matrix. These types of matrices has special names the matrix in Example $(1)$ is known as symmetric matrix and the matrix in Example $(2)$ is known as skew-symmetric matrix. Let us formally define symmetric matrix and skew-symmetric matrix.

## Symmetric Matrix :

A square matrix $A$ is called symmetric matrix if $A^T=A$.

In other words, a matrix $A=[a_{ij}]_{n\times n}$ is called symmetric matrix if $a_{ij}=a_{ji}\;\;\forall i,j$.

### Examples of Symmetric Matrices.

1. An example of a $2\times 2$ symmetric matrix.

 \begin{aligned} A&=\left[\begin{matrix}2&5\\5&3\end{matrix}\right]_{2\times 2} \end{aligned}

Transpose of A is

 \begin{aligned} A^T&=\left[\begin{matrix}2&5\\5&3\end{matrix}\right]_{2\times 2}\\ &=A \end{aligned}

1. An example of a $4\times 4$ symmetric matrix.

 \begin{aligned} B&=\left[\begin{matrix}1&2&-9&5\\2&2&3&9\\-9&3&9&8\\5&9&8&2\end{matrix}\right]_{4\times 4} \end{aligned}
The transpose of $B$ is

 \begin{aligned} B^T&=\left[\begin{matrix}1&2&-9&5\\2&2&3&9\\-9&3&9&8\\5&9&8&2\end{matrix}\right]_{4\times 4}\\ &=B \end{aligned}

## Skew-Symmetric Matrix:

A square matrix $A$ is called skew-symmetric matrix if $A^T=-A$.

In other words, a matrix $A=[a_{ij}]_{n\times n}$ is called symmetric matrix if $a_{ij}=-a_{ji}\;\;\forall i,j$.

### Examples of Skew-Symmetric Matrices.

1. An example of a $2\times 2$ skew-symmetric matrix.

 \begin{aligned} A&=\left[\begin{matrix}0&5\\-5&0\end{matrix}\right]_{2\times 2} \end{aligned}

Transpose of A is

 \begin{aligned} A^T&=\left[\begin{matrix}0&-5\\5&0\end{matrix}\right]_{2\times 2}\\ &=-\left[\begin{matrix}0&5\\-5&0\end{matrix}\right]_{2\times 2}\\ &=-A \end{aligned}

1. An example of a $4\times 4$ skew-symmetric matrix.

 \begin{aligned} B&=\left[\begin{matrix}0&2&-9&5\\-2&0&3&9\\9&-3&0&-8\\-5&-9&8&0\end{matrix}\right]\\ \Longrightarrow B^T&=\left[\begin{matrix}0&-2&9&-5\\2&0&-3&-9\\-9&3&0&8\\5&9&-8&0\end{matrix}\right]\\ &=-\left[\begin{matrix}0&2&-9&5\\-2&0&3&9\\9&-3&0&-8\\-5&-9&8&0\end{matrix}\right]\\ &=-B \end{aligned}

Symmetric and Skew-symmetric matrices has its own importance in the theory of matrices and in Liner Algebra. Let us see some of its properties.

## Properties of Symmetric Matrices and Skew-Symmetric Matrices

The scalar multiple of a symmetric matrix or a skew-symmetric matrix is a symmetric matrix or a skew-symmetric matrix, respectively. Let us put this in form of following properties $(1)$ and $(2)$.

### Property 1

If $A$ is a symmetric matrix, then $kA$, where $k\in \mathbb{R}$ is a symmetric matrix.

#### Proof

Let $A$ be a symmetric matrix. Since $A$ is a symmetric matrix, $A^T=A$. Now,

 \begin{aligned} (kA)^T&=kA^T\\ & \quad (\because \text{ by property of transpose of a matrix})\\ &=kA\\ & \quad (\because A^T=A) \end{aligned}

Hence $kA$ is a symmetric matrix.

Let us verify property 1 by an example.

#### Example

Let $A=\left[\begin{matrix}3&9\\9&-3\end{matrix}\right]$. Then

 \begin{aligned} A^T&=\left[\begin{matrix}3&9\\9&-3\end{matrix}\right]^T\\ &=\left[\begin{matrix}3&9\\9&-3\end{matrix}\right]\\ &=A. \end{aligned}

Thus $A$ is a symmetric matrix. Now we consider $5A$.

Then

 \begin{aligned} 5A&=5\left[\begin{matrix}3&9\\9&-3\end{matrix}\right]\\ &=\left[\begin{matrix}15&45\\45&-15\end{matrix}\right]. \end{aligned}

Now

 \begin{aligned} (5A)^T &=\left[\begin{matrix}15&45\\45&-15\end{matrix}\right]^T\\ &=\left[\begin{matrix}15&45\\45&-15\end{matrix}\right]\\ &=5A. \end{aligned}

Thus $5A$ is a symmetric matrix.

### Property 2

If $A$ is a skew-symmetric matrix, then $kA$, where $k\in \mathbb{R}$ is a skew-symmetric matrix.

#### Proof

Let $A$ be a skew-symmetric matrix. Since $A$ is a skew-symmetric matrix, $A^T=-A$. Now,

 \begin{aligned} (kA)^T&=kA^T\\ & \quad (\because \text{ by property of transpose of a matrix})\\ &=k(-A)\\ &\quad (\because A^T=-A)\\ &=-(kA) \end{aligned}

Hence $kA$ is a skew-symmetric matrix.

Let us verify property 2 by an example.

#### Example

Let $A=\left[\begin{matrix}0&-7\\7&0\end{matrix}\right]$. Then

 \begin{aligned} A^T&=\left[\begin{matrix}0&-7\\7&0\end{matrix}\right]^T\\ &=\left[\begin{matrix}0&7\\-7&0\end{matrix}\right]\\ &=-\left[\begin{matrix}0&-7\\7&0\end{matrix}\right]\\ &=-A. \end{aligned}

Thus $A$ is a skew-symmetric matrix. Now we consider $2A$. Then

 \begin{aligned} 2A&=2\left[\begin{matrix}0&-7\\7&0\end{matrix}\right]\\ &=\left[\begin{matrix}0&-14\\14&0\end{matrix}\right]. \end{aligned}

Now

 \begin{aligned} (2A)^T&=\left[\begin{matrix}0&-14\\14&0\end{matrix}\right]^T\\ &=\left[\begin{matrix}0&14\\-14&0\end{matrix}\right]\\ &=-\left[\begin{matrix}0&-14\\14&0\end{matrix}\right]\\ &=-(2A). \end{aligned}

Thus $2A$ is a skew-symmetric matrix.

With the help of a given square matrix $A$ we can always construct a symmetric matrix. Let us record this fact in the following property.

### Property 3

If $A$ is a square matrix, then $P=\frac{1}{2}(A+A^T)$ is always a symmetric matrix.

#### Proof

Let $A$ be a square matrix. Then $P=\dfrac{1}{2}(A+A^T)$ is also a square matrix. Now,

 \begin{aligned} P^{\;T}&=\left[\dfrac{1}{2}(A+A^T)\right]^T\\ &=\dfrac{1}{2}(A+A^T)^T\\ &\quad (\because (kA)^T=kA^T)\\ &=\dfrac{1}{2}(A^T+(A^T)^T)\\ &\quad (\because (A+B)^T=A^T+B^T )\\ &=\dfrac{1}{2}(A^T+A)\\ &\quad (\because (A^T)^T=A )\\ &=\dfrac{1}{2}(A+A^T)\\ &\quad (\because A+B=B+A )\\ &=P \end{aligned}

Hence $P$ is a symmetric matrix.

Let us verify property 3 by an example.

#### Example

If $A=\left[\begin{matrix}1&2&6\\1&-3&7\\-5&1&3\end{matrix}\right]$, then show that $P=\dfrac{1}{2}(A+A^T)$ is a symmetric matrix.

#### Solution

Here,

 \begin{aligned} A&=\left[\begin{matrix}1&2&6\\1&-3&7\\-5&1&3\end{matrix}\right]\\ \Longrightarrow A^T &=\left[\begin{matrix}1&1&-5\\2&-3&1\\6&7&3\end{matrix}\right]. \end{aligned}

So,

 \begin{aligned} P&=\dfrac{1}{2}(A+A^T)\\ &\\ &=\dfrac{1}{2}\left\{\left[\begin{matrix}1&2&6\\1&-3&7\\-5&1&3\end{matrix}\right]+\left[\begin{matrix}1&1&-5\\2&-3&1\\6&7&3\end{matrix}\right]\right\}\\ &\\ &=\dfrac{1}{2}\left[\begin{matrix}2&3&1\\3&-6&8\\1&8&6\end{matrix}\right]\\ &\\ &=\left[\begin{matrix}1&\frac{3}{2}&\frac{1}{2}\\\frac{3}{2}&-2&4\\\frac{1}{2}&4&3\end{matrix}\right]. \end{aligned}

Now

 \begin{aligned} P^T&=\left[\begin{matrix}1&\frac{3}{2}&\frac{1}{2}\\\frac{3}{2}&-2&4\\\frac{1}{2}&4&3\end{matrix}\right]^T\\ &=\left[\begin{matrix}1&\frac{3}{2}&\frac{1}{2}\\\frac{3}{2}&-2&4\\\frac{1}{2}&4&3\end{matrix}\right]\\ &=P. \end{aligned}

Thus $P$ is a symmetric matrix.

With the help of a given square matrix $A$ we can always construct a skew-symmetric matrix. Let us record this fact in the following property.

### Property 4

If $A$ is a square matrix, then $Q=\frac{1}{2}(A-A^T)$ is always a skew-symmetric matrix.

#### Proof

Let $A$ be a square matrix. Then $Q=\dfrac{1}{2}(A-A^T)$ is also a square matrix. Now,

 \begin{aligned} Q^T&=\left[\dfrac{1}{2}(A-A^T)\right]^T\\ &=\dfrac{1}{2}(A-A^T)^T&&\;\;\;(\because (kA)^T=kA^T)\\ &=\dfrac{1}{2}(A^T-(A^T)^T)&&\;\;\;(\because (A-B)^T=A^T-B^T )\\ &=\dfrac{1}{2}(A^T-A)&&\;\;\;(\because (A^T)^T=A )\\ &=-\dfrac{1}{2}(A-A^T)&&\;\;\;(\because A-B=-(B-A) )\\ &=-Q \end{aligned}

Hence $Q$ is a skew-symmetric matrix.

Let us verify property 4 by an example.

#### Example

If $A=\left[\begin{matrix}1&2&6\\1&-3&7\\-5&1&3\end{matrix}\right]$, then show that $Q=\dfrac{1}{2}(A-A^T)$ is a symmetric matrix.

#### Solution

Here,

 \begin{aligned} A&=\left[\begin{matrix}1&2&6\\1&-3&7\\-5&1&3\end{matrix}\right]\\ \Longrightarrow A^T&=\left[\begin{matrix}1&1&-5\\2&-3&1\\6&7&3\end{matrix}\right]. \end{aligned}

So,

 \begin{aligned} P&=\dfrac{1}{2}(A-A^T)\\ &\\ &=\dfrac{1}{2}\left\{\left[\begin{matrix}1&2&6\\1&-3&7\\-5&1&3\end{matrix}\right] -\left[\begin{matrix}1&1&-5\\2&-3&1\\6&7&3\end{matrix}\right]\right\}\\ &\\ &=\dfrac{1}{2} \left[\begin{matrix}0&1&11\\-1&0&6\\-11&-6&0\end{matrix}\right]\\ &\\ &=\left[\begin{matrix}0&\frac{1}{2}&\frac{11}{2}\\-\frac{1}{2}&0&3\\-\frac{11}{2}&-3&0\end{matrix}\right]. \end{aligned}

Now

 \begin{aligned} Q^T&=\left[\begin{matrix}0&\frac{1}{2}&\frac{11}{2}\\-\frac{1}{2}&0&3\\-\frac{11}{2}&-3&0\end{matrix}\right]^T\\ &=\left[\begin{matrix}0&-\frac{1}{2}&-\frac{11}{2}\\\frac{1}{2}&0&-3\\\frac{11}{2}&3&0\end{matrix}\right]\\ &=-\left[\begin{matrix}0&\frac{1}{2}&\frac{11}{2}\\-\frac{1}{2}&0&3\\-\frac{11}{2}&-3&0\end{matrix}\right]\\ &=-Q. \end{aligned}

Thus $Q$ is a skew-symmetric matrix.

If you have a square matrix then it may or may not be a symmetric matrix or a skew symmetric matrix. But it can be decompose into a symmetric matrix and a skew-symmetric matrix in the sense that; for a given square matrix we can always construct two matrices such that one of them is a symmetric matrix and the other one is a skew-symmetric matrix and its addition is the given matrix; and such two matrices are unique. This is a very important property of a square matrix associated with symmetric matrix and skew-symmetric matrix. Let us see this fact with proof in details in the form of following property.

### Property 5

Every square matrix can be expressed as a sum of a symmetric and a skew-symmetric matrix uniquely.

#### Proof

Let $A$ be a square matrix. Define $P=\dfrac{1}{2}(A+A^T)$ and $Q=\dfrac{1}{2}(A-A^T)$. Then by Properties (3) and (4), $P$ is a symmetric matrix and $Q$ is a skew-symmetric matrix. Now,

 \begin{aligned} P+Q&=\dfrac{1}{2}(A+A^T)+\dfrac{1}{2}(A-A^T)\\ &=\dfrac{1}{2}(A+A^T+A-A^T)\\ &=\dfrac{1}{2}(2A)\\ &=A \end{aligned}

Thus $A$ is a sum of $P$ and $Q$, where $P$ is a symmetric matrix and $Q$ is a skew-symmetric matrix. Hence any square matrix $A$ can be expressed as a sum of a symmetric and a skew-symmetric matrix. Now we will prove uniqueness.

Uniqueness. We have $A=P+Q$, where $P$ is a symmetric matrix and $Q$ is a skew-symmetric matrix. If possible, we also have $A=X+Y$ where $X$ is a symmetric matrix and $Y$ is a skew-symmetric matrix. Then in order to prove the uniqueness we need to prove that $P=X$ and $Q=Y$.

Since $X$ is a symmetric matrix any $Y$ is a skew-symmetric matrix, $$X^T=X\;\;\text{ and }\;\;Y^T=-Y.$$

So, $$A^T=(X+Y)^T=X^T+Y^T=X-Y.$$

Now,

 \begin{aligned} P&=\dfrac{1}{2}(A+A^T)\\ &=\dfrac{1}{2}(X+Y+X-Y)\\ &=\dfrac{1}{2}(2X)=X; \end{aligned}

and

 \begin{aligned} Q&=\dfrac{1}{2}(A-A^T)\\ &=\dfrac{1}{2}(X+Y-(X-Y))\\ &=\dfrac{1}{2}(2Y)=Y. \end{aligned}

This proves the uniqueness.

#### Example

Express the matrix $A=\left[\begin{matrix}1&5&6\\-1&-2&-4\\8&2&13\end{matrix}\right]$ as the sum of a symmetric matrix and a skew-symmetric matrix.

#### Solution

 \begin{aligned} A&=\left[\begin{matrix}1&5&6\\-1&-2&-4\\8&2&13\end{matrix}\right]\;\\ \Longrightarrow\; A^T& =\left[\begin{matrix}1&-1&8\\5&-2&2\\6&-4&13\end{matrix}\right]. \end{aligned}

Let

 \begin{aligned} P&=\dfrac{1}{2}(A+A^T)\\ &=\dfrac{1}{2}\left\{\left[\begin{matrix}1&5&6\\-1&-2&-4\\8&2&13\end{matrix}\right]+\left[\begin{matrix}1&-1&8\\5&-2&2\\6&-4&13\end{matrix}\right]\right\}\\ &=\dfrac{1}{2}\left[\begin{matrix}2&4&14\\4&-4&-2\\14&-2&26\end{matrix}\right]\\ &=\left[\begin{matrix}1&2&7\\2&-2&-1\\7&-1&13\end{matrix}\right] \end{aligned}

and

 \begin{aligned} Q&=\dfrac{1}{2}(A-A^T)\\ &=\dfrac{1}{2}\left\{\left[\begin{matrix}1&5&6\\-1&-2&-4\\8&2&13\end{matrix}\right]-\left[\begin{matrix}1&-1&8\\5&-2&2\\6&-4&13\end{matrix}\right]\right\}\\ &=\dfrac{1}{2}\left[\begin{matrix}0&6&-2\\-6&0&-6\\2&6&0\end{matrix}\right]\\ &=\left[\begin{matrix}0&3&-1\\-3&0&-3\\1&3&0\end{matrix}\right] \end{aligned}

Then

 \begin{aligned} P^T&=\left[\begin{matrix}1&2&7\\2&-2&-1\\7&-1&13\end{matrix}\right]^T\\ &=\left[\begin{matrix}1&2&7\\2&-2&-1\\7&-1&13\end{matrix}\right]\\ &=P \end{aligned}

Thus $P$ is a symmetric matrix. And

 \begin{aligned} Q^T&=\left[\begin{matrix}0&3&-1\\-3&0&-3\\1&3&0\end{matrix}\right]^T\\ &=\left[\begin{matrix}0&-3&1\\3&0&3\\-1&-3&0\end{matrix}\right]\\ &=-\left[\begin{matrix}0&3&-1\\-3&0&-3\\1&3&0\end{matrix}\right]\\ &=-Q \end{aligned}

Thus $Q$ is skew symmetric matrix. Also,

 \begin{aligned} A&=\left[\begin{matrix}1&5&6\\-1&-2&-4\\8&2&13\end{matrix}\right]\\ &=\underbrace{\left[\begin{matrix}1&2&7\\2&-2&-1\\7&-1&13\end{matrix}\right]}_{\text{ a symmetric matrix}}\;\;\;+\underbrace{\left[\begin{matrix}0&3&-1\\-3&0&-3\\1&3&0\end{matrix}\right]}_{\text{ a skew-symmetric matrix}}\\ &=P+Q \end{aligned}

If your matrix has at least one non zero diagonal entry, then that matrix can not be a skew-symmetric because all the diagonal entries of any skew-symmetric matrix must be zero. Let us prove this fact in the following property.

### Property 6

In a skew-symmetric matrix all the diagonal entries must be zero.

#### Proof

Let $A=[a_{ij}]_{n\times n}$ be a skew-symmetric matrix. In order to prove that all the diagonal entries are zero, we need to show that $a_{ii}=0$ for all $i$. Since $A=[a_{ij}]$ is a skew-symmetric matrix,

 \begin{aligned} &a_{ij}=-a_{ji}\;\;\;&&\text{ for all } i,j\\ \Rightarrow \;\;&a_{ii}=-a_{ii}\;\;\;&&\text{ in particular for } i=j\\ \Rightarrow\;\;&2a_{ii}=0\;\;\;&&\text{ for all } i\\ \Rightarrow\;\;&a_{ii}=0\;\;\;&&\text{ for all } i \end{aligned}

Thus all the diagonal entries are zero.

If $A$ is a square matrix then we can always construct a symmetric matrix with the help of matrix $A$, just you take $P=\dfrac{1}{2}(A+A^T)$, then $P$ is a symmetric matrix. But if $A$ is not a square matrix, then, can we always construct a symmetric matrix with the help of matrix $A$? The answer is YES. See the following property.

### Property 7

If $A$ is any matrix, then $A\cdot A^T$ and $A^T\cdot A$ are always symmetric matrices.

#### Proof

Let $A$ be any matrix of order $m\times n$. Then $A^T$ is a matrix of order $n\times m$. Thus $A\cdot A^T$ is defined and it is of the order $m\times m$. Similarly $A^T\cdot A$ is also defined and is of the order $n\times n$. Set $B=A\cdot A^T$ and $C=A^T\cdot A$. Now,

 \begin{aligned} B^T&=(A\cdot A^T)^T\\ &=(A^T)^T\cdot A^T\\ &=A\cdot A^T\\ &=B. \end{aligned}

 \begin{aligned} C^T&=(A^T\cdot A)^T\\ &=A^T\cdot (A^T)^T\\ &=A^T\cdot A\\ &=C. \end{aligned}

Hence $B=A\cdot A^T$ and $C=A^T\cdot A$ are symmetric matrices.

Let us verify property 7 by an example.

#### Example

Let $A=\left[\begin{matrix}3&9&1\\9&-3&5\end{matrix}\right]$. Then $A^T=\left[\begin{matrix}3&9\\9&-3\\1&5\end{matrix}\right]$.

Now

 \begin{aligned} AA^T&=\left[\begin{matrix}3&9&1\\9&-3&5\end{matrix}\right]\left[\begin{matrix}3&9\\9&-3\\1&5\end{matrix}\right]\\ &=\left[\begin{matrix}91&5\\5&115\end{matrix}\right] \end{aligned}

Clearly $AA^T$ is a symmetric matrix. Also

 \begin{aligned} A^TA &=\left[\begin{matrix}3&9\\9&-3\\1&5\end{matrix}\right]\left[\begin{matrix}3&9&1\\9&-3&5\end{matrix}\right]\\ &=\left[\begin{matrix}90&0&48\\0&90&-6\\48&-6&26\end{matrix}\right]\end{aligned}

It is easy to see that $A^TA$ is a symmetric matrix.

Following property tells about the sum and difference of two symmetric matrices.

### Property 8

Sum and difference of two symmetric matrices are symmetric matrix.

#### Proof

Let $A$ and $B$ be two symmetric matrices. Then $A^T=A$ and $B^T=B$. Let $X=A+B$ and $Y=A-B$. Then

 \begin{aligned} X^T&=(A+B)^T\\ &=A^T+B^T\\ &=A+B\\ &=X \end{aligned}

 \begin{aligned} Y^T&=(A-B)^T\\ &=A^T-B^T\\ &=A-B\\ &=Y \end{aligned}

Thus $X=A+B$ and $Y=A-B$ are symmetric matrices.

Following property tells about the sum and difference of two skew-symmetric matrices.

### Property 9

Sum and difference of two skew-symmetric matrices are skew-symmetric matrix.

#### Proof

Let $A$ and $B$ be two skew-symmetric matrices. Then $A^T=-A$ and $B^T=-B$. Let $X=A+B$ and $Y=A-B$. Then

 \begin{aligned} X^T&=(A+B)^T\\ &=A^T+B^T\\ &=-A+(-B)\\ &=-(A+B)\\ &=-X \end{aligned}

 \begin{aligned} Y^T&=(A-B)^T\\ &=A^T-B^T\\ &=-A-(-B)\\ &=-(A-B)\\ &=-Y \end{aligned}

Thus $X=A+B$ and $Y=A-B$ are skew-symmetric matrices.

Following examples shows that Property 8 and Property 9 are not true in the case of product of two matrices.

#### Examples

• Product of two symmetric matrices may not be symmetric matrix.

Let

 \begin{aligned} A&=\left[\begin{matrix}1&2\\2&1\end{matrix}\right]\;\;\text{ and }\;\;B=\left[\begin{matrix}2&3\\3&0\end{matrix}\right]. \end{aligned}

Then both $A$ and $B$ are symmetric matrices. Let $C=AB$. Then

 \begin{aligned} C&=\left[\begin{matrix}1&2\\2&1\end{matrix}\right]\left[\begin{matrix}2&3\\3&0\end{matrix}\right]\\ &=\left[\begin{matrix}8&3\\7&6\end{matrix}\right]\longleftarrow \text{ which is not a symmetric matrix. } \end{aligned}

• Product of two skew-symmetric matrices may not be symmetric matrix.

Let

 \begin{aligned} A&=\left[\begin{matrix}0&2\\-2&0\end{matrix}\right]\;\;\text{ and }\;\;B=\left[\begin{matrix}0&-3\\3&0\end{matrix}\right]. \end{aligned}

Then both $A$ and $B$ are skew-symmetric matrices. Let $C=AB$. Then

 \begin{aligned} C&=\left[\begin{matrix}0&2\\-2&0\end{matrix}\right]\left[\begin{matrix}0&-3\\3&0\end{matrix}\right]\\ &=\left[\begin{matrix}6&0\\0&6\end{matrix}\right]\longleftarrow \text{ which is not a skew-symmetric matrix. } \end{aligned}

## Summary of Properties of Symmetric Matrices

• If $A$ is a symmetric matrix, then $kA$, where $k\in \mathbb{R}$ is a symmetric matrix.
• If $A$ is a skew-symmetric matrix, then $kA$, where $k\in \mathbb{R}$ is a skew-symmetric matrix.
• If $A$ is a square matrix, then $P=\dfrac{1}{2}(A+A^T)$ is always a symmetric matrix.
• If $A$ is a square matrix, then $Q=\dfrac{1}{2}(A-A^T)$ is always a skew-symmetric matrix.
• Every square matrix can be expressed as a sum of a symmetric and a skew-symmetric matrix uniquely.
• In a skew-symmetric matrix all the diagonal entries must be zero.
• If $A$ is any matrix, then $A\cdot A^T$ and $A^T\cdot A$ are always symmetric matrices.
• Sum and difference of two symmetric matrices are symmetric matrix.
• Sum and difference of two skew-symmetric matrices are skew-symmetric matrix.