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.

Milan holds Ph.D. Degree in Mathematics, his area of research is Banach Algebra and Harmonic Analysis, a branch of pure mathematics. He has cleared NET and GPSC (Gujarat Public Service Commission) with first rank. Milan enjoys teaching and problem solving in Mathematics. He has good hand on Real Analysis, Complex Analysis, Linear Algebra, Abstract Algebra and Differential Equation. Apart from teaching Milan spends his time to guide the students for various competitive examinations in Mathematics like JAM, CSIR NET-JRF, GSET, GATE, NBHM, TIFR etc. More than 29 students had cleared such type of examinations under his guidance. He has teaching experience of more than 13 years in the field of Mathematics at graduate level in Engineering and Science College.

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