Hermitian matrix & Skew Hermitian Matrix

What is a Hermitian Matrix?

A Hermitian matrix is a complex square matrix that is equal to its own conjugate transpose. These matrices are widely used in linear algebra, quantum mechanics, and complex analysis. A key property is that the diagonal elements of a Hermitian matrix are always real. This section explains the definition, properties, and provides a Hermitian matrix example with detailed verification.

Definition of Hermitian Matrix

A matrix $A = [a_{ij}]$ is called a Hermitian matrix if:

$A = A^H$

where $A^H$ denotes the conjugate transpose of $A$. This condition means that:

$a_{ij} = \overline{a_{ji}}$

for all $i$ and $j$, where $\overline{a_{ji}}$ is the complex conjugate of the element at position $(j, i)$.

Properties of Hermitian Matrices

• A Hermitian matrix is always a square matrix, i.e., it has the same number of rows and columns.

• The diagonal elements of a Hermitian matrix are always real, i.e., $a_{ii} \in \mathbb{R}$.

• Off-diagonal elements satisfy $a_{ij} = \overline{a_{ji}}$.

• All eigenvalues of a Hermitian matrix are real.

• Hermitian matrices are diagonalizable by a unitary matrix.

• Hermitian matrices play a key role in representing real-valued observables in physics.

Diagonal Elements of a Hermitian Matrix

The condition $a_{ii} = \overline{a_{ii}}$ must hold for all diagonal elements. This is only possible if $a_{ii}$ is a real number. Therefore:

$\text{If } A \text{ is Hermitian, then } a_{ii} \in \mathbb{R}$

This is one of the most recognisable properties of Hermitian matrices.

Hermitian Matrix Example with Solution

Consider the matrix:

$A = \begin{bmatrix} 4 & 2 + i \\ 2 - i & 5 \end{bmatrix}$

Step 1: Find the transpose of $A$:

$A^T = \begin{bmatrix} 4 & 2 - i \\ 2 + i & 5 \end{bmatrix}$

Step 2: Take the complex conjugate of $A^T$ to get the conjugate transpose $A^H$:

$A^H = \begin{bmatrix} 4 & 2 - i \\ 2 + i & 5 \end{bmatrix}$

Step 3: Compare $A$ and $A^H$:

$A = A^H$

Therefore, $A$ is a Hermitian matrix.

We know that when we take the transpose of a matrix, its diagonal elements remain the same, and while taking conjugate we just change the sign from +ve to -ve and -ve to +ve for the imaginary part of all elements, So to satisfy the condition A' = A diagonal elements must not change, implies all diagonal element must be purely real,

e.g. Let, $A=\left[\begin{array}{ccc}3 & 3-4 i & 5+2 i \\ 3+4 i & 5 & -2+i \\ 5-2 i & -2-i & 7\end{array}\right]$
Then,

$
\begin{gathered}
\mathrm{A}^{\prime}=\left[\begin{array}{ccc}
3 & 3+4 i & 5-2 i \\
3-4 i & 5 & -2-i \\
5+2 i & -2+i & 7
\end{array}\right] \\
\therefore \mathrm{A}^H=\overline{\left(\mathrm{A}^{\prime}\right)}=\left[\begin{array}{ccc}
3 & 3-4 i & 5+2 i \\
3+4 i & 5 & -2+i \\
5-2 i & -2-i & 7
\end{array}\right]
\end{gathered}
$

here, A is Hermitian matrix as $\mathrm{A}=\mathrm{A}^H$

What is a Skew Hermitian Matrix?

A skew Hermitian matrix is a complex square matrix that satisfies a specific symmetry: it is equal to the negative of its conjugate transpose. These matrices are significant in advanced algebra, especially in fields like quantum mechanics and electrical engineering. A defining feature is that the diagonal elements of a skew Hermitian matrix are always purely imaginary or zero.

Definition of Skew Hermitian Matrix

A matrix $A$ is called a skew Hermitian matrix if:

$A = -A^H$

where $A^H$ is the conjugate transpose of $A$. This means that each element satisfies:

$a_{ij} = -\overline{a_{ji}}$

for all $i$ and $j$, where $\overline{a_{ji}}$ denotes the complex conjugate of the element in the $j^{\text{th}}$ row and $i^{\text{th}}$ column.

Properties of Skew Hermitian Matrices

• $A$ is always a square matrix.

• Diagonal elements satisfy: $a_{ii} = -\overline{a_{ii}} \Rightarrow a_{ii} \in \mathbb{C} \text{ such that } \text{Re}(a_{ii}) = 0$

So, each diagonal element is purely imaginary or zero.

• Off-diagonal elements satisfy: $a_{ij} = -\overline{a_{ji}}$

• All eigenvalues of a skew Hermitian matrix are either purely imaginary or zero.

• If $A$ is skew Hermitian, then the matrix $iA$ is Hermitian, where $i = \sqrt{-1}$.

Diagonal Elements of Skew Hermitian Matrix

For diagonal elements $a_{ii}$ in a skew Hermitian matrix:

$a_{ii} = -\overline{a_{ii}}$

This implies:

$a_{ii} + \overline{a_{ii}} = 0 \Rightarrow 2\text{Re}(a_{ii}) = 0$

Therefore, the real part of $a_{ii}$ must be zero, meaning:

$a_{ii} \in \{ bi \mid b \in \mathbb{R} \}$

So, diagonal elements are always purely imaginary or zero.

Skew Hermitian Matrix Example with Solution

Let us check whether the following matrix is skew Hermitian:

$A = \begin{bmatrix} 0 & 2+i \\ -2+i & 0 \end{bmatrix}$

Step 1: Transpose $A$:

$A^T = \begin{bmatrix} 0 & -2+i \\ 2+i & 0 \end{bmatrix}$

Step 2: Take complex conjugate:

$A^H = \begin{bmatrix} 0 & -2-i \\ 2-i & 0 \end{bmatrix}$

Step 3: Find $-A^H$:

$A^H = \begin{bmatrix} 0 & 2+i \\ -2+i & 0 \end{bmatrix}$

Since:

$A = -A^H$

The matrix is skew-Hermitian.

We know that when we take the transpose of a matrix, its diagonal elements remain the same, and while taking conjugate we just change the sign from +ve to -ve OR -ve to +ve in the imaginary part of all elements, So to satisfy the condition A? = - A, all diagonal element must be purely imaginary. As A' = - A so

$\begin{aligned} & \mathrm{a}_{\mathrm{ij}}=-\overline{\mathrm{a}_{\mathrm{ij}}} \forall \mathrm{i}, \mathrm{j} \\ & \text { if we put } \mathrm{i}=\mathrm{j} \text {, we have } \\ & \mathrm{a}_{\mathrm{ii}}=-\overline{\mathrm{a}_{\mathrm{ii}}} \Rightarrow \mathrm{a}_{\mathrm{ii}}+\overline{\mathrm{a}_{\mathrm{ii}}}=0 \\ & \Rightarrow \mathrm{a}_{\mathrm{ii}}=0\end{aligned}$

Hence all diagonal elements should be purely imaginary

e.g. Let, $\mathrm{A}=\left[\begin{array}{ccc}-3 i & -3-4 i & -5+2 i \\ 3-4 i & 5 i & i \\ 5+2 i & i & 0\end{array}\right]$
Then,

$
\begin{aligned}
\mathrm{A}^{\prime} & =\left[\begin{array}{ccc}
-3 i & 3-4 i & 5+2 i \\
-3-4 i & 5 i & i \\
-5+2 i & i & 0
\end{array}\right] \\
\therefore \mathrm{A}^\theta & =\overline{\left(\mathrm{A}^{\prime}\right)}=\left[\begin{array}{ccc}
3 i & 3+4 i & 5-2 i \\
-3+4 i & -5 i & -i \\
-5-2 i & -i & 0
\end{array}\right] \\
\quad & =-\left[\begin{array}{ccc}
-3 i & -3-4 i & -5+2 i \\
3-4 i & 5 i & i \\
5+2 i & i & 0
\end{array}\right]=-\mathrm{A}
\end{aligned}
$

Here, $A$ is Skew-Hermitian matrix as $A^\theta=-A$

Important Note:

1. For any square matrix $A$ with complex elements, the matrix $A - A^H$ is always a skew Hermitian matrix.

Proof:

Let us compute the conjugate transpose of $ A-A^H$:

$(A - A^H)^H = A^H - (A^H)^H = A^H - A = -(A - A^H)$

Hence, $A - A^H$ is skew Hermitian because:

$(A - A^H)^H = -(A - A^H)$

2. Every square matrix can be expressed as the sum of a Hermitian and a skew Hermitian matrix.

That is, if $A$ is any square matrix, then:

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

Here,

• $\frac{1}{2}(A + A^H)$ is a Hermitian matrix

• $\frac{1}{2}(A - A^H)$ is a skew Hermitian matrix

Thus, any complex square matrix can be decomposed into a Hermitian part and a skew Hermitian part.

Proofs of Properties of Hermitian and Skew-Hermitian Matrices

Understand the logical foundations behind key properties of Hermitian and Skew-Hermitian matrices. This section includes step-by-step proofs using transpose and conjugate rules, helping reinforce conceptual clarity.

i) If $A$ is a square matrix, then both $A A^H$ and $A^H A$ are Hermitian matrices.

Proof:

We check the conjugate transpose of $A A^H$:

$(A A^H)^H = (A^H)^H A^H = A A^H$

Therefore, $A A^H$ is Hermitian. Similarly,

$(A^H A)^H = (A)^H (A^H)^H = A^H A$

Hence, $A^H A$ is also Hermitian.

ii) If $A$ is a Hermitian matrix, then $i A$ is a skew Hermitian matrix, where $i = \sqrt{-1}$.

Proof:

We need to show that:

$(i A)^H = -i A$

Now,

$(i A)^H = A^H \cdot \overline{i} = A \cdot (-i) = -i A$

Since $A$ is Hermitian, $A^H = A$. Hence, $ iA$ is skew Hermitian.

iii) If $A$ is a skew Hermitian matrix, then $i A$ is a Hermitian matrix, where $i = \sqrt{-1}$.

Proof:

We need to show that:

$(i A)^H = i A$

Now, $(i A)^H = A^H \cdot \overline{i} = (-A) \cdot (-i) = i A$

Since $A$ is skew Hermitian, $A^H = -A$. Hence, $i A$ is Hermitian.

iv) If $A$ and $B$ are Hermitian matrices of the same order, then:

(a) $c A$ and $d B$ are also Hermitian, where $c, d \in \mathbb{R}$

Explanation:

If $A^H = A$ and $B^H = B$, then:

$(c A)^H = \overline{c} A^H = c A$

since $c$ is real, $\overline{c} = c$. Hence, $c A$ is Hermitian. Similarly, $d B$ is Hermitian.

Therefore,

$(c A + d B)^H = (c A)^H + (d B)^H = c A + d B$

So, $c A + d B$ is Hermitian.

(b) $A B$ is Hermitian if $A B = B A$

Proof:

Check:

$(A B)^H = B^H A^H = B A = A B$

Since $A$ and $B$ are Hermitian ($A^H = A$, $B^H = B$), and $A B = B A$, the product $A B$ is Hermitian.

(c) $A B + B A$ is Hermitian

Explanation:

From part (b), if $A B = B A$, then both $A B$ and $B A$ are Hermitian. Their sum is:

$(A B + B A)^H = (A B)^H + (B A)^H = A B + B A$

Hence, $A B + B A$ is Hermitian.

(d) $A B - B A$ is skew Hermitian

Proof:

We compute:

$(A B - B A)^H = (A B)^H - (B A)^H = B^H A^H - A^H B^H = B A - A B = - (A B - B A)$

Therefore, $A B - B A$ is skew Hermitian.

v) If $A$ and $B$ are skew Hermitian matrices of the same order, then $c A + d B$ is also skew Hermitian, where $c, d \in \mathbb{R}$

Proof:

Since $A^H = -A$ and $B^H = -B$, we have:

$(c A + d B)^H = \overline{c} A^H + \overline{d} B^H = c (-A) + d (-B) = - (c A + d B)$

Hence, $ cA + dB$ is skew Hermitian.

Difference Between Hermitian and Skew Hermitian Matrix

Solved Examples Based on Hermitian and Skew-Hermitian Matrices

Example 1: If $A$ is Hermitian such that $A^2=0$, Then

1) $A=I$

2) $\mathrm{A}=0$

3) $A^3=A$

4) none of these

Solution: Hermitian matrices –

$A^\theta = A \ \text{where } A^\theta \ \text{is the complex conjugate transpose of } A$

$A = \left[\begin{array}{cc} a & b+ic \\ b-ic & d \end{array}\right]$

$B = \left[\begin{array}{ccc} 3 & 3-4i & 5+2i \\ 3+4i & 5 & -2+i \\ 5-2i & -2-i & 2 \end{array}\right]$

$\mathrm{A} = \left[\begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \ldots & \ldots & \ldots & \ldots \\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{array}\right]$

$= \left[\begin{array}{cccc} \bar{a}_{11} & \bar{a}_{21} & \ldots & \bar{a}_{n1} \\ \bar{a}_{12} & \bar{a}_{22} & \ldots & \bar{a}_{n2} \\ \ldots & \ldots & \ldots & \ldots \\ \bar{a}_{1n} & \bar{a}_{2n} & \ldots & \bar{a}_{nn} \end{array}\right]$

Since $\mathrm{A}^2=0$, each element of $A A^\theta$ is zero.

$\Rightarrow \mathrm{a}_{\mathrm{i} 1} \bar{a}_{i 1}+\mathrm{a}_{\mathrm{i} 2} \bar{a}_{i 2^{+}} \ldots \ldots+\mathrm{a}_{\mathrm{in}} \bar{a}_{i n}=\left|\mathrm{a}_{\mathrm{i} 1}\right|^2+\left|\mathrm{a}_{\mathrm{i} 2}\right|^2+\ldots .+\left|\mathrm{a}_{\mathrm{in}}\right|^2=0$

$\left|a_{i 1}\right|=\left|a_{i 2}\right|=\ldots .=\left|a_{i n}\right|=0 \Rightarrow a_{i 1}=a_{i 2}=\ldots .=a_{i n}=0 \text {. Hence } A=0 \text {. }$

Hence, the correct answer is option 2.

Example 2: Find the Hermitian matrix of the matrix $A=\left[\begin{array}{ccc}3 & 4-2 i & 5+3 i \\ 4+2 i & 4 & 4+5 i \\ 5-3 i & 4-5 i & 5\end{array}\right]$

1) $A^\theta=\left[\begin{array}{ccc}3 i & -2 i & 5-3 i \\ 4+2 i & 4 & 4+5 i \\ 5+3 i & 4-5 i & 5 i\end{array}\right]$

2) $A^\theta=\left[\begin{array}{ccc}3 & 4-2 i & 5+3 i \\ 4+2 i & 4 & 4+5 i \\ 5-3 i & 4-5 i & 5\end{array}\right]$

3) $A^\theta=\left[\begin{array}{ccc}3 & 4-2 i & 5+3 i \\ 4+2 i & 4 & 4-5 i \\ 5-3 i & 4+5 i & 5\end{array}\right]$

4) $A^\theta=\left[\begin{array}{ccc}3 & 4-2 i & 5-3 i \\ 4+2 i & 4 & 4+5 i \\ 5+3 i & 4-5 i & 5\end{array}\right]$

Solution: For the matrix to be Hermitian, $A^\theta=A$
So we find $A^\theta$ and verify that it is equal to A or not
To find $A^\theta$, we first take the transpose of A and then its conjugate
So, taking the transpose of A, we have
$
A^{\prime}=\left[\begin{array}{ccc}
3 & 4+2 i & 5-3 i \\
4-2 i & 4 & 4-5 i \\
5+3 i & 4+5 i & 5
\end{array}\right]
$
Taking its conjugate now
$
\overline{\mathrm{A}^{\prime}}=\mathrm{A}^\theta=\left[\begin{array}{ccc}
3 & 4-2 i & 5+3 i \\
4+2 i & 4 & 4+5 i \\
5-3 i & 4-5 i & 5
\end{array}\right]=\mathrm{A}
$

Hence, the answer is option 2.

Example 3: Find the skew-hermitian matrix of matrix $\left[\begin{array}{ccc}i & 1-i & 2 \\ -1-i & 3 i & i \\ -2 & i & 0\end{array}\right]$.

1) $\left[\begin{array}{ccc}-i & -1+i & -2 \\ 1+i & -3 i & -i \\ 2 & -i & 0\end{array}\right]$
2) $\left[\begin{array}{ccc}i & 1-i & 2 \\ -1-i & 3 i & i \\ -2 & i & 0\end{array}\right]$
3) $\left[\begin{array}{ccc}i & -1+i & -2 \\ 1+i & 3 i & -i \\ 2 & -i & 0\end{array}\right]$
4) $\left[\begin{array}{ccc}-i & -1+i & 2 \\ 1+i & 3 i & -i \\ -2 & -i & 0\end{array}\right]$

Solution: First, we take the transpose and then it's conjugate and equate it to -A.
$
\mathrm{A}^{\prime}=\left[\begin{array}{ccc}
i & -1-i & -2 \\
1-i & 3 i & i \\
2 & i & 0
\end{array}\right]
$
now taking conjugate of the transpose
$
\overline{\mathrm{A}^{\prime}}=\left[\begin{array}{ccc}
-i & -1+i & -2 \\
1+i & -3 i & -i \\
2 & -i & 0
\end{array}\right]=-\mathrm{A}
$

Hence, the answer is the option 1.

Example 4: If $A=\left[\begin{array}{cc}2+i & 3 i \\ -3 i & x\end{array}\right]$ is a skew-Hermitian matrix, then find the value of x :
1) $2-\mathrm{i}$
2) $2+\mathrm{i}$
3) 0
4) All of these

Solution: we know that Skew hermitian matrices - $A^{\theta}=-A$

$A^{\theta}=-A$ is a complex conjugate transpose matrix of matrix $A$

since there is only restriction on the elements such as $a_{12}$ and $a_{21}$ not $a_{11}$ and $a_{22}$.

Hence, the answer is the option 4.

Example 5: Which of the following statements is true?
1) If $A$ is a hermitian matrix then, $i A$ is a skew hermitian matrix, where $i=\sqrt{-1}$
2) If $A$ is a skew-hermitian matrix then $i A$ is a hermitian matrix, where $i=\sqrt{-1}$
3) If $A$ and $B$ are hermitian matrices of the same order, then $A B-B A$ will be skew-hermitian.
4) All of the above

Solution: Let A be a matrix of order $2 \times 2$
$
A=\left[\begin{array}{cc}
a & b+i c \\
b-i c & d
\end{array}\right]
$
then $i A=\left[\begin{array}{cc}a i & b i-c \\ b i+c & d i\end{array}\right]=\left[\begin{array}{cc}a i & -c+b i \\ c+b i & d i\end{array}\right]$ which is a skew hermitian matrix.
$
\begin{aligned}
& \text{now,} \space i A=\left[\begin{array}{cc}
a i & b i-c \\
b i+c & d i
\end{array}\right]=\left[\begin{array}{cc}
a i & -c+b i \\
c+b i & d i
\end{array}\right] \\
& i^2 A=\left[\begin{array}{cc}
a i^2 & b i^2-c i \\
b i^2+c i & d i^2
\end{array}\right]=\left[\begin{array}{cc}
-a & -b-c i \\
-b+c i & -d
\end{array}\right]=-A \quad \text { which is a hermitian matrix.}
\end{aligned}
$

Thus, options 1 and 2 are true.
Let $A$ and $B$ be a hermitian matrix of order $2 \times 2$
$
A=\left[\begin{array}{cc}
a & i c \\
-i c & d
\end{array}\right] \quad B=\left[\begin{array}{cc}
e & i g \\
-i g & h
\end{array}\right]
$

Now taking transpose
$
A^{\prime}=\left[\begin{array}{cc}
a & -i c \\
i c & d
\end{array}\right] \quad B^{\prime}=\left[\begin{array}{cc}
e & -i g \\
i g & h
\end{array}\right]
$

Now taking conjugate
$
\begin{aligned}
& A^\theta=\left[\begin{array}{cc}
a & i c \\
-i c & d
\end{array}\right] \quad B^\theta=\left[\begin{array}{cc}
e & i g \\
-i g & h
\end{array}\right] \\
& A \cdot B=\left[\begin{array}{cc}
a & i c \\
-i c & d
\end{array}\right] \cdot\left[\begin{array}{cc}
e & i g \\
-i g & h
\end{array}\right] \\
& A \cdot B=\left[\begin{array}{cc}
a e+c g & i a g+i c h \\
-i e c-i d g & g c+d h
\end{array}\right] \\
& A \cdot B=\left[\begin{array}{cc}
a e+c g & i(a g+c h) \\
-i(e c+d g) & g c+d h
\end{array}\right] \\
& B \cdot A=\left[\begin{array}{cc}
e & i g \\
-i g & h
\end{array}\right] \cdot\left[\begin{array}{cc}
a & i c \\
-i c & d
\end{array}\right] \\
& B \cdot A=\left[\begin{array}{cc}
a e+g c & i e c+i g d \\
-i a g-i h c & c g+d h
\end{array}\right]
\end{aligned}
$

$
A \cdot B-B \cdot A=\left[\begin{array}{cc}
0 & i(a g+c h+e c+g d) \\
-i(e c+d g+a g+h c) & 0
\end{array}\right] \space {\text {this matrix is a skew hermitian matrix}}
$
Therefore, statement (3) is also correct
Hence, the answer is option 4.

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NCERT Resources

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Practice Questions based on Hermitian and Skew Hermitian Matrices

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