Hermitian matrix & Skew Hermitian Matrix

A matrix (plural: matrices) is a rectangular arrangement of symbols along rows and columns that might be real or complex numbers. Thus, a system of m x n symbols arranged in a rectangular formation along m rows and n columns is called an m by n matrix (which is written as m x n matrix). In real life, we use the Hermitian matrix and the Skew Hermitian Matrix in quantum mechanics and signal processing.

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1. Hermitian matrix
2. Skew-hermitian matrix
3. Properties of hermitian and skew-hermitian matrices
4. Solved Examples Based on Hermitian and skew-Hermitian matrices

In this article, we will cover the concept Hermitian matrix and the Skew Hermitian Matrix. This category falls under the broader category of Matrices, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of five questions have been asked on this topic in JEE MAINS(2013 to 2023).

Hermitian matrix

A square matrix $\mathrm{A}={\left[{\mathrm{a}}_{\mathrm{ij}}\right]}_{\mathrm{n}\times \mathrm{n}}$ is said to be a Hermitian matrix if $a_{ij}={\bar{a}}_{ji}\mathrm{\forall }\mathrm{i},\mathrm{j}$,
i.e. $A=A^{\theta },[$ where $A^{\theta }$ is conjugate transpose of matrix $A]$

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, ⇒ all diagonal element must be purely real,

E.g.

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

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

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

Note :

For any square matrix say A, with complex number entries,

$A+A^{\theta }$

Skew-hermitian matrix

$\mathrm{A}+{\mathrm{A}}^{\theta }$ is a Hermitian matrix

$[\because {(\mathrm{A}+{\mathrm{A}}^{\theta })}^{\theta }={\mathrm{A}}^{\theta }+{\left({\mathrm{A}}^{\theta }\right)}^{\theta }={\mathrm{A}}^{\theta }+\mathrm{A}]$
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

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

Hence all diagonal elements should be purely imaginary

E.g

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

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

here, $A$ is Skew - Hermitian matrix as $A^{\theta }=-A$

Important Note:

1. for any square matrix A with elements containing complex numbers, then A-A' is a skew hermitian matrix.

Proof : (A-A')'= A' - (A')'= A' - A = -(A-A'), hence skew-hermitian.

2. Every square matrix can be written as the sum of hermitian and skew-hermitian matrices i.e.

If A is a square matrix, then we can write $A=\frac{1}{2}(A+A^{\theta })+\frac{1}{2}(A-A^{\theta })$

Properties of hermitian and skew-hermitian matrices

i) If A is a square matrix then AA' and A' A are hermitian matrices.

Proof: for the hermitian matrix A' = A, so we check the condition on AA'

(AA')' = (A')'A' = AA' hence it is hermitian, and in the same way, A'A will also be hermitian.

ii) If A is a hermitian matrix then:

A is a skew hermitian matrix, where i = √-1

Proof: we need to show (i A)'= -iA

(i A)'= A' i'= A' (-i) = -i A'

Since A is hermitian A' = A

Hence we have

-i A' = -i A. Proved.

iii) if A is a skew-hermitian matrix, then:

i A is a hermitian matrix, where i = √-1

Proof: we need to show (i A)' = iA

(i A)' = A' i'= A'(-i)

A'(-i) = Ai = iA (since A is skew-hermitian, so A' = -A)

iv) if A and B are hermitian matrices of the same order, then

a. cA and dB are also hermitian matrices of the same order when c and d are scalar real constants.

Since A and B are of the same order, hence they are conformable for addition and by multiplying through a scalar we are just magnifying their values and nothing else, hence they will hold their property of hermitian matrices and cA + dB will be a hermitian matrix.

b. AB is also hermitian if AB = BA

Proof: (AB)' = B'A'= BA = AB (Since A, B are hermitian so A' = A, B' =B)

c. AB + BA will also be Hermitian

Proof: from part (b) AB and BA are hermitian and from part (c) AB + BA will also be hermitian.

d. AB - BA will be skew-hermitian

Proof: we need to show (AB-BA)* = -(AB-BA)

(AB-BA)* = (AB)* - (BA)* = B*A* - A*B* = BA - AB = -(AB - BA)

Using A' = A and B' = B, proved.

v) if A and B are skew hermitian matrix then cA +dB will be skew-hermitian

Proofs are similar to those above, just verify the basic condition, using the given conditions of A and B.

Recommended Video Based on Hermitian and Skew-Hermitian Matrices

Solved Examples Based on Hermitian and skew-Hermitian matrices

Example 1: Which of the following matrices is Hermitian?

1) $A=\left(\begin{array}{ll}1& 2\\ 2& 3\end{array}\right)$
2) $B=\left(\begin{array}{ll}1& 2\\ 3& 4\end{array}\right)$
3) $C=\left(\begin{array}{cc}1& 2+i\\ 2-i& 3\end{array}\right)$
4) Both $A$ and $C$

Solution:
Matrix $A$ is symmetric and has real entries, thus it is Hermitian. Matrix $C$ is Hermitian because it equals its conjugate transpose.

Hence, the answer is option 4.

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

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

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

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

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

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
$A^{\prime }=\left[\begin{array}{ccc}3& 4+2i& 5-3i\\ 4-2i& 4& 4-5i\\ 5+3i& 4+5i& 5\end{array}\right]$
taking its conjugate now
$\overline{{\mathrm{A}}^{\prime }}={\mathrm{A}}^{\theta }=\left[\begin{array}{ccc}3& 4-2i& 5+3i\\ 4+2i& 4& 4+5i\\ 5-3i& 4-5i& 5\end{array}\right]=\mathrm{A}$

Hence, the answer is the option 2.

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

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

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

Hence, the answer is the option 1.

Example 4: If $A=\left[\begin{array}{cc}2+i& 3i\\ -3i& x\end{array}\right]$ 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, $iA$ is a skew hermitian matrix, where $i=\sqrt{-1}$
2) If $A$ is a skew-hermitian matrix then $iA$ is a hermitian matrix, where $i=\sqrt{-1}$
3) If $A$ and $B$ are hermitian matrices of the same order, then $AB-BA$ will be skew-hermitian.
4) All of the above

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

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

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

Now taking conjugate
$\begin{array}{rl}& A^{\theta }=\left[\begin{array}{cc}a& ic\\ -ic& d\end{array}\right]B^{\theta }=\left[\begin{array}{cc}e& ig\\ -ig& h\end{array}\right]\\ & A\cdot B=\left[\begin{array}{cc}a& ic\\ -ic& d\end{array}\right]\cdot \left[\begin{array}{cc}e& ig\\ -ig& h\end{array}\right]\\ & A\cdot B=\left[\begin{array}{cc}ae+cg& iag+ich\\ -iec-idg& gc+dh\end{array}\right]\\ & A\cdot B=\left[\begin{array}{cc}ae+cg& i(ag+ch)\\ -i(ec+dg)& gc+dh\end{array}\right]\\ & B\cdot A=\left[\begin{array}{cc}e& ig\\ -ig& h\end{array}\right]\cdot \left[\begin{array}{cc}a& ic\\ -ic& d\end{array}\right]\\ & B\cdot A=\left[\begin{array}{cc}ae+gc& iec+igd\\ -iag-ihc& cg+dh\end{array}\right]\end{array}$

$A\cdot B-B\cdot A=\left[\begin{array}{cc}0& i(ag+ch+ec+gd)\\ -i(ec+dg+ag+hc)& 0\end{array}\right]\text{this matrix is a skew hermitian matrix}$
Therefore, statement (3) is also correct
Hence, the answer is the option 4.