Conjugate of a Matrix

The conjugate transpose, also known as the Hermitian transpose, of a m x n complex matrix A is a n x m matrix obtained by transposing A and applying complex conjugation to each entry (the complex conjugate of a+ bi being a-bi, for real numbers a and b ). In real life, we can use the conjugate of a matrix in computing the quantum mechanical adjoint of an operator. In quantum systems, the Hermitian conjugate of a matrix is determined to represent observable quantities. In signal processing, the inverse Fourier transform is computed using the conjugate transpose of a matrix.

This Story also Contains

1. Conjugate
2. Conjugate of a matrix
3. The transpose conjugate of a matrix
4. Solved Examples Based on Conjugate Matrix

In this article, we will cover the concept of Conjugate of Matrix. This category falls under the broader category of Matrices, which is a crucial Chapter in class 11 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. Over the last ten years of the JEE Main Exam (from 2013 to 2023), a total of twelve questions has been asked on this concept, including one in 2021.

Conjugate

A pair of binomials that have the same terms but separate arithmetical operators in the center are called conjugates. Below are some further instances of conjugate pairs:

2+3i , 2-3i

P+qi , P-qi

Conjugate of a matrix

If a matrix A has complex numbers as its elements, then the matrix obtained by replacing those complex numbers with their conjugates is called the conjugate of the matrix A and it is denoted by $\overline{A}$. (If the element of a matrix is a + ib, then it is replaced by a - ib .)

Example, $ \begin{aligned} \mathrm{A} & =\left[\begin{array}{ccc}2 i & 3+4 i & 7 \\ 3 i & 9 & 4+5 i \\ 4+5 i & 4 i & 3+7 i\end{array}\right] \text { then, } \\ \overline{\mathrm{A}} & =\left[\begin{array}{ccc}-2 i & 3-4 i & 7 \\ -3 i & 9 & 4-5 i \\ 4-5 i & -4 i & 3-7 i\end{array}\right]\end{aligned} $

The transpose conjugate of a matrix

The transpose of a conjugate matrix A is called the transposed conjugate of A and is denoted by A^’. The conjugate of the transpose of A is the same as the transpose of the conjugate of A

$\begin{aligned} & \text { i.e. } \mathrm{A}^\theta=(\overline{\mathrm{A}})^{\prime}=\overline{\left(\mathrm{A}^{\prime}\right)} \\ & \mathrm{A}=\left[\begin{array}{ccc}1+2 i & 3 i & 5+4 i \\ 2 i-1 & 1-i & 0 \\ 3+i & 1+i & 12\end{array}\right] \\ & \overline{\mathrm{A}}=\left[\begin{array}{ccc}1-2 i & -3 i & 5-4 i \\ -2 i-1 & 1+i & 0 \\ 3-i & 1-i & 12\end{array}\right] \\ & (\overline{\mathrm{A}})^{\prime}=\left[\begin{array}{ccc}1-2 i & -2 i-1 & 3-i \\ -3 i & 1+i & 1-i \\ 5-4 i & 0 & 12\end{array}\right]\end{aligned}$

Properties of the transpose conjugate matrix:

If A and B are two matrices of the same order then

i) conjugate of a conjugate of matrix is the same as the original matrix itself,

In mathematical language (A^T)^T= A, which is quite obvious as we are reversing back to the things that we did while taking conjugate the first time.

ii) (A + B)^T = A^T + B^T, this is obvious if a matrix is conformable, as addition is done element-wise.

Iii )(kA)^T = kA^T, since multiplication in a matrix is elementwise, hence this is also obvious, as all elements multiplied before conjugate will be amplified in the way as after taking conjugate.

iv) (AB)^T = B^TA^T, here A and B should be conformable for matrix multiplication.

Recommend Video Based on Trace of a Matrix and Properties:

Solved Examples Based on Conjugate Matrix

Example 1: Find the conjugate of matrix $
\mathrm{A}=\left[\begin{array}{ccc}
2 i & 3+4 i & 7 \\
3+4 i & 9 & 4+5 i \\
7 & 4+5 i & 3+7 i
\end{array}\right]
$

Solution:
$
\begin{aligned}
& \mathrm{A}=\left[\begin{array}{ccc}
2 i & 3+4 i & 7 \\
3+4 i & 9 & 4+5 i \\
7 & 4+5 i & 3+7 i
\end{array}\right] \text { then, } \\
& \overline{\mathrm{A}}=\left[\begin{array}{ccc}
-2 i & 3-4 i & 7 \\
3-4 i & 9 & 4+5 i \\
7 & 4-5 i & 3-7 i
\end{array}\right]
\end{aligned}
$

Hence, the required answer is $\left[\begin{array}{ccc}-2 i & 3-4 i & 7 \\ 3-4 i & 9 & 4+5 i \\ 7 & 4-5 i & 3-7 i\end{array}\right]$

Example 2: If matrix $B=\bar{A}, C=\bar{B}, D=\bar{C}$, so on then $\bar{G}=$
Solution
Property of Conjugate - $\overline{(\bar{A})}=A$
Conjugate of a matrix $A$ is denoted by $\bar{A}$
$
\begin{aligned}
& B=\bar{A} \\
& C=\bar{B}=(\overline{\bar{A}})=A
\end{aligned}
$

Similarly $G=A$
and $\bar{G}=\bar{A}=B$
$\bar{G}=\mathrm{B}$

Hence, the required answer is B.

Example 3: If $A_{2 \times 2}$ is a matrix such that $a_{i j}=(\omega)^{i+j}$ where $\omega$ is the cube root of unity then $B=A+\bar{A}$. Find $\bar{B}$.
Solution:
Property of Conjugate
$
\overline{A+B}=\bar{A}+\bar{B}
$

The conjugate of matrix $A$ is $\bar{A}$
$
\begin{aligned}
& A=\left[\begin{array}{ll}
\omega^{(1+1)} & \omega^{(1+2)} \\
\omega^{(2+1)} & \omega^{(2+2)}
\end{array}\right] \\
& A=\left[\begin{array}{ll}
\omega^2 & \omega^3 \\
\omega^3 & \omega^4
\end{array}\right] \\
& A=\left[\begin{array}{cc}
\omega^2 & 1 \\
1 & \omega
\end{array}\right]
\end{aligned}
$

$\omega$ is the conjugate of $\omega^2$
Hence,
$
\begin{aligned}
& \bar{A}=\left[\begin{array}{cc}
\omega & 1 \\
1 & \omega^2
\end{array}\right] \\
& \text { Thus } A+\bar{A}=\left[\begin{array}{cc}
\omega^2+\omega & 2 \\
2 & \omega+\omega^2
\end{array}\right] \\
& =\left[\begin{array}{cc}
-1 & 2 \\
2 & -1
\end{array}\right] \\
&
\end{aligned}
$

Therefore
$
B=\left[\begin{array}{cc}
-1 & 2 \\
2 & -1
\end{array}\right] \text { and } \bar{B}=\left[\begin{array}{cc}
-1 & 2 \\
2 & -1
\end{array}\right]
$

Hence, the required answer is $\left[\begin{array}{cc}-1 & 2 \\ 2 & -1\end{array}\right]$

Example 4: What is the transpose conjugate of $\left[\begin{array}{cc}i+3 & i-3 \\ 0 & i+1\end{array}\right]$

Solution

The transpose conjugate of a Matrix -

It is denoted by $\mathrm{A}^\theta$ and $\mathrm{A}^\theta=(\overline{\mathrm{A}})^{\prime}$

$\begin{aligned} & A=\left[\begin{array}{cc}i+3 & i-3 \\ 0 & i+1\end{array}\right] \\ & \bar{A}=\left[\begin{array}{cc}-i+3 & -i-3 \\ 0 & -i+1\end{array}\right]\end{aligned}$
$(\bar{A})^{\prime}=\left[\begin{array}{cc}-i+3 & 0 \\ -i-3 & -i+1\end{array}\right]$

Hence, the required answer is $\left[\begin{array}{cc}-i+3 & 0 \\ -i-3 & -i+1\end{array}\right]$