by James
Unitary Matrix is a non-singular matrix with complex numbers. The product of conjugation of unitary matrix and transpose of unitary matrix results in the Identity matrix. Students can find detailed information regarding the unitary matrix likewise definition, formula, properties, and examples from here. Thus have a look at this page to learn in detail about the unitary matrix.
Unitary Matrix Definition
A unitary matrix is a matrix whose inverse is equal to the conjugate transpose. Conjugate transpose is referred to as the Hermitian matrix. The unitary matrices may also be non-square matrices but the matrices have orthonormal columns. The unitary matrix is denoted by U.
Unitary Matrix Formula
Formulas of a unitary matrix are
U*U = UU* = I
where I is an identity matrix
Properties of Unitary Matrix
- The two complex vectors are x and y, this x and y are multiplied by U that is, (Ux, Uy) = (x, y).
- U is normal (U*U = UU*)
- U is diagonalizable, that is, U is unitarily similar to a diagonal matrix, as a consequence of the spectral theorem.
Therefore U is a decomposition of the form U=VDV*
where V is unitary
D is diagonal and unitary. - |det(U)| = 1
- Its eigenspaces are orthogonal.
- U can be written as U = e power iH, where
e indicates the matrix exponential.
i is the imaginary unit.
H is a Hermitian matrix.
Unitary Matrix Solved Problems
Example 1.
Find the matrix \( A =\left[\begin{matrix}
1/√2 & 1/√2 \cr
1/√2 i & -1/√2 i \cr
\end{matrix}
\right]
\) is a unitary matrix or not.
Solution:
Given that the matrix is
\( A =\left[\begin{matrix}
1/√2 & 1/√2 \cr
1/√2 i & -1/√2 i \cr
\end{matrix}
\right]
\)
The conjugate transpose of a matrix A is A*
A*\(=\left[\begin{matrix}
1/√2 & -1/√2 i \cr
1/√2 & 1/√2 i \cr
\end{matrix}
\right]
\)
A A* \(=\left[\begin{matrix}
½ – ½ i² & ½ + ½ i² \cr
½ + ½ i² & ½ – ½ i² \cr
\end{matrix}
\right]
\)
A A* \(=\left[\begin{matrix}
1 & 0 \cr
0 & 1 \cr
\end{matrix}
\right]
\)
Therefore A A* = I
Therefore the given matrix is unitary
Example 2.
Show that the matrix \( A =\left[\begin{matrix}
1+i & 1-i \cr
1-i & 1+i \cr
\end{matrix}
\right]
\) × ½ is Unitary or not
Solution:
Given that the matrix is
\( A =\left[\begin{matrix}
1+i & 1-i \cr
1-i & 1+i \cr
\end{matrix}
\right]
\) × ½
The Conjugate transpose of a matrix A is A*.
\( A* =\left[\begin{matrix}
1-i & 1+i \cr
1+i & 1-i \cr
\end{matrix}
\right]
\) × ½
\( A A* =\left[\begin{matrix}
4 & 0 \cr
0 & 4 \cr
\end{matrix}
\right]
\) × ¼.
Therefore the given matrix is Unitary
Example 3.
If the matrix A = ½ \( =\left[\begin{matrix}
1 & -i & -1+i \cr
i & 1 & 1+i \cr
1+i & -1+i 0 \cr
\end{matrix}
\right]
\) Is unitary or not.
Solution:
Given that the matrix is
A = ½ \( \left[\begin{matrix}
1 & -i & -1+i \cr
i & 1 & 1+i \cr
1+I -1+I 0 \cr
\end{matrix}
\right]
\)
The Conjugate transpose of a matrix A = A*
A*= ½ \(=\left[\begin{matrix}
1 & -i & -1-i \cr
i & 1 & 1+i \cr
1+i & -1+i & 0 \cr
\end{matrix}
\right]
\)
Then A A*
A A* = 1/4 \( = \left[\begin{matrix}
1 & 0 & 0 \cr
0 & 1 & 0\cr
0 & 0 & 1 \cr
\end{matrix}
\right]
\)
Hence the given matrix is Unitary.
Example 4.
Find the matrix A = ⅓ \( =\left[\begin{matrix}
2+i & 2i \cr
2i & 2-i \cr
\end{matrix}
\right]
\) is Unitary or not.
Solution:
Given that the matrix is
A = ⅓ \( =\left[\begin{matrix}
2+i & 2i \cr
2i & 2-i \cr
\end{matrix}
\right]
\)
The Conjugate transpose of the matrix A is A*
A* = ⅓\( =\left[\begin{matrix}
2-i & -2i \cr
-2i & 2+i \cr
\end{matrix}
\right]
\)
A A* = ⅓ \( =\left[\begin{matrix}
2+i & 2i \cr
2i & 2-i \cr
\end{matrix}
\right]
\) × ⅓ \( =\left[\begin{matrix}
2-i & -2i \cr
-2i & 2+i \cr
\end{matrix}
\right]
\)
A A* = 1/9 \( =\left[\begin{matrix}
9 & 0 \cr
0 & 9 \cr
\end{matrix}
\right]
\)
Hence the given matrix is Unitary.
Example 5.
If the matrix \( A =\left[\begin{matrix}
0 & 1+2i \cr
-1+2i & 0 \cr
\end{matrix}
\right]
\) then shows that (I-A)(I+A) inverse is unitary or not.
Solution:
Given that the matrix is
\( A =\left[\begin{matrix}
0 & 1+2i \cr
-1+2i & 0 \cr
\end{matrix}
\right]
\)
Let \( I =\left[\begin{matrix}
1 & 0 \cr
0 & 1 \cr
\end{matrix}
\right]
\)
\( I+A =\left[\begin{matrix}
1 & 1+2i \cr
-1+2i & 1 \cr
\end{matrix}
\right]
\)
\( I-A =\left[\begin{matrix}
1 & -1+2i \cr
1-2i & 1 \cr
\end{matrix}
\right]
\)
I-A inverse = 1/|I+A| \( =\left[\begin{matrix}
0 & 1+2i \cr
-1+2i & 0 \cr
\end{matrix}
\right]
\)
(I-A) inverse = ⅙ \( A =\left[\begin{matrix}
0 & 1+2i \cr
-1+2i & 0 \cr
\end{matrix}
\right]
\)
(I+A)(I-A) inverse = 1/36 \( =\left[\begin{matrix}
36 & 0 \cr
0 & 36 \cr
\end{matrix}
\right]
\)
Hence the given matrix is Unitary.
Related Topics:
- Non-singular matrix
- Solve matrices
FAQs on Unitary Matrix
1. What is the Order of a Unitary Matrix?
The order of the unitary matrix is n × n which means the same number of rows and the same number of columns. It is a square matrix.
2. What are the Properties of the Unitary Matrix?
The properties of a unitary matrix are
- A unitary matrix is a nonsingular matrix and an Invertible matrix.
- The product of two unitary matrices is a unitary matrix.
- The addition and subtraction of two unitary matrices is also a unitary matrix.
3. What is a Unitary Matrix?
A unitary matrix is a square matrix of complex figures. And the product of the antipode of a unitary matrix, and the conjugate transpose of a unitary matrix is equal to the identity matrix. U power H = U power-1.
