To determine if the given matrix is unitary, we need to verify if the matrix multiplied by its conjugate transpose results in the identity matrix.
1. Given Matrix [tex]\( A \)[/tex]:
[tex]\[
A = \frac{1}{\sqrt{3}} \begin{pmatrix}
1+i & -1 \\
1 & 1-i
\end{pmatrix}
\][/tex]
2. Calculate the Conjugate Transpose [tex]\( A^\dagger \)[/tex]:
The conjugate transpose of matrix [tex]\( A \)[/tex], denoted as [tex]\( A^\dagger \)[/tex], is found by taking the transpose of [tex]\( A \)[/tex] and then taking the complex conjugate of each element.
[tex]\[
A^\dagger = \frac{1}{\sqrt{3}} \begin{pmatrix}
1-i & 1 \\
-1 & 1+i
\end{pmatrix}
\][/tex]
3. Compute [tex]\( A \cdot A^\dagger \)[/tex]:
We need to multiply matrix [tex]\( A \)[/tex] by its conjugate transpose [tex]\( A^\dagger \)[/tex]:
[tex]\[
A \cdot A^\dagger = \left( \frac{1}{\sqrt{3}} \begin{pmatrix}
1+i & -1 \\
1 & 1-i
\end{pmatrix} \right)
\times
\left( \frac{1}{\sqrt{3}} \begin{pmatrix}
1-i & 1 \\
-1 & 1+i
\end{pmatrix} \right)
\][/tex]
4. Result of the Product [tex]\( A \cdot A^\dagger \)[/tex]:
After performing the matrix multiplication and simplifying, the product will be:
[tex]\[
A \cdot A^\dagger = \begin{pmatrix}
1.00000000e+00+0.j & 2.94613014e-18+0.j \\
2.94613014e-18+0.j & 1.00000000e+00+0.j
\end{pmatrix}
\][/tex]
5. Compare with the Identity Matrix [tex]\( I \)[/tex]:
For [tex]\( A \)[/tex] to be unitary, the product [tex]\( A \cdot A^\dagger \)[/tex] should equal the identity matrix [tex]\( I \)[/tex]:
[tex]\[
I = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\][/tex]
The result matrix is numerically very close to the identity matrix, considering small numerical precision errors. The diagonal elements are [tex]\( 1 \)[/tex], and the off-diagonal elements are approximately zero.
6. Conclusion:
Since the product [tex]\( A \cdot A^\dagger \)[/tex] is extremely close to the identity matrix, we conclude that the given matrix [tex]\( A \)[/tex] is indeed a unitary matrix.
