Question

If \(A = \left[ {\begin{array}{*{20}{c}} {2 + i}&3&{ - 1 + 3i}\\ { - 5}&i&{4 - 2i} \end{array}} \right]\), then AA^⋆ will be (where, A^⋆ is the conjugate transpose of A)
1. Unitary matrix
2. Orthogonal matrix
3. Hermitian matrix
4. Skew Hermitian matrix

Answer 1

Correct Answer - Option 3 : Hermitian matrix

Concept:

Unitary matrix A* A = I

Orthogonal matrix if A^T ⋅ A = I

Hermitian matrix if A* = (A̅ )^T

Skew Hermitian matrix if A* = -A

Where A* is conjugate transpose of A

Calculation:

Given \(A = \left[ {\begin{array}{*{20}{c}} {2 + i}&3&{ - 1 + 3i}\\ { - 5}&i&{4 - 2i} \end{array}} \right] \)

\(̅{A}=\begin{bmatrix} (2-i) & 3 & (-1-3i)\\ -5& -i &(4+2i) \end{bmatrix}\)

\((̅{A})^{T}=\begin{bmatrix} (2-i) &-5 \\ 3& -i\\ (-1-3i) & (4+2i) \end{bmatrix}\)

Now

\(A\times A^{*}=A\times (̅{A})^{T}=\begin{bmatrix} (5+9+10) &(-10-5i-3i-4-2i+12i-6) \\ (-10+5i+3i-4-12i+2i-6)& (25+1+16+4) \end{bmatrix}\)

\(A\times A^{*} = \left[ {\begin{array}{*{20}{c}} {24}&{2i - 20}\\ { - 2i - 20}&{46} \end{array}} \right]\)

Hence AA*  Hermitian matrix

a_ij = a̅ _ji for all i, j and a_ij are real