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    Skew Hermitian Matrix - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

    Quick Facts

    • 4 Questions around this concept.

    Solve by difficulty

    If  matrix   $A=\begin{bmatrix} 1 & 0&w \\ 1& w^{2}& 1\\ 1& w^{2}& w \end{bmatrix}$  where $\omega$ is a cube root of unity. Then matrix $A-A^{\theta }$ is

    Concepts Covered - 2

    Skew-hermitian matrix

    Skew-hermitian matrix

    A square matrix $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{n} \times \mathrm{n}}$ is said to be Skew-Hermitian matrix if $a_{i j}=-\overline{a_{i j}} \forall \mathrm{i}, \mathrm{j}_{\text {}}$
    i.e. $\mathrm{A}^\theta=-\mathrm{A}, \quad\left[\right.$ where $\mathrm{A}^\theta$ 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 sign from + ve to -ve $O R-v e$ to + ve in imaginary part of all elements, So to satisfy the condition $A$ ? $=-\mathrm{A}$, all diagonal element must be purely imaginary. As $A^?=-\mathrm{A}$ so
    $\mathrm{a}_{\mathrm{ij}}=-\overline{\mathrm{a}_{\mathrm{ij}}} \forall \mathrm{i}, \mathrm{j}$
    if we put $\mathrm{i}=\mathrm{j}$, 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
    $


    Hence all diagonal element should be purely imaginary
    E.g

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

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

    Then,

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

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

    Note:
    1. for any square matrix $A$ with elements containing complex numbers, then $A-A$ ? is a skew hermitian matrix.

    Proof: $\left(A-A^?\right)^?=A^?-\left(A^?\right)^?=A^?-A=-\left(A-A^?\right)$, hence skew-hermitian.
    2. Every square matrix can be written as the sum of hermitian and skew-hermitian matrix i.e.

    If $A$ is a square matrix, then we can write

    $
    A=\frac{1}{2}\left(A+A^\theta\right)+\frac{1}{2}\left(A-A^\theta\right)
    $
     

    Properties of hermitian and skew-hermitian matrices

    Properties of hermitian and skew-hermitian matrices

    i) If A is a square matrix then AA𝛩 and A𝛩 A are hermitian matrix.

       Proof: for hermitian matrix A𝛩 = A, so we check the condition on AA𝛩 

       (AA𝛩)𝛩 = (A𝛩)𝛩A𝛩 = AA𝛩 hence it is hermitian, and in the same way, A𝛩A will also be hermitian.

    ii) If A is hermitian matrix then:

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

        Proof: we need to show (iA)𝛩 = -iA

        (iA)𝛩= A𝛩i𝛩 = A𝛩 (-i) = -iA𝛩

        Since A is hermitian so A𝛩 = A

        Hence we have

        -iA𝛩 = -iA. Proved.

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

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

         Proof: we need to show (iA)𝛩 = iA

        (iA)𝛩 = A𝛩i𝛩 = A𝛩(-i) 

        A𝛩(-i) = Ai = iA   (since A is skew-hermitian, so A𝛩 = -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 constant. 

           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)𝛩 = B𝛩A𝛩 = BA = AB (Since A, B are hermitian so A𝛩 = A, B𝛩 =B)

      c.  AB + BA will also we hermitian

           Proof: from part (b) AB and BA is 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 A𝛩 = A and B𝛩 = B, proved.

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

        Proof are similar as above, just verify the basic condition, using the given condition of A and B.

    Study it with Videos

    Skew-hermitian matrix
    Properties of hermitian and skew-hermitian matrices

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