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Skew Hermitian Matrix - Practice Questions & MCQ

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

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  • 3 Questions around this concept.

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Skew-hermitian matrix

Skew-hermitian matrix

A square matrix \mathrm{A=[a_{ij}]_{n\times n}} is said to be Skew-Hermitian matrix if \mathrm{\mathit{a_{ij}=-\overline{a_{ij}}}} ∀ i, j,  

i .e.\mathrm{A^\theta=-A,\;\;[where\;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 OR -ve to +ve in imaginary part of all elements, So to satisfy the condition A?  = - A, all diagonal element must be purely imaginary. As A?  = - A so 

 

\\\mathrm{a_{ij}=-\overline{a_{ij}} \;\; \forall \;i,j} \\\mathrm{if \; we \; put \; i=j, \; we\; have} \\\mathrm{a_{ii}= -\overline{a_{ii}} \Rightarrow a_{ii}+\overline{a_{ii}}=0} \\\mathrm{\Rightarrow a_{ii}=0}

Hence all diagonal element should be purely imaginary

E.g 

\\\mathrm{Let,\;\;A=\begin{bmatrix} -3i &-3-4i & -5+2i\\ 3-4i& 5i &i \\ 5+2i&i &0 \end{bmatrix}}\\\mathrm{Then,}\\\mathrm{\;\;\;\;\;\;\;\;A'=\begin{bmatrix} -3i&3-4i &5+2i \\-3-4i &5i &i \\ -5+2i & i &0 \end{bmatrix}}\\\\\\\mathrm{\therefore A^\theta=\overline{(A')}=\begin{bmatrix} 3i&3+4i &5-2i \\-3+4i &-5i &-i \\ -5-2i & -i &0 \end{bmatrix}}\\\\\\\mathrm{\;\;\;\;\;\;\;\;=-\begin{bmatrix} -3i &-3-4i & -5+2i\\ 3-4i& 5i &i \\ 5+2i&i &0 \end{bmatrix}=-A}\\\\\mathrm{here,\;A\;is\;Skew-Hermitian\;matrix\;as\;A^\theta=-A}


Note:

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

        Proof : (A-A?)? = A? - (A?)? = A? - A = -(A-A?), 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    \\\mathrm{A = \frac{1}{2}(A + A^{\theta})+\frac{1}{2}(A-A^{\theta})}  

Properties of hermitian and skew-hermitian matrices

Properties of hermitian and skew-hermitian matrices

i) If A is a square matrix then AA? and A? A are hermitian matrix.

   Proof: for hermitian matrix A? = A, so we check the condition on AA? 

   (AA?)? = (A?)?A? = AA? hence it is hermitian, and in the same way, A?A 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)? = -iA

    (iA)?= A?i? = A? (-i) = -iA?

    Since A is hermitian so A? = A

    Hence we have

    -iA? = -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)? = iA

    (iA)? = A?i? = A?(-i) 

    A?(-i) = Ai = iA   (since A is skew-hermitian, so A? = -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)? = B?A? = BA = AB (Since A, B are hermitian so A? = A, B? =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 A? = A and B? = 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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