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Question

Diagonal elements of a skew-Hermitian matrix are:

The correct answer is
Purely imaginary or zero.

Skew-Hermitian Diagonal Elements Property

A matrix '$A$' is called skew-Hermitian if its conjugate transpose '$A^\dagger$' is equal to its negative, i.e., $A^\dagger = -A$.

Let the elements of the matrix '$A$' be denoted by '$a_{ij}$'. The conjugate transpose '$A^\dagger$' has elements $(A^\dagger)_{ij} = \overline{a_{ji}}$ (where $\overline{z}$ denotes the complex conjugate of $z$).

The condition for a skew-Hermitian matrix is $A^\dagger = -A$. This means $(A^\dagger)_{ij} = -a_{ij}$ for all $i$ and $j$. Specifically, for the diagonal elements where $i = j$, the condition becomes:

$ (A^\dagger)_{ii} = -a_{ii} $ $ \overline{a_{ii}} = -a_{ii} $

Now, let's represent a diagonal element '$a_{ii}$' in terms of its real part '$x$' and imaginary part '$y$':

$ a_{ii} = x + iy $

The complex conjugate is:

$ \overline{a_{ii}} = x - iy $

Substituting these into the condition $\overline{a_{ii}} = -a_{ii}$:

$ x - iy = -(x + iy) $ $ x - iy = -x - iy $

Adding '$iy$' to both sides gives:

$ x = -x $

This equation implies:

$ 2x = 0 $ $ x = 0 $

Since the real part '$x$' must be zero, the diagonal element '$a_{ii}$' must be of the form $0 + iy$, which simplifies to '$iy$'.

  • If $y = 0$, then $a_{ii} = i(0) = 0$.
  • If $y \neq 0$, then $a_{ii}$ is a non-zero purely imaginary number.

Therefore, the diagonal elements of a skew-Hermitian matrix must be purely imaginary or zero.

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