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Attempted solution:
Say that we have matrix A, that is anti-hermitian. $$A= - A^{\dagger}$$ Taking the negative: $$-A= - (-A)^{\dagger}$$ $$-A= A^{\dagger}$$ which shows that it's still anti-Hermitian. So multiplying a negative doesn't effect the state of the matrix A being anti-Hermitian. Does my argument check out?

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  • $\begingroup$ Yes and in general, if $A$ is skew-Hermitian, $cA$ will also be skew-Hermitian for any real scalar $c$. $\endgroup$ – user1551 Aug 4 at 11:00
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Yes, your argument is correct. The negative of an antihermitian matrix $A$ remains antihermitian. For an example, observe that if

$$A= \begin{bmatrix} -i & 2+i\\ -2+i & 0 \end{bmatrix}$$

then $A$ is anti-Hermetian since

$$-A = \begin{bmatrix} i & -2 - i \\ 2 - i & 0 \end{bmatrix} = \begin{bmatrix} \overline{-i} & \overline{-2 + i} \\ \overline{2 + i} & \overline{0} \end{bmatrix} = \begin{bmatrix} \overline{-i} & \overline{2 + i} \\ \overline{-2 + i} & \overline{0} \end{bmatrix}^\mathsf{T} = A^\mathsf{H} $$

and writing

$$-A= \begin{bmatrix} i & -2-i\\ 2-i & 0 \end{bmatrix}$$

would produce

$$-(-A) = A= \begin{bmatrix} -i & 2 + i \\ -2 + i & 0 \end{bmatrix} = \begin{bmatrix} \overline{i} & \overline{2 - i} \\ \overline{-2 - i} & \overline{0} \end{bmatrix} = \begin{bmatrix} \overline{i} & \overline{-2 - i} \\ \overline{2 - i} & \overline{0} \end{bmatrix}^\mathsf{T} = -A^\mathsf{H} $$

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