# Construct Real Skew Symmetric Matrix given eigenvalue and eigenvector

Is there a way to construct a real skew-symmetric matrix given a single purely imaginary eigenvalue and corresponding eigenvector? The other eigenvalues and eigenvectors can be chosen at will. Thanks

I don't think so. Consider $0\in\mathbb R$ is a skew-symmetric matrix, but it always has eigenvalue $0$. Furthermore it is easy to show that every skew-symmetric matrix $A\in\mathbb R^{2\times 2}$ with $$A = \begin{bmatrix}0 & a\\ -a & 0\end{bmatrix}$$ for $a\in\mathbb R$ has always the eigenvalues $ia$ and $-ia$ with eigenvectors $$\begin{bmatrix}-i\\1\end{bmatrix},\qquad \begin{bmatrix}i\\1\end{bmatrix}.$$ This means you cannot construct a skew-symmetric $2\times 2$ matrix to any given eigenvector, for example $$\begin{bmatrix}1\\1\end{bmatrix}.$$
Let me assume that you want to construct a skew-symmetric matrix $A \in M_{2n}(\mathbb{R})$ with eigenvalues $a_1, -a_1 \ldots, a_n, -a_n$ for pure imaginary, non-zero and distinct $a_i$. First, as Wauzl observed, you can't specify the eigenvectors for $A$ arbitrary. If $v_i$ is an eigenvector of $A$ corresponding to the eigenvalue $a_i$, then $\overline{v_i}$ must be an eigenvector of $A$ corresponding to the eigenvalue $-a_i$. In particular, since $a_i \neq -a_i$, the vectors $v_i$ and $\overline{v_i}$ must be linearly independent. In fact, since $A$ is normal (considered as a matrix over the complex numbers), the eigenvalues $(v_1, \overline{v_1}, \ldots, v_n, \overline{v_n})$ must be an orthonormal basis of $\mathbb{C}^{2n}$ (with respect to the standard Hermitian inner product).
In the other direction, if you are given a list $(v_1, \ldots, v_n)$ of complex vectors such that $\mathcal{B} = (v_1, \overline{v_1}, \ldots, v_n, \overline{v_n})$ is an orthonormal ordered basis for $\mathbb{C}^{2n}$, you can construct a unitary matrix $U \in M_{2n}(\mathbb{C})$ with the vectors of $\mathcal{B}$ as columns (in the same order as the are given in $\mathbb{B}$). Choose any pure imaginary list of numbers $(a_1, \ldots, a_n)$ and let $D = \mathrm{diag}(a_1, \ldots, a_n)$. Then $UDU^{*}$ will be a real, skew-symmetric matrix with eigenvalues $(a_1, -a_1, \ldots, a_n, -a_n)$ and corresponding eigenvectors $(v_1, \overline{v_1}, \ldots, v_n, \overline{v_n})$.