Vtyjkyuk

We have definition that $c in F$, field, is called eigenvalue of matrix $A$ if there exist a vector $X$ such that $AX = cX$.
I have doubt that as skew hermitian matrices form vector space over $Bbb R$ not over $Bbb C$. So why do we say that skew hermitian matrices have eigenvalues either $0$ or purely imaginary complex numbers ?
I want to get clarity in this context.

I hope I understand the question correctly. This is what I understand: Consider a $ntimes n$ skew-hermitian matrix $A$ with entries all in $mathbbR$. It is said that the eigenvalues of $A$ are pure imaginary. How come the eigenvalues are complex while the matrix is real?

What is happening here would become most apparent with an example. Consider the following matrix
$$
A=beginpmatrix
0 & 1\
-1 & 0
endpmatrix
$$
the chracteristic polynomial of $A$ is $p(x)=det(A-xI)=x^2+1$. Since the eigenvalues are the roots of this polynomial, we encounter a problem. In the field $mathbbR$ the polynomial $p(x)$ has not roots, so what does that mean for matrix $A$?

The matrix $A$ as a real matrix has no eigenvalues and no eigenvectors.

Yes, if $A$ is a square matrix with entries in a field $mathbbF$, then $A$ does not necessarily admit eigenvalues unless the field $mathbbF$ is algebraically closed (a field is called algebraically closed if every polynomial of degree $d$ has exactly $d$ roots, counting the multiplicities).

If you instead think of $A$ as a complex matrix (noting that $mathbbC$ is algebraically closed), then you find two purely imaginary eigenvalues $pm i$.

What is meant by the statement "The eigenvalues of a square real skew-hermitian matrix $M$ are purely imaginary eigenvalues" is:

1. If one regards $M$ as a complex matrix (using the usual $mathbbRsubset mathbbC$), then the eigenvalues are all purely imaginary.


2. If one regards $M$ as a real matrix, then $M$ has no eigenvalues/eigenvectors (not even one!).