We have the definition that $c \in F$, field, is called eigenvalue of the matrix $A$ if there exists 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 $n\times n$ skew-hermitian matrix $A$ with entries all in $\mathbb{R}$. 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 $$ A=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} $$ 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 $\mathbb{R}$ 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 $\mathbb{F}$, then $A$ does not necessarily admit eigenvalues unless the field $\mathbb{F}$ 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 $\mathbb{C}$ 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:
- If one regards $M$ as a complex matrix (using the usual $\mathbb{R}\subset \mathbb{C}$), then the eigenvalues are all purely imaginary.
- If one regards $M$ as a real matrix, then $M$ has no eigenvalues/eigenvectors (not even one!).
