# Why do we say that the eigenvalues of skew-Hermitian matrices are either $0$ or purely imaginary?

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.

• Aug 18, 2018 at 2:50
• Is it possible to do this with mobile? I don't know this Aug 18, 2018 at 3:01
• Does this answer help ? Aug 18, 2018 at 7:46

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:

1. If one regards $M$ as a complex matrix (using the usual $\mathbb{R}\subset \mathbb{C}$), 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!).
• Thanks for your reply. I got it. But my problem was that eigenvalues of a matrix should belongs to that field by which entries of matrices are being taken or the field on which it forms vector space. eg -As to form skew hermitian matrix we take entries from complex field but it form vector space over it's subfield set of real numbers. So my doubt is from which field we should take eigenvalues? Aug 21, 2018 at 2:31