Let \(M\) be a skew-symmetric matrix (with real entries).

Let \(\lambda\) be an eigenvalue of \(M\). This means, there exists vector \(v\) such that \(Mv=\lambda v\).

To relate with ``skew-symmetricness'' of \(M\), we apply transpose on both sides of previous equation, to get \(v^TM^T=\lambda v^T\).

As \(M\) is skew-symmetric, we see that \(v^TM^T=\lambda v^T\) is equivalent to \(-v^TM=\lambda v^T\).

Now, multiply by \(v\) on both sides of the above equation to get \(-v^TMv=\lambda v^Tv\).

As \(v\) is eigenvector associated to eigenvalue \(\lambda\), we get \(-v^T\lambda v=\lambda v^Tv\) which is samething as \(-\lambda v^Tv=\lambda v^Tv\).

But, we know \(v^Tv\neq 0\). So, cancelling \(v^Tv\) of both sides tells that \(-\lambda=\lambda\) which means \(\lambda =0\).

There is no assurance that \(\lambda\) is a real number to begin with, but, it turns out \(\lambda\) has to be real number.

But, we have heard multiple times and repeated multiple times in high pitch that skew-symmetric matrices has eigenvalues as \(0\) or purely imaginary number.

What went wrong here??

Even if we start with \(\lambda\) is a complex eigenvalue, the above observation can be used line by line.

Is it?

Are we sure that \(v^Tv\neq 0\) irrespective of what kind of \(v\) is, as long as \(v\neq 0\)?

Let \(v=(1,i)\). Then, \(v^Tv=1+i^2=0\). So, we can not cancel the component \(v^Tv\) as per our interest. That is the issue in above discussion.

We can not say that \(v^Tv\) is nonzero if \(v\neq 0\), but, we can always say \(\bar{v}^Tv\neq 0\) as far as \(v\neq 0\). So, instead of applying \(v^T\), it may be smart idea to apply \(\bar{v}^T\). This is done by considering conjugate and transport.

As \(M\) is a real matrix, conjugate does not any effect. We have

\[Mv=\lambda v\]

\[\bar{v}^T\bar{M}^T=\bar{\lambda}\bar{v}^T\]

\[\bar{v}^TM^T=\bar{\lambda}\bar{v}^T\]

\[\bar{v}^T(-M)=\bar{\lambda}\bar{v}^T\]

\[-\bar{v}^TM=\bar{\lambda}\bar{v}^T\]

\[-\bar{v}^TMv=\bar{\lambda}\bar{v}^Tv\]

\[-\bar{v}^T\lambda v=\bar{\lambda}\bar{v}^T\]

\[-\lambda \bar{v}^Tv=\bar{\lambda}\bar{v}^T\]

As \(v\) is such that \(\bar{v}^Tv\neq 0\). So, we have \(\lambda =-\bar{\lambda}\) that is, real part of \(\lambda\) is zero.

So, any eigenvalues of real skew symmetric matrix is zero or with a complex number with real part zero (purely imaginary complex number).