Real Skew Symmetric $3\times 3$ matrix has one eigen value $2i$ so another eigen value is $-2i$ another eigen value must be $0$ right? as we know these kind of matrix has eigen values $0$ or purely imaginary.

  • $\begingroup$ Yes. $\quad\mu^{*} = \mu\quad$ and $\quad\mu^{*} = -\mu\quad$ $\Longrightarrow\quad\mu = 0$. $\endgroup$ – Felix Marin Sep 13 '13 at 19:50

Yes, the characteristic polynomial of a real $3\times 3$ matrix will be a real cubic polynomial. If not identically zero, the real skew matrix will have one pair of conjugate imaginary eigenvalues and a zero eigenvalue, each of multiplicity one.

  • 1
    $\begingroup$ Does this eliminate the possibility of eigenvalues $2i,2i,-2i$? $\endgroup$ – Rebecca J. Stones Sep 13 '13 at 16:48
  • $\begingroup$ @Rebecca: Yes, it does (for a 3x3 real skew matrix). $\endgroup$ – hardmath Sep 13 '13 at 17:17
  • $\begingroup$ In case someone has doubts, the characteristic polynomial being a real cubic "eliminates the possibility" of three purely imaginary nonzero roots. Since these must occur in conjugate pairs, divding out a monic quadratic with one such pair as roots gives as quotient a real degree 1 polynomial having the third purely imaginary nonzero root (contradiction). $\endgroup$ – hardmath Sep 13 '13 at 21:50

In general, a real skew-symmetric $3 \times 3$ matrix $K$ looks like

$K = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}, \tag{1}$

so that the characteristic polynomial is

$\det(\lambda - K) = \lambda^3 + (a^2 + b^2 + c^2) \lambda, \tag{2}$

which is easily seen to have roots

$\lambda = 0, \pm i\sqrt{a^2 + b^2 + c^2}, \tag{3}$

which follow the general pattern the OP suggested; apparently the matrix she/he was thinking of satisfies $\sqrt{a^2 + b^2 + c^2} = 2$; but $0$ will always be an eigenvalue for such $K$ in any case.

Hope this helps. Cheers.


Since a real matrix $A$ is skew if, and only if, $A=-A^T$, since $\text{spec} (A)=\text{spec}(A^T)=-\text{spec}(-A^T)$ and since the characteristic polynomial has degree $3$ and non-real roots come in pairs, it follows that $0\in\text{spec}(A)\cap \text{spec}(A^T)$.


We know that a skew symmetric matrix $A=(a_{ij})_{3 \times 3}$ satisfies $A^T=-A$. This means that $a_{ii}=-a_{ii}$ or equivalently $a_{ii}=0$ for all $i \in \{1,2,3\}$.

We know the trace of $A$ is the sum of its eigenvalues, which is $0$ since the main diagonal of $A$ comprises of zeroes.

Since a $3 \times 3$ matrix has $3$ (not necessarily distinct) eigenvalues (given by the roots of the characteristic polynomial [a degree $3$ polynomial with leading coefficient either $1$ or $(-1)^3$, depending on how it's defined]), there is precisely one other eigenvalue $\lambda$ and we must have $2i-2i+\lambda=0$, which implies $\lambda=0$.


Here's another way of looking at it, which I thought of after posting my previous answer.

The approach here seemed sufficiently different that I felt a separate answer was merited. Hope I'm not overdoing it. Anyway . . .

First, note that for any skew-symmetric, real matrix, the only possible real eigenvalue is zero. For, if $K$ is real skew-symmetric, so that $K^T = -K$, and if $v \ne 0$ is an eigenvector with real eigenvalue $\lambda$, so that

$Kv = \lambda v, \tag{1}$

we have

$\langle Kv, v \rangle = \lambda \langle v, v \rangle, \tag{2}$


$\langle Kv, v \rangle = \langle v, K^Tv \rangle = \langle v, -Kv \rangle = -\langle Kv, v \rangle, \tag{3}$


$\lambda \langle v, v \rangle = -\lambda \langle v, v \rangle, \tag{4}$

and since $\langle v, v \rangle \ne 0$ this shows that

$\lambda = - \lambda, \tag{5}$


$\lambda = 0. \tag{6}$

Here I take $\langle \cdot, \cdot \rangle$ to be the standard inner product on the real vector space on which $K$ operates.

Now argue as follows: since $2i$ is an eigenvalue of $K$, and the characteristic polynomial of $K$ is real, $-2i$ is an eigenvalue as well. This means

$\lambda^2 + 4 \mid p_K(\lambda), \tag{7}$


$p_K(\lambda) = det(\lambda - K) \tag{8}$

is the characteristic polynomial of $K$. But $\deg p_K(\lambda) = 3$ in the present case, so the quotient polynomial $q(\lambda)$ is of degree $1$, i.e. we must have

$p_K(\lambda) = (\lambda^2 + 4)q(\lambda) \tag{9}$


$q(\lambda) = \lambda - a, \tag{10}$

$a$ real. But this of course implies $a$ is a real eigenvalue of $K$, whence we must have

$a = 0. \tag {11}$

Well, I hope this answer is more than mere mathematical logorrhea, and it sheds a little more light on the subject at hand.



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