[Math] Symmetric and Skew Symmetric Matrices

linear algebra

$V_0=$ the set of $2\times2$ skew symmetric matrices

I know that any element of $V_0$ has a characteristic polynomial that will not factor over the real numbers, and therefore has no eigenvectors. Why is this so? How would I go about showing this?

Best Answer

Let us examine the chacteristic polynomial $p_S(\lambda)$ of $S$, a skew-symmetric $2 \times 2$ matrix. As is well-known, and in fact easily derived, we have

$p_S(\lambda) = \lambda^2 - \text{Tr}(S)\lambda + \det (S). \tag{1}$

Writing

$S = [s_{ij}], \tag{2}$

we see that the skew-symmetry of $S$ implies the diagonal entries $s_{ii} = 0$, since $S^T = -S$ implies $s_{ii} =-s_{ii}$ for $i = 1,2$. Thus the trace of $S$ vanishes, and furthermore we have

$\det(S) = -s_{12}s_{21} = s_{12}^2 >0 \tag{3}$

if $S \ne 0$. Thus

$p_S(\lambda) = \lambda^2 + s_{12}^2; \tag{4}$

but the roots of (4) are $\pm i\vert s_{12} \vert$, both purely imaginary, so $p_A(\lambda)$ is irreducuble over $\Bbb R$.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

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