How does this term in the Majorana mass not vanish?

Question

This is classical field theory. In the Majorana mass term, we have the expression $$\nu_L^T\sigma_2\nu_L \tag{1}$$ where the left-handed spinor field $\nu$ has a Grassmann-valued amplitude, i.e., $\nu = \mathcal{N} \text{(blah)}$ where $\mathcal{N}$ is a Grassmann number. How does $(1)$ not vanish? For, we can write $(1)$ as $$\nu_L^T\sigma_2\nu_L = \mathcal{N}^2\text{(blah)}=0$$ by antisymmetry of $\mathcal{N}$ with itself.

Now, say that the above problem is resolved. Then, it seems like $$\nu_L^T\sigma_2\nu_L = (\nu_L^T\sigma_2\nu_L)^T = \nu_L^T\sigma_2^T\nu_L = -\nu_L^T\sigma_2\nu_L,$$ which would again suggest that this expression does equal $0$. How does this expression not vanish given these two observations?

Answer 1

Remember that $\nu_L$ is a two-component column vector with Grassmann entries $\nu_L^1$ and $\nu_L^2$, so with $$ \sigma_2 =\left[\matrix{0&-i\cr i&0}\right], $$ we have $$ \nu_L^T \sigma_2 \nu_L = \left[\matrix{\nu_L^1, &\nu_L^2}\right] \left[\matrix{0&-i\cr i&0}\right] \left[\matrix{\nu_L^1 \cr \nu_L^2}\right]= -i(\nu_L^1\nu_L^2- \nu_L^2 \nu_L^1) =-2i\nu_L^1\nu_L^2\ne0. $$