# Show that conic C has rank 2

Given the conic $C=lm^T+ml^T$ defined by two distinct lines $l$ and $m$. $C$ is a symmetric 3x3 matrix. How can i show that $C$ has rank 2? The rank of a matrix $M$ can be calculated using gaussian elimination. But for $C$ i get: $$C=\begin{pmatrix} 2\cdot l_1m_1 & l_1m_2+l_2m_1 & l_1m_3+l_3m_1 \\ l_1m_2+l_2m_1 & 2\cdot l_2m_2 & l_2m_3+l_3m_2 \\ l_1m_3+l_3m_1 & l_2m_3+l_3m_2 & 2\cdot l_3m_3 \end{pmatrix}$$

Can i see what the rank of $C$ is? I don't know where to start. Can anyone give me some hints please?

## 2 Answers

from $$x^TCx = 0$$ and $$C = l^Tm + m^Tl$$

we know $$x$$ lying on two lines, means $$R(Cx) =2$$

from $$R(AB) \le \min(R(A),R(B))$$

we know $$R(C) = 2$$

I tried to calculate the rank of the conic $C=ll^T$ from which i know it has rank 1. I used the characteristical polynom to calculate the eigenvalues of $C$. I got a polynom of the form $$-\lambda^3 +(l_1^2+l_2^2+l_3^2)\lambda^2$$ The polynom has the value 0 for $\lambda_1=\lambda_2=0$ and $\lambda_3=(l_1^2+l_2^2+l_3^2)$. Because just one of three lambdas are different from zero $C$ has rank 1. I assume it will be very similar for $C=lm^T+ml^T$