# Equivalent Definitions of Lines in Projective Space

I’ve been working with two definitions of lines in $$\mathbb{P}_\mathbb{R}^2$$, and tried to show their equivalence.

The first is that, given two points $$a=(a_0:a_1:a_2)$$ and $$b=(b_0:b_1:b_2)$$, the line between them is given by $$\{ua+vb:u,v\in\mathbb{R}\}$$.

The second is $$\{(X_0:X_1:X_2):k_0X_0+k_1X_1+k_2X_2=0\}$$ for some $$k_i\in\mathbb{R}$$ not all $$0$$.

Given two distinct points $$a$$ and $$b$$, we can use simple linear algebra to find $$k_i$$ such that our points lie on that line.

However I've been struggling to show the converse. That is, given two distinct points $$a=(a_0:a_1:a_2)$$ and $$b=(b_0:b_1:b_2)$$ such that $$k_0a_0+k_1a_1+k_2a_2=0$$ and $$k_0b_0+k_1b_1+k_2b_2=0$$ for some $$k_i\in\mathbb{R}$$ not all $$0$$, then for any point $$c=(c_0:c_1:c_2)$$ such that $$k_0c_0+k_1c_1+k_2c_2=0$$ we should be able to write $$c=ua+vb$$ for some $$u,v\in\mathbb{R}$$.

Perhaps I'm overlooking something simple, but any help would be much appreciated.

• I don't think your definitions are equivalent as stated: take $k_0 = k_1 = k_2 = 0$ and you see that all of $\mathbb{P}^2$ lies in your second 'line'. I'm sure that can be easily corrected, however. – bounceback Jan 10 at 18:01
• Edited, thanks. – Dave Jan 10 at 18:07
• Hints: if $a$ and $b$ are distinct, then their coordinate tuples are linearly independent. What is the dimension of the orthogonal complement of $(k_0,k_1,k_2)$? – amd Jan 10 at 18:51

Since $$(k_0,k_1,k_2$$) isn't in its own orthogonal complement it can't be of dimension $$3$$. Then because both $$a$$ and $$b$$ are, and are linearly independent, it must be of dimension $$2$$ with $$a$$ and $$b$$ as a basis. Then since $$c$$ is in the orthogonal complement it must be expressible in terms of $$a$$ and $$b$$, and we can use simple linear algebra to find the coefficients.