# Show that the set of rank two matrices in $M_{2\times3}(\Bbb R)$ is open.

Consider the map $$f : M_{2\times3}(\Bbb R) \to \Bbb R^3$$ given by sending a matrix to the triple of its $$2\times 2$$ minors.

The set of rank $$2$$ matrices is the inverse image of the set $$\{f(x_1, x_2, x_3)\in \Bbb R^3 \mid (x_1,x_2,x_3) \neq (0,0,0)\}$$ This set is open in $$\Bbb R^3$$, hence the set of rank $$2$$ matrices is open if the map $$f$$ is continuous.

But how to prove that $$f$$ is continuous?

• I have MathJax-ed up your question. For next time, please refer to, for instance, this guide on how to write readable math on this site. – Arthur May 6 '19 at 7:36
• As to the problem itself, before you can talk about continuous, you have to be aware of the topology on both your spaces. Presumably you have the standard topology on $\Bbb R^3$. Is the topology on $M_{2\times3}(\Bbb R)$ given basically by the standard topology on $\Bbb R^6$ by way of the six entries of the matrices? – Arthur May 6 '19 at 7:38
• @Arthur, I suspect, the OP may not yet be familiar with the concept of topology. But I understand you wanted to ask them, what "continuous" means in that context. – avs May 6 '19 at 7:50

How about the map $$f : M_{2\times3}(\mathbb{R}) \to \mathbb{R}$$ that maps a matrix A to sum of determinant squares of all the possible $$2\times2$$ minors. Then, set of matrices of rank 2 is just the inverse image of $$\mathbb{R}\setminus{0}$$, which is open in $$\mathbb{R}$$. Note that this map is continuous because it is a polynomial in the entries of the matrix.