# Showing that the solution set of a system of linear equations is an affine set

I'm currently studying convex optimization using the textbook written by Boyd. I came across an example where the author demonstrates how an equivalent way to express affine sets is that they are a solution set for a system of linear equations.

The example given in the book is in two variables (i.e. $$x_1$$ and $$x_2$$). I was attempting to prove this using the same method used by the author, but this time in $$k$$ variables. I'm not sure if my approach is correct, however as it seems a bit naive and was hoping to get some feedback on it.

My Approach:

Suppose that $$C = \{x\ |\ Ax = b\}$$, $$x_1, \cdots , x_k \in C$$, and $$\theta_1, \cdots , \theta_k \in \Bbb{R}$$ with $$\sum_{i = 1}^k \theta_i = 1$$. Since each $$x_i$$ is in $$C$$, we can say that $$Ax_k = b$$. We want to show that $$A(\theta_1 x_1 + \theta_2 x_2 + \cdots + \theta_k x_k) = b$$.

We could rewrite the last equation as

$$A(\theta_1 x_1 + (1 - \theta_1)(\frac{\theta_2}{1 - \theta_1}x_2 + \cdots + \frac{\theta_k}{1 - \theta_1} x_k))$$

which is

$$\theta_1 Ax_1 + (1 - \theta_1)A(\frac{\theta_2}{1 - \theta_1}x_2 + \cdots + \frac{\theta_k}{1 - \theta_1}x_k)$$

We can conclude that $$A(\frac{\theta_2}{1 - \theta_1}x_2 + \cdots + \frac{\theta_k}{1 - \theta_1}x_k) \in C$$ because each $$x_i$$ is in $$C$$ and the coefficients are greater than $$0$$ and sum up to $$1$$. Thus, $$A(\frac{\theta_2}{1 - \theta_1}x_2 + \cdots + \frac{\theta_k}{1 - \theta_1}x_k) = b$$.

If we put these values into the previous equation we get

$$\theta_1 Ax_1 + (1 - \theta_1)A(\frac{\theta_2}{1 - \theta_1}x_2 + \cdots + \frac{\theta_k}{1 - \theta_1}x_k) = \theta_1 b + (1 - \theta_1)b = b$$

Thus we can conclude that this set is affine.

Is my approach correct? I'm not sure if the conclusion I drew that $$A(\frac{\theta_2}{1 - \theta_1}x_2 + \cdots + \frac{\theta_k}{1 - \theta_1}x_k) = b$$ is acceptable or not.

Thank you.

• You are using mathematical induction, so I suggest that you mention that explicitly at an appropriate place in the proof. But let me give you a hint for what could be a simpler proof: In $A(\theta_1 x_1 + \cdots \theta_k x_k)$, All the $\theta_i$ are scalars, and each $x_i \in C$. So how can you rewrite this expression using the former, and what can you conclude from the latter (look at the definition of $C$ again if it helps)? – M. Vinay Apr 2 at 2:14