Show that no such vector exists.

Assume you have $p+1$ real numbers $\beta_0,\beta_1,\ldots\beta_p$. Let $i \in \{1\ldots,p\}.$ I want to show that there does not exist a vector $a$ so that

$$\beta_i=a^T\begin{bmatrix}\beta_0+\beta_1 \\\beta_0+\beta_2 \\\vdots\\\beta_0+\beta_p \end{bmatrix},$$

for all $\beta_0,\beta_1,\ldots,\beta_p$.

attempt:

One way to show this is to find a counterexample. But I think that when we introduce specific numbers for the $\beta$'s we are able to solve it. We must use that fact that it should hold for all $\beta$'s?

Another way is that it seems likely that if the property should hold we must have that $a_i=1$ and we can work from there, however this is not a stringent proof that $a_i=1$.

Any tips?

It can be rewritten as \begin{align} \left[\begin{array}{c} \beta_0\\ \vdots \\\beta_p \end{array}\right] = \left[\begin{array}{ccc} \beta_0+\beta_1 & \dots & \beta_0+\beta_p \\ \vdots & \vdots & \vdots \\ \beta_0+\beta_1 & \dots & \beta_0+\beta_p \\ \end{array}\right] \left[\begin{array}{c} a_1\\ \vdots \\a_p \end{array}\right] \end{align} where the matrix is rectangular $(p+1)\times p$ and has rank $1$. This can have solution only if $\beta_0=\beta_1=\dots=\beta_p$. Otherwise you can substract two rows with different values of $\beta_i$ and you immediately get a contradiction. If $\beta_0=\beta_1=\dots=\beta_p$ then it has an infinite number of solutions.