$y=Xb+e$ is a (General) linear model with $n=p=3$. Given that $b_3$ is non estimable, and $b_1+ 2b_2 + 2b_3$ is estimable. How can I find an $X$ which is consistent with the above?
I think I have a partial solution to it but I do not understand the methodology of solving it.
My questions
1) How can we show/prove that $\lambda_1\notin C(X^T)$ but $\lambda_2 \in C(X^T)$?
and lastly,
2) How does the design matrix $X$ looks and why ?
So I know
$$b = \begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix}$$
Now since $b_3$ is non estimable : $b_3 = λ_1^T b$
$\lambda_1 = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$ and $\lambda_1\notin C(X^T)$ (the column space of $X^T$).
and since $b_1 +2b_2 + 2b_3$ is estimable then $$b_1 +2b_2 + 2b_3 = \lambda_2^T b$$
$\lambda_2 = \begin{bmatrix} 1\\ 2\\ 2 \end{bmatrix}$ and $\lambda_2$ is in $C(X^T)$.
We know probably that $X$ is not full column rank.
I was thinking to use $X^T b = \lambda_1$ , $X^T b = \lambda_2 $ to find $X$ matrix but I am not sure how to proceed.