Proving linear in $\langle a_a, a_a \rangle =
\begin{cases}
\frac{1}{9},  & \text{if $a$ $\ne$ $a$}\
\end{cases}$
 A: To begin with, let us consider the linear combination
\begin{align*}
\alpha_{1}a_{1} + \alpha_{2}a_{2} + \ldots + \alpha_{k}a_{k} = 0
\end{align*}
If we take the inner product of such linear combination with $a_{1}$ we get the equation
\begin{align*}
2\alpha_{1}b + \alpha_{2} + \alpha_{3} + \ldots + \alpha_{k} = 0\\\\
\end{align*}
If we apply such inner product to $a_{j}$, where $2\leq j\leq k$, we get the following system of equations:
\begin{align*}
\begin{bmatrix}
2b & 1 & \cdots & 1\\
1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 2b
\end{bmatrix}
\begin{bmatrix}
\alpha_{1}\\
\alpha_{2}\\
\vdots\\
\alpha_{k}
\end{bmatrix} =
\begin{bmatrix}
0\\
0\\
\vdots\\
0
\end{bmatrix}
\end{align*}
Such system of linear equations has only the trivial solution iff the determinant of the coefficient matrix is different from zero. Based on such result, we need to determine its value in terms of $b$ and $k$:
\begin{align*}
\begin{vmatrix}
2b & 1 & \cdots & 1\\
1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 2b
\end{vmatrix} = 
\begin{vmatrix}
2b + k - 1 & 1 & \cdots & 1\\
2b + k - 1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
2b + k - 1 & 1 & \cdots & 2b
\end{vmatrix} & =
(2b + k - 1)
\begin{vmatrix}
1 & 1 & \cdots & 1\\
1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 2b
\end{vmatrix}\\\\
& =(2b + k - 1)
\begin{vmatrix}
1 & 0 & \cdots & 0\\
1 & 2b - 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
1 & 0 & \cdots & 2b - 1
\end{vmatrix}\\\\
& = (2b+k-1)(2b-1)^{k-1} \neq 0
\end{align*}
and we are done.
Hopefully this helps!
A: *

*Let $A$ be the $n \times k$ matrix whose columns are $a_i$. Linear independence of the columns of $A$ is equivalent to $\text{rank}(A)=k$.

*You are given that
$$A^\top A = \frac{1}{2} J_k + \left(b-\frac{1}{2}\right) I_k,$$
where $I_k$ is the $k \times k$ identity matrix, and $J_k$ is the $k \times k$ "all ones" matrix. (Phrased differently, $A^\top A$ has diagonal entries equal to $b$, and all other entries equal to $1/2$.)

*$\text{rank}(A) = \text{rank}(A^\top A)$

*Thus, "$\text{rank}(A) = k$" is equivalent to "$A^\top A$ is invertible," which is equivalent to "zero is not an eigenvalue of $A^\top A$."

*The eigenvalues of $J_k$ are $0$ and $k$, so the eigenvalues of $A^\top A$ are $b - \frac{1}{2}$ and $b - \frac{1}{2} + \frac{k}{2}$.

