Let $I_n, J_n \in M_n(\mathbb{R})$ where $I_n$ is the identity and $(J_n)_{ij} = 1$ for all $i,j$.
(1) Calculate the determinant, eigenvalues and eigenvectors of $I_n$ and $J_n$.
(2) Calculate the eigenvalues and eigenvectors of $A_{a,b} = aI_n + bJ_n$, where $a,b \in \mathbb{R}$. Note that $A_{a,b}$ is symmetric.
(3) Calculate the determinant of $A_{a,b}$.
If $X = \{x_1,...,x_m\}$ and $A = \{A_1,...,A_n\}$ is a family of subsets of $X$, the Incidence Matrix $B$ of $A$ is defined by $B_{ij} = 1$ if $x_j \in A_i$ and $B_{ij} = 0$ if $x_j \not\in A_i$. What is the meaning of $BB^T$? Calculate the eigenvalues of $BB^T$.
(1) For $I_n$:
$\det I_n = 1$;
The only eigenvalue is $1$, since $I_nh = h$ for all $h$;
The eigenvectors are $e_1,...,e_n$, since for each $h$, $h = a_1e_1 + \cdots + a_ne_n$.
For $J_n$:
$\det J_n = 0$;
We have that $\det(J_n - \lambda I_n) = \pm\lambda^{n} \mp n\lambda^{n-1} = \pm\lambda^{n-1}(\lambda - n)$, so the eigenvalues are $0$ with multiplicity $n-1$ and $n$ with multiplicity $1$.
If $\lambda = 0$, then
$$(J_n - 0I_n)h = (0,...,0)^T \Longrightarrow (h_1 + \cdots + h_n,0,...,0)^T = (0,0,...,0)^T.$$
So, $h = (h_1,...,h_n)^T = (-h_2 - \cdots -h_n, h_2,...,h_n)^T$. Thus, the eigenvalues are $\{(-1,1,0,...,0)^T, (-1,0,1,0,...,0)^T,...,(-1,0,...,1,0)^T,(-1,0,...,0,1)^T\}$.
If $\lambda = n$, then
$$(J_n - nI_n)h = (0,...,0)^T \Longrightarrow h_1 = \cdots = h_n$$
from where the eigenvector is $(1,...,1)^T$.
(2) By definition $(A_{a,b})_{ij} = a + b$ if $i=j$ and $(A_{a,b})_{ij} = b$ if $i \neq j$. We have that
$$\det(A_{a,b} - \lambda I_n) = \pm(\lambda - a)^{n-1}(\lambda - a - nb).$$
Then, the eigenvalues are $a$ with multiplicity $n-1$ and $a+nb$ with multiplicity $1$.
For $\lambda = a$:
$$(A_{a,b} - aI_n)h = 0 \Longrightarrow h_1 = -h_2 - \cdots - h_n.$$
Thus the eigenvectors are $\{(-1,1,0,...,0)^T, (-1,0,1,0,...,0)^T,...,(-1,0,...,1,0)^T,(-1,0,...,0,1)^T\}$
For $\lambda = a + nb:$
$$(A_{a,b}-(a+nb)I_n)h = 0 \Longrightarrow h_1 = \cdots = h_n.$$
Thus the eigenvalue is $(1,...,1)^T$.
(3) $\det A_{a,b} = a^{n-1}(a + nb)$.
Question 1. I cannot prove that $$\det(J_n - \lambda I_n) = \pm \lambda^{n-1}(\lambda - n)$$ $$\det(A_{a,b} - \lambda I_n) = \pm(\lambda - a)^{n-1}(\lambda - a - nb)$$ for the general case. I only check for some small $n$. My first attempt was induction and cofactor expansion, but I cannot conclude.
Question 2. Taking $\lambda = 0$ we get $\det(A_{a,b}) = \pm(-a)^{n-1}(-a-nb)$, but some calculations show that $\det(A_{a,b}) = a^{n-1}(a + nb)$. Why this difference? Is there some mistake?
Question 3. I don't know the meaning of $BB^T$. Moreover, since $B$ depends on the connections between $x_j$ and $A_i$, I cannot see a viable way to calculate the eigenvalues. I think that there is a connection with the previuous itens, but I cannot see.
I apologize for the long question.
Question 1: Consider the matrix $J_n$. It has rank 1, thus $0$ is an eigenvalue with multiplicity at least $n-1$.
Since the $n=tr(J_n)=\lambda_1+...+\lambda_n$ it follows that the last eigenvalue of $J_n$ is $n$.
Then $$\det(xI_n-J_n)=(x-\lambda_1)...(x-\Lambda_n)=x^{n-1}(n-n)$$
Alternate solution $$\det(J_n - \lambda I_n)= \det \begin{bmatrix} 1-\lambda & 1 & 1 &1 ...&1 \\ 1& 1-\lambda & 1 &1 ...&1 \\ ...&...&...&...&...&\\ 1& 1 & 1 &1 ...&1-\lambda \\ \end{bmatrix} $$
Add all rows to the first $$\det(J_n - \lambda I_n)= \det \begin{bmatrix} n-\lambda & n-\lambda & n-\lambda &n-\lambda ...&n-\lambda \\ 1& 1-\lambda & 1 &1 ...&1 \\ ...&...&...&...&...&\\ 1& 1 & 1 &1 ...&1-\lambda \\ \end{bmatrix}\\= (n-\lambda) \det \begin{bmatrix} 1 &1 & 1 &1 ...&1\\ 1& 1-\lambda & 1 &1 ...&1 \\ ...&...&...&...&...&\\ 1& 1 & 1 &1 ...&1-\lambda \\ \end{bmatrix} $$
Now, subtract the first row from the rest: $$\det(J_n - \lambda I_n)=(n-\lambda) \det \begin{bmatrix} 1 &1 & 1 &1 ...&1\\ 0& -\lambda & 0 &0 ...&0 \\ ...&...&...&...&...&\\ 0& 0 & 0 &0 ...&-\lambda \\ \end{bmatrix}=(n-\lambda)(-\lambda)^{n-1} $$
Then $$\det(A_{a,b} - \lambda I_n) = \det(b \left(J_n -\frac{(\lambda-a)}{b} I_n\right)=b^n (n-\frac{(\lambda-a)}{b})(-\frac{(\lambda-a)}{b})^{n-1}$$ by the above with $\lambda$ replaced by $\frac{(\lambda-a)}{b}$.
Question 2: $$(-a)^{n-1}(-a-nb)=(-1)^{n-1} a^{n-1} (-1)(a+nb)= (-1)^n a^{n-1}(a+nb)$$ which is consistent with what you get in (a) is you write properly the choice of $\pm$.
Question 3 $B^T$ means the transpose matrix. $BB^T$ mean $B$ multiplied with its transpose. The problem is expecting you to realize what this means in terms of vertices/edges of the graph, if you figure this you'll likely know how to solve the last part.