Calculating $\det(A+I)$ for matrix $A$ defined by products Let $b_1,\ldots,b_n\in\mathbb{R}$. I have an $n\times n$ matrix $A$ whose entry is given by $a_{ij}=b_ib_j$, and I'd like to show that $\det(A+I)=\sum_{i=1}^nb_i^2+1$.
Define $b=(b_1,\ldots,b_n)$. I know that $Ab=\left(\sum_{i=1}^nb_i^2\right)b$, and $Ac=0$ for all $c$ such that $b\cdot c=0$. So all the eigenvalues of $A$ are $\sum_{i=1}^nb_i^2, 0, 0, \ldots, 0$. What can I do next?
 A: This is an example of the frequently recurring (on this site) questions of the form "what is the determinant of the sum of a square rank$~1$ matrix and a multiple of the identity matrix". In the question it is clear that $A$ has rank${}\leq1$; this would even be true if one had put $a_{i,j}=b_ic_j$ for different vectors $b,c$.
This general question is best replaced by the question by "what is the characteristic polynomial of a rank${}\leq1$ square matrix", which is easily seen to be equivalent by the definition of characteristic polynomial. The answer is very simple: if $A$ is a $n\times n$ matrix of rank${}\leq1$, then it's eigenspace for eigenvalue$~0$ has dimension at least $n-1$, so its characteristic polynomial is divisible by$~X^{n-1}$, and the sum of all eigenvalues is $\def\tr{\operatorname{tr}}\tr(A)$, so the characteristic polynomial of$~A$ is necessarily equal to $X^{n-1}(X-\tr(A))$.
Back to your concrete question, which asks to evaluate the characteristic polynomial of $-A$ at $X=1$. Since $\tr(-A)=-\sum_{i=1}^nb_i^2$, this gives $1^{n-1}(1-\tr(-A))=1+\sum_{i=1}^nb_i^2$. 
A: The matrix $A=b b^T$. It is easy to see that $A b = \|b\|^2 b$, so $A$ has an eigenvalue of $\|b\|^2$ corresponding to the eigenvector $b$.
Now suppose $b^T x = 0$. Then we see that $A x = 0$.
Hence $A$ has eigenvalues $0,....,0,\|b\|^2$.
It follows that $A+I$ has eigenvalues $1,...,1,
\|b\|^2+1$.
Since $\det C = \Pi_k \lambda_k$, where $\lambda_k$ are the eigenvalues of $C$, we see that $\det (A+I) = 1 + \|b\|^2 = 1+\sum_k b_k^2$.
