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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?

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  • $\begingroup$ My first thought might be to do a diagonalization since $A$ is symmetric. $\endgroup$ – Cameron Williams Nov 23 '13 at 7:31
  • $\begingroup$ Consider the characteristic polynomial $\det(A-xI)$ and set $x=-1$? I suppose that would give $\pm$ of what you want though. $\endgroup$ – Casteels Nov 23 '13 at 7:41
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    $\begingroup$ Put aside the issue of continuing your argument, note that the problem statement is a special case of the formula that $\det(I_m+A_{m\times n}B_{n\times m})=\det(I_n+B_{n\times m}A_{m\times n})$, which follows from the fact that $$ \pmatrix{I_m&0\\ B&I_n}\pmatrix{I_m+AB&A\\ 0&I_n}\pmatrix{I_m&0\\ -B&I_n}=\pmatrix{I_m&A\\ 0&I_n+BA}. $$ $\endgroup$ – user1551 Nov 23 '13 at 7:49
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    $\begingroup$ This result also follows from the more general fact that $AB$ and $BA$ have the same non-zero eigenvalues. $\endgroup$ – copper.hat Nov 23 '13 at 7:58
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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 $\det (A+I) = 1 + \|b\|^2 = 1+\sum_k b_k^2$.

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  • $\begingroup$ For the final "note the parentheses", they are unnecessary. The precedence (to the right) of "$\sum$" is higher than that of "$+$" and "$-$" (and lower than that of "$\times$" in its explicit of suppressed form, and of "$/$"). The same goes (although this is less universally agreed upon) for all large operators like "$\prod$", "$\int$", "$\bigoplus$", "$\coprod$". See (the final part of) this answer. $\endgroup$ – Marc van Leeuwen Nov 25 '13 at 7:54
  • $\begingroup$ @MarcvanLeeuwen: Thanks for pointing that out, I have rearranged the text and side-stepped the precedence issue. (Actually, my initial answer was incorrect and the parentheses were relevant, and I never deleted the comment after the fix.) $\endgroup$ – copper.hat Nov 25 '13 at 8:03
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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 $1^{n-1}(1-\tr(-A))=1+\sum_{i=1}^nb_i^2$.

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