Finding an odd determinant I'm looking for the determinant of a rather odd looking matrix, for scalars $x_i,y_i$ with $ 1\leq i \leq n$ the matrix $A_{n\times n}$ defined this way : $A_{ij} = x_i\cdot y_j$ except for when $i=j$ then $A_{ii} = 1+x_i\cdot y_i$.
Now simple testing seems to suggest that the determinant would turn out to be $1+\sum_{i=1}^n x_i\cdot y_i$. I also have a strong intuition towards proving an equality by induction. What I tried to do was to seperate the first row into two vectors and solving two separate determinants of the resulting matrices (in one I'm left with $1$ and a bunch of zero's and the other with $x_1\cdot y_i$ for $1 \leq i \leq n$.
Any tips would be appreciated.
Edit: I worked out the solution. It is indeed very viable by induction.
 A: Here's another solution which, from the sound of the OP's description of his/her work, at least appears to give a different view of things.
Write
$\mathbf x = \begin{pmatrix} x_1 \\ x_2 \\ . \\ . \\ . \\ x_n \end{pmatrix} \tag{1}$
and
$\mathbf y = \begin{pmatrix} y_1 \\ y_2 \\ . \\ . \\ . \\ y_n \end{pmatrix}; \tag{2}$
then
$A = I + \mathbf x \mathbf y^T, \tag{3}$
and we seek to show that $\det A = 1 + \mathbf y^T \mathbf x = 1 + \sum_1^n y_i x_i$.  Note that if $\mathbf y = 0$, then $A = I$, so we have $\det A = 1$ and we are done; a similar argument applies if $\mathbf x = 0$; we thus assume that $\mathbf x \ne 0 \ne \mathbf y$.  We examine the eigen-structure of $\mathbf x \mathbf y^T$.  Observe that
$(\mathbf x \mathbf y^T) \mathbf x = (\mathbf y^T \mathbf x) \mathbf x, \tag{4}$
showing that $\mathbf x \ne 0$ is an eigenvector of $\mathbf x \mathbf y^T$ with eigenvalue $\mathbf y^T \mathbf x$.  Since $\mathbf y \ne 0$, we must have $y_i \ne 0$ for some $i$.  Consider the vectors $\mathbf z_j$, $j \ne i$, with components $\mathbf z_{jk} = -\delta_{jk} + y_i^{-1}y_j \delta_{ik}$; then $\mathbf z_j \ne 0$ for all $j$ and $\mathbf y^T \mathbf z_j = \mathbf z_j^T \mathbf y = 0$ for all $j$ as well, since
$\mathbf z_j^T \mathbf y = \sum_{k = 1}^{k = n} (-\delta_{jk} + y_i^{-1}y_j \delta_{ik})y_k = -y_j + y_j = 0. \tag{5}$
Finally, the $\mathbf z_j$ are linearly independent, for if we take $n - 1$ scalars $\alpha_j$, $j \ne i$, we see that the components of $\sum_{j = 1, j \ne i}^n \alpha_j \mathbf z_j$ are given by $(\sum_{j = 1, j \ne i}^n \alpha_j \mathbf z_j)_k = -\alpha_k$ as long as $k \ne i$, so that $\sum_{j = 1, j \ne i}^n \alpha_j \mathbf z_j = 0$ forces $\alpha_j = 0$ for all $j \ne i$.  These arguments also establish that
$(\mathbf x \mathbf y^T) \mathbf z_j = 0, \; j \ne i, \tag{6}$
whence the $\mathbf z_j$ are a set of $n - 1$ linearly independent eigenvectors of $\mathbf x \mathbf y^T$, all with eigenvalue $0$.  Thus we see that $\mathbf x \mathbf y^T$ has $n - 1$ eigenvalues $0$ and one eigenvalue $\mathbf y^T \mathbf x$.  Things are finished off by exploiting the simple fact that for any square matrix $B$, $B \mathbf v= \lambda \mathbf v  \Leftrightarrow (B + \mu I)\mathbf v = (\lambda + \mu)\mathbf v$.  Applying this to $A = I + \mathbf x \mathbf y^T$ we see that the eigenvalues of $A$ are $1$ ($n - 1$ times) and $1 + \mathbf y^T \mathbf x$ (one time).  Multiplying the eigenvalues together shows that
$\det A = 1 + \mathbf y^T \mathbf x, \tag{7}$
as was hypothesized.  
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
