Find the determinant of the following matrix Find the determinant of the following matrix:
$$A = \begin{bmatrix}
1+x_1^2 &x_1x_2  & ... & x_1x_n \\ 
 x_2x_1&1+x_2^2  &...  & x_2x_n\\ 
 ...& ... & ... &... \\ 
 x_nx_1& x_nx_2  &...  & 1+x_n^2
\end{bmatrix}$$
I computed for the case $n=2$, and $n=3$ and guessed that $\det(A)$ should be $ 1+\sum_{i=1}^n x_i^2 $  but not sure how to proceed for any $n$. 
 A: To expand on darij grinberg's comment, let
$$
X=A-I_n = \begin{bmatrix}
x_1^2 &x_1x_2  & ... & x_1x_n \\ 
 x_2x_1&x_2^2  &...  & x_2x_n\\ 
 ...& ... & ... &... \\ 
 x_nx_1& x_nx_2  &...  & x_n^2
\end{bmatrix}=(x_ix_j)_{1\leq i,j\leq n}
$$
Then all the lines of $X$ are multiples of $(x_1,x_2,\ldots,x_n)$ ; so
${\textsf{rank}}(X)\leq 1$. The eigenvalues of $X$ (counted with multiplicity)
are therefore $0,0,\ldots,0$ ($n-1$ times), plus some $\lambda\in{\mathbb R}$.
Since the trace of ​​$X$​​ equals the sum of its eigenvalues, we must have $\lambda={\textsf{trace}}(X)=\sum_{i=1}^n x_i^2$. Then
$X$ is similar to a triangular matrix with diagonal ${\textsf{diag}}(0,0,0,\ldots, 0,\sum_{i=1}^n x_i^2)$, so that
$A$ is similar to a triangular matrix with diagonal ${\textsf{diag}}(1,1,1,\ldots, 1,1+\sum_{i=1}^n x_i^2)$, whence
$$
{\textsf{det}}(A)=1+\sum_{i=1}^n x_i^2
$$
A: Here is a method without using (at least explicitely) the notion of eigenvalue. 
Call $f(x_1,\dots,x_n)$ the wanted determinant. 
View the last column as $$\pmatrix{0\\\vdots\\ 0\\1 }+x_n\pmatrix{x_1\\\vdots\\ x_{n-1}  \\x_n }.$$
This gives by linearity with respect to the last column,
$$f(x_1,\dots,x_n)=f(x_1,\dots,x_{n-1})+x_n\det\begin{bmatrix}
1+x_1^2 &x_1x_2  & ...&x_1x_{n-1}  & x_1 \\ 
 x_2x_1&1+x_2^2  &...  &x_2x_{n-1}& x_2\\ 
 ...& ... & ... &&... \\ 
 x_nx_1& x_nx_2  &...& x_nx_{n-1} & x_n
\end{bmatrix}$$
(the first determinant is computed by expanding with respect to the last column).
In the last determinant, do $C_i\leftarrow C_i-x_iC_n$, $1\leqslant i\leqslant n-1$ in order to show that 
$$\det\begin{bmatrix}
1+x_1^2 &x_1x_2  & ...&x_1x_{n-1}  & x_1 \\ 
 x_2x_1&1+x_2^2  &...  &x_2x_{n-1}& x_2\\ 
 ...& ... & ... &&... \\ 
 x_nx_1& x_nx_2  &...& x_nx_{n-1} & x_n
\end{bmatrix}=x_n.$$
Now we can conclude by induction.
A: Consider the eigenvalues of 
$x \cdot x^T + I$  
$x \cdot x^T v + v = \lambda v$
$x \cdot x^T v = \lambda v - v$
$v$ must be parallel to $x$ or ($\lambda$ = 1). wlog $v = x$  
$||x|| ^2 x= (\lambda - 1) x$
$\lambda = ||x||^2 + 1$  
You can use the fact that the determinant is the product of eigenvalues (there should be $n$ of them)
