TITLE: Find the determinant of the following matrix
QUESTION [6 upvotes]: 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 $n$.

REPLY [3 votes]: 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.