Prove the following determinant formula

i need to prove the following

$$\begin{bmatrix} 1+ x_1y_1 & x_1y_2 & \cdots & x_1y_n \\ x_2y_1 & 1+ x_2y_2 & \cdots & x_2y_n \\ \vdots & \vdots & \ddots & \vdots \\ x_ny_1 & x_ny_2 & \cdots & 1+x_ny_n \\ \end{bmatrix} = 1 + \sum_{i=1}^n{x_iy_i}$$

can anyone point me the first move?

• I think the first element on the second column is $x_1y_2$, isn't it? Feb 13, 2014 at 16:57
• Should it be $x_1y_2$ in the second column, first row? Feb 13, 2014 at 16:57
• yes, i fixed it Feb 13, 2014 at 16:58

Your matrix can be written as $I+xy^T$ where $I$ is the identity matrix and $x=(x_1,\dots,x_n)^T, y=(y_1,\dots,y_n)^T$. Then apply here.

The matrix $$A= \begin{bmatrix} x_1y_1 & x_1y_2 & \cdots & x_1y_n \\ x_2y_1 & x_2y_2 & \cdots & x_2y_n \\ \vdots & \vdots & \ddots & \vdots \\ x_ny_1 & x_ny_2 & \cdots & x_ny_n \\ \end{bmatrix}$$ can be written as $$A=\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix} \begin{bmatrix}y_1 & y_2 & \dots & y_n\end{bmatrix}$$ so it has rank $1$. It also has the eigenvalue $\lambda=x_1y_1+\dots+x_ny_n$, so its characteristic polynomial is $$p_A(X)=(0-X)^{n-1}(\lambda-X)$$ The determinant of $A+I$ is thus $$p_A(-1)=\lambda+1=1+\sum_{1\le i\le n}x_iy_i.$$

• Could you please explain to me how you got $p_A(X)$? Nov 23, 2017 at 17:35
• @user444042 The characteristic polynomial is the product of all factors $\mu-X$, for $\mu$ in the eigenvalues counted with their algebraic multiplicity. Nov 23, 2017 at 17:50
• Thank you. I got it now. Nov 23, 2017 at 17:54

In this answer, it is shown that if $A$ is an $n\times m$ matrix and $B$ is an $m\times n$ matrix, then $$\lambda^m\det(\lambda I_n-AB)=\lambda^n\det(\lambda I_m-BA)$$ Setting $\lambda=-1$ gives $$\det(I_n+AB)=\det(I_m+BA)$$ Setting $A=\left[\begin{array}{c}x_1\\x_2\\\vdots\\x_n\end{array}\right]$ and $B=\left[\begin{array}{c}y_1&y_2&\dots&y_n\end{array}\right]$ gives your result.

Hint: Just use induction, and use cofactor expansion along last row, along with elementary row operations

• assuming the statement is true for $n=k$ and trying to prove it for $n=k+1$, i can easily calculate the the determinant of the cofactor $x_{k+1} y_{k+1}$, but i dont know how to calculate the other grid positions Feb 13, 2014 at 17:19