Prove the following determinant formula

Question

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? — Jlamprong, Feb 13 '14 at 16:57

score 2 · Answer 1 · edited Apr 13 '17 at 12:20

2

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.

edited Apr 13 '17 at 12:20

Community

1

answered Feb 13 '14 at 17:13

Jlamprong

1,837

score 2 · Answer 2 · answered Feb 13 '14 at 17:16

2

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 $$ p_A(-1)=\lambda+1=1+\sum_{1\le i\le n}x_iy_i. $$

answered Feb 13 '14 at 17:16

egreg

238,574

Could you please explain to me how you got $p_A(X)$? – user444042 Nov 23 '17 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. – egreg Nov 23 '17 at 17:50
Thank you. I got it now. – user444042 Nov 23 '17 at 17:54

score 2 · Answer 3 · edited Apr 13 '17 at 12:20

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.

meta_warrior · Answer 4 · 2014-02-13T17:12:02.957

0

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

edited Feb 13 '14 at 17:12

answered Feb 13 '14 at 17:01

meta_warrior

3,288
18
34

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 – guynaa Feb 13 '14 at 17:19

Prove the following determinant formula

4 Answers4