# Determinant calculation

Prove: $$\det\left[ \begin{array}{cccc} 1+x_1y_1 & x_1y_2 & \cdots & x_1y_n\\ x_2y_1 & 1+x_2y_2 & \cdots & x_2y_n \\ \vdots & \vdots & \ddots & \vdots \\ x_ny_1 & x_ny_2 & \cdots & 1+x_ny_n \\ \end{array} \right]=1+\sum_{i=1}^{n}x_iy_i$$

I tried to do some elementary operations, and develop by first row, but couldn't get further.

Need to be proven without using eigenvalues.

Any help appreciated.

• I am fairly sure you can modify math.stackexchange.com/a/2112473 to solve the more general question of computing $\det \left( \begin{array}{cccc} a_1 & x_1 y_2 & \cdots & x_1 y_n \\ x_2 y_1 & a_2 & \cdots & x_2 y_n \\ \vdots & \vdots & \ddots & \vdots \\ x_n y_1 & x_n y_2 & \cdots & a_n \end{array} \right)$. – darij grinberg Feb 26 '17 at 15:34

The determinant can be written under the form:

$\det(I_n+UV^T)$ where $U=\left(\begin{array}{c}x_1\\x_2\\ \vdots \\x_n\end{array}\right)$ and $V=\left(\begin{array}{c}y_1\\y_2\\ \vdots \\y_n\end{array}\right).$

Let us recall that the matrix determinant lemma (https://en.wikipedia.org/wiki/Matrix_determinant_lemma) says that

$$\det(A+UV^T)=(1+V^TA^{-1}U) \det(A)$$

Taking $A=I$ gives:

$$\det(I_n+UV^T)=1+V^TU$$

which is the desired result.

Use $n$-linearity with respect to the first column to put the first $1$ alone on the first column.

• Could you please explain? – Itay4 Feb 26 '17 at 15:54

The matrix $A=(x_1,\dots,x_n)^T(y_1,\dots,y_n)$ is rank 1, so its characteristic polynomial is $X^n-\text{tr}(A)X^{n-1}$. Put $X=-1.$

Question has changed since this answer!

• You should delete this answer as you have given another one. – Jean Marie Feb 27 '17 at 8:11

If you just take a deep breath and calculate it will come out!

For after all, the answer is clearly going to be a sum of terms each of which is a product of $k$ $x$'s and $k$ $y$'s, for $k=0,1,\dots,n$. Every such term will arise as a product of $n-k$ $1$'s from the diagonal, multiplied by the top degree term of its cofactor, that is, multiplied by $\det\hat{x}^T\hat{y}$, where $\hat{x}$ is just the $k$-vector got from some $k$ of the indices. But now this little matrix is rank $1$ and so singular for all $k\not=0,1$. The only non-zero terms are then $1$ and $\sum x_iy_i$.