1
$\begingroup$

Assume that $A$ is an $n\times n$ real matrix whose entries are all $1$. Then how can we show the following determinant equality for any $x$?

$\det(A-xI)$=\begin{vmatrix} 1 -x & 1 & 1 & \cdots & 1 \\ 1 & 1 -x & 1 & \cdots & 1 \\ 1 & 1 & 1 -x & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & 1 -x \end{vmatrix}=

\begin{vmatrix} n-x & n-x & n-x & \cdots & n -x \\ 1 & 1 -x & 1 & \cdots & 1 \\ 1 & 1 & 1 -x & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & 1 -x \end{vmatrix}

Thanks.

$\endgroup$

closed as off-topic by Did, Davide Giraudo, Dragonemperor42, hardmath, Henrik May 20 '17 at 13:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Davide Giraudo, Dragonemperor42, hardmath, Henrik
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Presenting a bare problem statement without context is frowned upon. The context might consist of where you found a difficulty or what results in your studies you think are connected to the problem. This allows Readers to respond more cogently and expeditiously to your Question. $\endgroup$ – hardmath May 20 '17 at 13:43
8
$\begingroup$

Recall that if an $n\times n$ matrix $B$ is obtained from an $n\times n$ matrix $A$ by adding a scalar multiple of one row of $A$ to another row, then $\det(B)=\det(A)$.

In your problem the second matrix is obtained from the first by adding rows $2$ through $n$ to row $1$. So the determinants are the same.

$\endgroup$
5
$\begingroup$

Just add row 2, row 3,... up to row $n$ to row 1.

$\endgroup$
1
$\begingroup$

Note that $$ \begin{bmatrix} n-x & n-x & n-x & \cdots & n -x \\ 1 & 1 -x & 1 & \cdots & 1 \\ 1 & 1 & 1 -x & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & 1 -x \end{bmatrix} = \\ \begin{bmatrix} 1&1&1&\cdots&1\\ &1&0&\cdots&0\\ &&\ddots&\ddots&\vdots\\ &&&&0\\ &&&&1 \end{bmatrix} \begin{bmatrix} 1 -x & 1 & 1 & \cdots & 1 \\ 1 & 1 -x & 1 & \cdots & 1 \\ 1 & 1 & 1 -x & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & 1 -x \end{bmatrix} $$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.