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Factorize the polynomial

$$P(x)= \begin{vmatrix} a_1^2-x & a_{1}a_2 & a_1a_3 & \cdots & a_1a_n \\ a_2a_1 & a_2^2-x & a_{2}a_3 & \cdots & a_2a_n\\ a_3a_1 & a_3a_2 & a_3^2-x & \cdots & a_3a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_na_1 & a_na_2 & a_na_3 & \cdots & a_n^2-x\\ \end{vmatrix}$$

I was thinking of getting the eigenvalues for the matrix , however i couldn't find a way to determine them.

I also tried to get the determinant for n=2 :

determinant = $X(X-(a_1^2+a_2^2))$

However that alone is not sufficient to generalize the determinant

Any hints or advices appreciated.

I would also be grateful if anyone would know somewhere where i can tackle similar problems .

Thanks in advance.

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  • $\begingroup$ Does it help that the matrix could be symmetric? $\endgroup$ May 2, 2018 at 19:02
  • $\begingroup$ @DietrichBurde my bad ,was a typo $\endgroup$
    – Raku
    May 2, 2018 at 19:04
  • $\begingroup$ For $n=3$, we get $x^2 (a_1^2 + a_2^2 + a_3^2 - x)$. $\endgroup$
    – lhf
    May 2, 2018 at 19:06
  • $\begingroup$ @MichaelMcGovern , A symmetric matrix is always diagonalizable is all i can think of. $\endgroup$
    – Raku
    May 2, 2018 at 19:06
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    $\begingroup$ If you put $x=0$, then the matrix is equal to $(a_1,...,a_n)^T\cdot(a_1,...,a_n)$, which is a rank $1$ matrix. Therefore, it has an $(n-1)$-dimensional kernel. Hence the eigenspace of $0$ is $(n-1)$-dimensional. Therefore $x^{n-1}$ is a factor of the determinant. The other factor must be linear. So, you just need to prove that $a_1^2+...+a_n^2$ is a root too. $\endgroup$
    – user551819
    May 2, 2018 at 19:23

1 Answer 1

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The determinant is the characteristic polynomial of the matrix $$ \begin{pmatrix} a_1^2 & a_{1}a_2 & a_1a_3 & \cdots & a_1a_n \\ a_2a_1 & a_2^2 & a_{2}a_3 & \cdots & a_2a_n\\ a_3a_1 & a_3a_2 & a_3^2 & \cdots & a_3a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_na_1 & a_na_2 & a_na_3 & \cdots & a_n^2\\ \end{pmatrix} =(a_1,...,a_n)^T\cdot(a_1,...,a_n) $$ As mentioned in the comments, this matrix has rank $1$ and so has a kernel of dimension $n-1$. Thus, $0$ is an eigenvalue of multiplicity $n-1$.

The cases $n=1,2,3$ suggest that $a_1^2+a_2^2+\cdots+a_n^2$ is an eigenvalue with eigenvector $(a_1, a_2, \dots, a_n)$, which is easy to verify.

Therefore, the characteristic polynomial of the matrix is $(-1)^{n}x^{n-1}(x-(a_1^2+a_2^2+\cdots+a_n^2))$.

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  • $\begingroup$ I think there shouldn't be an x in the 3rd column of the matrix. $\endgroup$
    – Raku
    May 2, 2018 at 19:50
  • $\begingroup$ And i'm not quite sure where did the $(-1)^{n-1}$ come from , is it because $-x$ is multiplied $n-1$ times when calculating the determinant? $\endgroup$
    – Raku
    May 2, 2018 at 20:12
  • $\begingroup$ @Raku, all fixed, thanks. $\endgroup$
    – lhf
    May 2, 2018 at 20:17

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