# Factorizing a polynomial.

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

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

• Does it help that the matrix could be symmetric? May 2, 2018 at 19:02
• @DietrichBurde my bad ,was a typo
– Raku
May 2, 2018 at 19:04
• For $n=3$, we get $x^2 (a_1^2 + a_2^2 + a_3^2 - x)$.
– lhf
May 2, 2018 at 19:06
• @MichaelMcGovern , A symmetric matrix is always diagonalizable is all i can think of.
– Raku
May 2, 2018 at 19:06
• 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.
– user551819
May 2, 2018 at 19:23

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 $(-1)^{n}x^{n-1}(x-(a_1^2+a_2^2+\cdots+a_n^2))$.

• I think there shouldn't be an x in the 3rd column of the matrix.
– Raku
May 2, 2018 at 19:50
• 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?
– Raku
May 2, 2018 at 20:12
• @Raku, all fixed, thanks.
– lhf
May 2, 2018 at 20:17