# How to tackle this polynomial given as a determinant? [closed]

Let $$p (x) = \begin{vmatrix} 1 & x & x & \dots & x & x \\ x & 1 & x & \dots & x & x \\ x & x & 1 & \dots & x & x \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ x & x & x & \dots & 1 & x \\ x & x & x & \dots & x & 1 \\ \end{vmatrix} .$$

How to find the (multiple) zeroes, the degree of the polynomial and the initial coefficient, all depending on the natural number $b$?

• What $b$?There is no $b$ in your question. – Alex M. Mar 28 '16 at 11:33

Add all columns (from second one) to the first one and then substract first row from all the rest:

$$\begin{vmatrix}1+(n-1)x&x&x&\ldots&x\\0&1-x&0&\ldots&0\\0&0&1-x&\ldots&0\\..&..&..&..&..\\0&0&0&\ldots&1-x\end{vmatrix}=\left(1+(n-1)x\right)(1-x)^{n-1}$$

Hint: You can determine the characteristic polynomial of the corresponding matrix. Note that the matrix is symmetric and therefore diagonalizable. Now $-x + 1$ is an eigenvalue with a geometric multiplicity of at least $n - 1$ and the sum of all eigenvalues [according to their algebraic multiplicity] is equal to the trace of the matrix.

Once you've calculated the characteristic polynomial you can determine $p(x)$ by plugging in $\lambda = 0$.

• n-1 zeros are only 1 ? – Algebra 2015 Mar 28 '16 at 11:06
• I don't understand your question. – Dominik Mar 28 '16 at 11:08
• I do not understand, which numbers are zeros ? – Algebra 2015 Mar 28 '16 at 11:11
• You first need to calculate the characteristic polynomial, then you will be able to calculate the zeroes of $p$. – Dominik Mar 28 '16 at 11:14
• Actually, you don't really need to exhibit the characteristic polynomial, as the determinant is simply the product of eigenvalues (but both ways to view this are equivalent, of course). @Algebra2015 see also this: math.stackexchange.com/questions/507641/… – xxx Mar 28 '16 at 11:51