Calculating the determinant of this complicated matrix

Question

I am calculating the characteristic polynomial for this matrix:

$$A = \begin{pmatrix} 1 & 2 & \cdots & n \\ 1 & 2 & \cdots & n \\ \vdots & \vdots & \cdots & \vdots \\ 1 & 2 &\cdots & n \end{pmatrix}$$ First I was asked to figure out that $0$ is an eigenvalue, and since it is not invertible then $0$ is an eigenvalue, and its' geometric multiplicity is $n-1$. Now I need to calculate the characteristic polynomial but I am finding this determinant hard!

$$\mbox{det}\begin{pmatrix} \lambda - 1 & -2 & \cdots & -n \\ -1 & \lambda - 2 & \cdots & -n \\ \vdots & \vdots & \cdots & \vdots \\ -1 & -2 & \cdots & \lambda -n \end{pmatrix} = ? $$

Answer 1

The last eigenvalue $\lambda$ is the trace of the matrix $A$ so $$\lambda=1+2+\cdots+n=\frac{n(n+1)}{2}$$ hence the chararcteristic polynomial is $$\chi_A=x^{n-1}\left(x-\frac{n(n+1)}{2}\right)$$

Answer 2

Suggestion: add to the first column all the others. We get $\lambda-\frac{n(n+1)}2$ for each entry. Then use linearity with respect to this column, and finally do $L_j\leftarrow L_j+jL_1$.