Determinant of a magic square using eigen values

Question

Let M be a $3\times3$ integer matrix whose columns, rows, diagonal, and anti-diagonal each total $15$. $$\begin{bmatrix} 8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2 \end{bmatrix}$$ I intend to find the determinant using eigen values. Since row sum is $15$ each, I took one eigen value as $15$. But here trace is $15$. Does this mean that other two eigen values are $0$? I know that this is not possible since determinant is either $+/-360$. Where did I go wrong?

Answer 1

Since the sum of the eigenvalues is $15$ and one of the eigenvalues is $15$, the conclusion that you reach is that the sum of the remaing eigenvalues is $0$, not that each of them is $0$.

Answer 2

Yes, the other eigenvalues are not zeros because of the determinant! In general you do not conclude anything about the other eigenvalues, but your matrix, we know only the others are add up to zero.

For this matrix, you know $15$ is one eigenvalue, so we can say something about its characteristic polynomial $$\rho_A(x)=(x-15)(x^2+ax+b)=x^3-(\underbrace{\text{trace}}_{15} )x^2+(\underbrace{A_{11}+A_{22}+A_{33}}_{-24}))x-\underbrace{\det}_{-360} $$

From this we have $b=-24$ and $a=0$ and we have an equation $x^2-24=0$ to see the other eigenvalues are $\pm 2 \sqrt{6}$