Let $\alpha, \beta,\gamma,\delta $ be the eigenvalues of the matrix find $\alpha ^2+\beta^2+\gamma^2+\delta^2 $

Question

Let $\alpha, \beta,\gamma,\delta $ be the eigen values of the matrix $$ A=\begin{pmatrix} 0 & 0 & 0& 0\\ 1 & 0 & 0 &-2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 &1 & 2 \end{pmatrix} $$

then $\alpha ^2+\beta^2+\gamma^2+\delta^2$ is ?

Matrix has determinant $|A|=0$

Thus one eigen value is zero. I assumed $\delta=0$.

$trace(A)=\alpha +\beta+\gamma=2$

$\alpha ^2+\beta^2+\gamma^2=(\alpha +\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\alpha \gamma)=4-2(\alpha \beta+\beta \gamma+\alpha \gamma)$

I am not sure how to figure out $2(\alpha \beta+\beta \gamma+\alpha \gamma)$ term.I am not going for long method of finding eigen values by characteristic polynomial since this question came in $2$ marks so I am trying to find out quick way to solve this problem.

Answer 1

You did most of the work!

Let $P(\lambda)$ be $A$'s characteristic polynomial. Then $\alpha \beta+\beta \gamma+\alpha \gamma$ is the coefficient of $\lambda^2$. To see this, write $$P(\lambda)=\lambda(\lambda-\alpha)(\lambda-\beta)(\lambda-\gamma)$$ If I'm correct, $P(\lambda)=\lambda^4-2\lambda^3-\lambda^2+2\lambda$. So $\alpha \beta+\beta \gamma+\alpha \gamma=-1$. So $$\alpha ^2+\beta^2+\gamma^2+\delta^2=6$$