Determine The Sign Of Eigenvalue? Consider a matrix of the form
$\begin{bmatrix}0 & a & b\\
c & 0 & 0\\
0 & d & 0
\end{bmatrix}$
Where a,b,c and d are positive real numbers. Suppose the matrix A has three distinct real eigenvalues. What can you say about the signs of the eigenvalues? (How many of them are positive, negative, zero?) Is the eigenvalue with the largest absolute value positive or negative?
$\mathbf{What}$ $\mathbf{I}$ $\boldsymbol{Know}$$\Longrightarrow$I
found the solution In a book .
Author conluded Trace is Zero from nowhere and made His easy way.I
dont know how ??.Please Help.
 A: The product of the eigenvalues is $\det A=bcd$. It's negative, so there's an even number of negative roots ($0$ or $2$).
Their sum is $\operatorname{Tr}A=0$, so the eigenvalues cannot all have the same sign.
As a conclusion, one eigenvalue is  positive and two negative.
A: Determinant of a matrix is the product of its eigenvalues and Trace of a matrix is the sum of its eigenvalues.
If you observe $Det(A) = bcd >0$ and $Trace(A) =0 $ implying that product of three eigenvalues is $>0$ implying that there can be all three positive eigenvalues or there can be one positive and two other negative eigenvalues.
Also see that none of the eigenvalues is zero(from the determinant value).
Now the former case(three positive eigenvalues) cannot happen since from $trace(A) = 0$ or sum of eigenvalues equal to zero and if the three eigevalues are positive(former case ) then their sum cannot be zero so first case is excluded.
Next we see that the remaining case is two eigenvalues negative and one eigenvalue positive and there is a chance that $trace(A) =0$ ,so $A$ has two negative eigenvalues and one positive eigenvalue.
Sinc ethe trace is zero and there are two negative eigenvalues and one positive eigenvalue so the positive eigenvalue is equal to the sum of the absolute values of each negative eigenvalue implying the positive eigenvalue is the largest one .   
A: Hint: Let $A$ be a $N\times N$ matrix and $\lambda_n$ be the $n^{\text{th}}$ eigenvalue, then we have the following equalities (they are related to Vietas formulas and proving them for $N=3$ is simple):
$$\det A =\prod_{n=1}^{N}\lambda_n$$
$$\operatorname{trace} A=\sum_{n=1}^{n}\lambda_n.$$
