How do you quickly find the eigenvalues of this matrix? I have a final exam tomorrow, am sure a 3x3 eigen value problem like the one below is there.
But I find it very hard to find eigen values without zeros in the matrix
Show me how you do it quickly so that I can apply it tomorrow;
thanks

 A: I would not exactly call this quick, but a few tricks could simplify the steps:


*

*We have $\det(A)=\det(A^T)$

*If you add/subtract multiples of one row to another the determinant remains unchanged

*(1)+(2) means that you can add/subtract multiples of one column to another "for free" too

*You may want to modify the signs of a row or column which will change the sign of the determinant, but since we are trying to determine when $\det(A-\lambda I)=0$ we do not mind whether the sign has changed or not

*We do not need it for this example, but you could also easily scale a row or column by any factor $r\neq 0$ since $r\cdot\det(A-\lambda I)=0\iff\det(A-\lambda I)=0$. In fact (4) follows directly from this using $r=\pm 1$


so with all this in place we have
$$
\begin{align}
\begin{vmatrix}32-\lambda&-12&-2\\66&-25-\lambda&-4\\54&-21&-2-\lambda\end{vmatrix}&=\begin{vmatrix}32-\lambda&12&2\\66&25+\lambda&4\\54&21&2+\lambda\end{vmatrix}\\
\quad\\
&=\begin{vmatrix}32-\lambda&0&2\\66&1+\lambda&4\\54&9-6\lambda&2+\lambda\end{vmatrix}
\end{align}
$$
where the first equality was produced changing signs of the last two columns and the last equality is due to having subtracted $6$ times the last column from the middle column. Thus we got rid of annoying negative signs and are down to having to compute only two minors:
$$
\begin{vmatrix}1+\lambda&4\\9-6\lambda&2+\lambda\end{vmatrix}=\lambda^2+27\lambda-34
$$
and
$$
\begin{vmatrix}66&1+\lambda\\54&9-6\lambda\end{vmatrix}=540-450 \lambda
$$
to finally arrive at
$$
\begin{align}
\det(A-\lambda I)&=(32-\lambda)(\lambda^2+27\lambda-34)+2(540-450 \lambda)\\
&=-\lambda^3+5\lambda^2-2\lambda-8
\end{align}
$$
This is still tedious work! BTW the zeros for the above expression are $\lambda=-1,2,4$.
A: You could (specially if you have a calculator at hand to crunch numbers) do this:


*

*Get the trace (or sum of the diagonal elements). It is $32-25-2=5$

*Compute the determinant. It is $-8$

*Compute the square of the matrix and get the trace of it. It is $21$


You now know that:
$$\lambda_{1} \lambda_{2} \lambda_{3} = -8 \space\space\space\space\space(1)$$
$$\lambda_{1} + \lambda_{2} + \lambda_{3} = 5 \space\space\space\space\space(2)$$
$$\lambda_{1}^{2} + \lambda_{2}^{2} + \lambda_{3}^{2} = 21 \space\space\space\space\space(3)$$
Where $\lambda_{1}, \lambda_{2}, \lambda_{3}$ are the eigenvalues to work out. Now, let's say you feel lucky and want to assume that all the eigenvalues are integer. Then, from equation $(3)$ you know the largest one could be only $3$ or $4$ in absolute value, in which case the second largest would have to be $\pm 2$ and then $\pm 1$. If the largest were $3$, the second would have to be $3$ and the third couldn't be an integer. Now, from $(1)$ you know either one is negative or all are. Now, at once you see from $(2)$ that $4$ got to be positive, and the last choice of whether $1$ or $2$ is negative is solved by mentally trying, to find that the answer is indeed $-1$, $2$, $4$.
Yes, we assumed the answer to be made of integers, but, from experience, this is a good assumption to make to give it a first go in exams questions, as it's usually the case!
