Suppose a $3×3$ matrix A has only two distinct eigenvalues. Suppose that $\operatorname{tr}(A)=−1$ and $\det(A)=45$. Find the eigenvalues of $A$.
I have solved a similar problem with a 2×2 matrix by using the properties of trace and determinant (trace = a + d and det = ad-bc). I tried to take the same approach for the 3×3 matrix to no success, as expressing the characteristic polynomial is much more complex. Is there any other approach I could take?
Best Answer
Suppose your eigenvalues are $x$ and $y$. your matrix $A$ is similar to a diagonal matrix $B$ which has it's eigenvalues on its diagonal.
Now, similar matrices have the same determinant and the same trace, thus we can get to the following equations: $$2x+y = -1$$ $$x^2y=45$$ The first one is the sum of the diagonal (we know that there are 2 unique eigenvalues thus, one of them will show up 2 times on the diagonal).
The second one is the product of the diagonal (determinant of diagonal matrix).
$$... y=\frac{45}{x^2}$$ $$... x=-3 \space\space\space$$
if $x=-3 => y=5$
$x^2y=45$ and $2x+y=-1$. And that's our answer :)