# Finding eigenvalues of a 3x3 matrix given determinant and trace

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 2x2 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 3x3 matrix to no success, as expressing the characteristic polynomial is much more complex. Is there any other approach I could take?

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 :)

• I see how to get from your system of equations to $45+x^2+2x^3=0$. I am not sure where does $x=\frac54$ come from. WA returns this for the system of questions and this for the cubic equation. Dec 6, 2020 at 15:57
• Resolved it and noticed where my mistake was, thank you.
– NirF
Dec 6, 2020 at 16:07
• Why is $A$ diagonalizable ?? Dec 6, 2020 at 16:46

There holds for a matrix $$A$$ that $$\sum_i \lambda_i = \operatorname{tr}(A), \quad \prod_i \lambda_i = \det(A)$$ Since you have one eigenvalue twice (i assume $$\lambda_1$$) this results in: $$2 \lambda_1 + \lambda_2 = -1, \quad \lambda_1^2 \cdot \lambda_2 = 45$$

// Edit: corrected result: You can solve this and get to:

$$\lambda_1 = -3, \quad \lambda_2 = 5$$