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?


2 Answers 2


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 :)

  • 2
    $\begingroup$ 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. $\endgroup$ Dec 6, 2020 at 15:57
  • $\begingroup$ Resolved it and noticed where my mistake was, thank you. $\endgroup$
    – NirF
    Dec 6, 2020 at 16:07
  • 1
    $\begingroup$ Why is $A$ diagonalizable ?? $\endgroup$ 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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.