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 :)

  • 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. $\endgroup$ – Martin Sleziak Dec 6 '20 at 15:57
  • $\begingroup$ Resolved it and noticed where my mistake was, thank you. $\endgroup$ – NirF Dec 6 '20 at 16:07
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    $\begingroup$ Why is $A$ diagonalizable ?? $\endgroup$ – TheSilverDoe Dec 6 '20 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:

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


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