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Let $A\in \text{Mat}_3(\mathbb{R})$ be a symmetric singular $3\times 3 $ matrix, so with $\det(A)=0$. We know that it must have one zero eigenvalue (Is it correct to say that we don’t know its multiplicity, so it could be 1, 2 or 3?), say $λ_1=0$.

Is there a way to calculate the rest of the eigenvalues without having to calculate the charateristic polynomial $\det(A-I_3)$ and finding its real roots? Maybe this theorem is useful?

  • For example, for a symmetric 2x2 matrice $B$, we have that, if $\det(B)=0$, then the other eigenvalue is $\text{tr}(B)$.
  • Note that I don’t know if the “symmetric” hypothesis on the matrix is really useful, but since symmetric equals (orthogonally) diagonalizable in $\mathbb{R}$, I thought it might make things easier.
Sebastiano
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selenio34
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    You could compute $\ker A$ and study the restriction of $A$ to $(\ker A)^\bot,$ but it is not simpler than calculating $\det(A-I_3).$ – Anne Bauval Jan 11 '23 at 11:13
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    The remaining two eigenvalues are roots of the quadratic equation $\lambda^2+\left(\operatorname{tr}(A)-\lambda\right)^2=\operatorname{tr}(A^2)$. – user1551 Jan 11 '23 at 11:15
  • @user1551 Perfect, that was what I was searching for. Could you show how you got to the formula in an answer? I’ll accept it. – selenio34 Jan 11 '23 at 11:16
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    Thanks but no thanks. It's just saying that the two eigenvalues sum up to $\operatorname{tr}(A)$ and their squares sum up to $\operatorname{tr}(A^2)$. One may also derive the quadratic equation from Cayley-Hamilton theorem, because the characteristic polynomial of a general 3x3 matrix $A$ is $x^3-\operatorname{tr}(A)x^2+\frac12\left(\operatorname{tr}(A)^2-\operatorname{tr}(A^2)\right)x-\det(A)$. – user1551 Jan 11 '23 at 12:07

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I’m responding to my question because of this question, which says

Encourage them to post an answer. If they don't, don't be afraid to post your own answer to your question explaining how you solved your problem and mark it as the answer when you can. You can make sure to credit the user who helped you right inside the answer itself if you like.

My question has been solved by @user1551: we can calculate the remaining eigenvalues of the matrix by solving $$λ^2 + (tr(A)-λ)^2 - tr(A^2)=0$$

selenio34
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