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If I have a $4\times 4$ matrix $A$ with real entries that has all $1$'s on the main diagonal, $A$ is singular and we know one eigenvalue $k_{1}=3+2i$. What about the others three eigenvalues?

I think one should be $k_{2}=3-2i$ because they always come in pairs, right?

Then, since $A$ is singular I think one should be $k_{3}=0$

And I don't know what about $k_{4}$ and I don't know how to use the fact that the diagonal has all $1$'s.

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3 Answers 3

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You have the sum of all eigenvalues being equal to the trace; hence $k_1+k_2+k_3+k_4=4$. This'll give you $k_4$.

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  • $\begingroup$ So k4 is equal to -2? $\endgroup$
    – billy
    Jun 26, 2013 at 9:30
  • $\begingroup$ Yes — which also implies, as a by-product, that your matrix can be diagonalized in $\mathbb{C}$ (all roots are distinct). $\endgroup$
    – Clement C.
    Jun 26, 2013 at 12:49
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Hints:

Yes, the the roots of real polynomials appear in pair of conjugate complex ones, so you already have three roots of the char. polynomial:

$$3\pm 2i\;,\;0$$

Now, we also have that

$$\text{tr.}(A)=4$$

and since

$$+3+2i+3-2i+\alpha=\;-\text{cubic coefficient in the char. pol.}$$

we get that the last root $\;\alpha\;$ is...

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    $\begingroup$ So k4 is equal to -2? $\endgroup$
    – billy
    Jun 26, 2013 at 9:28
  • $\begingroup$ Yes @billy, I think so. Nice!+1 $\endgroup$
    – DonAntonio
    Jun 26, 2013 at 9:30
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Hint: Note that the trace is invariant under similarity transformations, that is for any $S \in GL(4)$ we have $$ \mathrm{tr}(S^{-1}AS) = \mathrm{tr}(A) = 4 $$ Now choose an $S$ such that $S^{-1}AS$ has the eigenvalues on the diagonal.

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