# Finding eigenvalues if trace and determinant of the matrix is given

Let $A$ be a $3\times 3$ matrix with real entries such that $\det(A)=6$ and $tr(A)=0$. If $\det(A+I)=0$ ($I$ denotes $3\times 3$ identity matrix), then the eigenvalues of $A$ are:
(i) $-1,2,3$;
(ii) $-1,2,-3$;
(iii) $1,2,-3$;
(iv) $-1,-2,3$.

If a,b,c are 3 eigenvalues then a+b+c=0 and abc=6 because sum of eigen values is trace and product is the determinant value. Then how to apply $\det(A+I)$?

• Sos de la facultad de ingeniería? – Cure May 3 '14 at 1:26

Eigen values of $A+I$ are obtained by adding $1$ to the eigenvalues of $A$. So $\det(A+I)=0$ gives a third condition on them (besides $\det A = 6, \ \mathrm{tr\,}A=0$) and that should enable you to find the answer.

• If a,b,c are eigen values of A then a+1,b+1,c+1 are eigen values of A+I.Am i right? then we have a+b+c=0, abc=6 and (a+1)(b+1(c+1)=0.IS it possible to solve these equation to find a,b,and c. – mercy May 3 '14 at 3:07
• Expand the product $(a+1)(b+1)(c+1)=0$ fully then the solution will emerge. – P Vanchinathan May 3 '14 at 5:51

In the above question option (iv) is correct. Sum of eigenvalues is $$0$$ product of eigenvalues is $$6$$ (ie) $$a+b+c =0$$ and $$abc=6$$ and also $$(a+1)(b+1)(c+1)=0$$ since $$\det(A+I)=0$$ Solving we get the roots.

Hint:

Check the following: for a $\;3\times 3\;$ matrix $\;A\;$, and putting $\;\Delta:=\det A\;,\;\;\mathcal T:=tr. A\;$ , we have that its characteristic polynomial is

$$x^3-\mathcal T x^2+\left(\mathcal T^2-tr.\left(A^2\right)\right)x-\Delta$$

• @ DonAnatonio: Is the coefficient of $x^2$ really the determinant? – P Vanchinathan May 3 '14 at 2:44
• Good catch, @PVanchinathan: it is minus the trace. Thanks. – DonAntonio May 3 '14 at 10:32

It is much simpler than this; $$\det(A+I) = 0$$ means that one of the eigenvalues must equal $$-1$$, so without loss of generality we can say $$a = -1$$; the other two bits of information then become $$bc = -6, b+c=1$$. So: $$a=-1, b=-2, c=3$$