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I'm stuck solving this exercise and I need some help. Given the matrix $$ A = \begin{pmatrix} \alpha^2 & \alpha & 0 \\ \alpha & 2\alpha + \alpha^2 & 2 \\ 0 & 2 & \alpha^2-2\alpha+2 \end{pmatrix} $$ i'm asked to find the values of $\alpha>0$, using Gersghorin's circles, so that $A$ is positive-definite matrix.

As $A=A^t$, we know that all eigenvalues must be real. On the other hand, according to the Gersghorin criterion, $$ sp(A)=\{\text{eigenvalues of A}\} \subset D(\alpha^2,\alpha)\cup D(2\alpha+\alpha^2,\alpha+2)\cup D(\alpha^2-2\alpha+2,2) $$ And finally, $A$ will be positive definite if and only if all of its eigenvalues are positive. So I know I need to find all the values of $\alpha >0$ for the above to be true, but I don't know how to continue. Any help is welcome.

Feliciano
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  • So you need to find all $α$ with $α^2>|α|$, $α(2+α)>|2+α|$ and $(1-α)^2+1>2$. – Lutz Lehmann Apr 19 '21 at 16:36
  • You can use Gerschgorin discs to obtain some values of $\alpha$ that make $A$ positive definite, but I'm afraid they cannot give you all such values. The correct way to find all feasible values of $\alpha$, I believe, is to use matrix congruence rather than Gerschgorin discs. Apparently, the answer is ${\alpha\in\mathbb R:(\alpha^2+2\alpha-1)(\alpha^2-2\alpha+2)-4>0}\approx(-\infty,-2.5163)\cup(1.3559,+\infty)$. – user1551 Apr 19 '21 at 16:58
  • note: you could adapt the interlacing argument here: https://math.stackexchange.com/questions/3872707/show-that-all-the-eigenvalues-of-m-are-positive-real-numbers/3872865#3872865 which tells you that $A\succ \mathbf 0$ iff $\alpha^2-2\alpha+2 \gt 0$ and $\det\big(A\big)>0$ – user8675309 Apr 19 '21 at 17:47
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    @user8675309 Applying Sylvester's criterion to $A([3,1,2],[3,1,2])$ also gives the same condition you've got. – user1551 Apr 19 '21 at 18:03
  • @user1551 Agreed. I guess it felt like an Interlacing day today. – user8675309 Apr 19 '21 at 20:08

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