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Consider the following problem, that was part of an old exam I am studying for:

Let $$ A = \begin{pmatrix} 4 & 0 & 2\\ -2 & 8 & 2\\ 0 & 2 & -4 \end{pmatrix}$$

  1. Using the Gershgorin circle theorem, show that $A$ has exactly one eigenvalue with a negative real part.
  2. Find three disjunct circles, such that each contains exactly one eigenvalue.
  3. Give an approximation for the biggest eigenvalue that is as close as possible.

Hint: For 2. and 3., consider $\hat{A} = D^{-1}AD$ with $D = diag(1,c,1),\: c>0$. For 3., choose $c$ to yield an optimal approximation.

Now, we dealt with the Gershgorin circle theorem in class, and I can apply it well in the first part of the problem. My issue is with the second and last parts - I have never seen the use of similarity transforms in combination with the theorem. I only know that such transformations keep the eigenvalues intact, and that in some cases, it can be used to extract more information from the Gershgorin circles.

My question is though: How do I select such a $c$ to retrieve an optimal approximation? What is the strategy here?

What I have tried so far:

I calculated the Matrix $D^{-1}AD$ parameterized by $c$: $$D^{-1}AD = \begin{pmatrix}4 & 0 & 2 \\ -\frac{2}{c} & 8 & \frac{2}{c} \\ 0 & 2c & -4\end{pmatrix}$$ The three resulting Gershgorin circles are therefore:

  • $C_1 = (4, 2)$
  • $C_2 = (8, \frac{4}{c})$
  • $C_3 = (-4, 2c)$

$C_3$ immediately solves the first part of the problem, since for $c=1$ it's the only circle that is on the negative side of the real axis. For the second part, I need to find a $c$ such that $C_1$ and $C_2$ dont overlap anymore (which they do for $A$), which means that: $$\frac{4}{c} < 2$$ This is easily the case for $c > 2$, for example $c = 2.1$.

But how do I pick an optimal $c$ for the last part of the problem? The circle with the biggest "magnitude" $C_2 = (8, \frac{4}{c})$ just gets smaller in radius for bigger $c$.

Rodrigo de Azevedo
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Katharina
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    I can address 3 if you'd like, but first I'd like to suggest a better route for 2. If you instead apply G.discs to $^T$ you get disjoint* discs of (4,2), (8,2) and (-4,4). If you look at or $^T$ there is a single communicating class in the underlying graph. Due to Taussky re: boundaries of G.Discs -- with an irreducible matrix, a weakly diagonally dominant matrix (strict on at least 1 row) is nonsingular -- so e.g. consider $(A^T - 6I)$ and $(A^T - 0I)$ which are nonsingular. This e.g. implies (4,2), (8,2) and (-4,4) are all in effect open discs and hence disjoint. – user8675309 Jan 21 '20 at 21:48
  • @user8675309 That's an interesting way to solve part 2. Thank you! And yes, I would love to see your take on part 3. – Katharina Jan 21 '20 at 22:35
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    I think there's a desired choice of $c$ by the test writer. Main idea suppose you know the dominant eigenvalue $\lambda $ is $\geq 8$-- so as $c$ increases in magnitude you can 'pinch' the disc around 8 and get a tighter and tighter approximation around that eigenvalue... this tightening stops when your disc around 8 is completely inside the expanding disc around -4. (note: The disc for the top row is invariant to choice of c.) Is this enough? note: at the moment it isn't immediately obvious to me that $\lambda \geq 8$ without looking at the charpolynomial, and evaluating it at 8 ( – user8675309 Jan 22 '20 at 02:36
  • @user8675309 Thank you! With that idea, I was able to construct an optimal $c$ - I found the exact point where the right bound of the disk around 8 and the one around -4 meet, which provides the best possible approximation for $\lambda_\max$. I evaluated this $c$ to be $3 + \sqrt{11}$. Thanks a lot for your help! – Katharina Jan 22 '20 at 13:24
  • @user8675309 wouldn't you post your comments as an answer? – Viera Čerňanová Jan 26 '20 at 22:11
  • @user376343 fair enough -- I dropped in my final comment as an answer – user8675309 Jan 27 '20 at 23:47

1 Answers1

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posting my comment as an answer:
Part 1 tells you that each disc has one eigenvalue -- hence they are all real.

check characteristic polynomial $p$
$p(8) = \det\big(8I - A\big) \lt 0$
but $p(x) \gt 0$ for x large enough, so you know you know by Intermediate Value Theorem that there is a zero in $(8,x)$. I.e. the dominant eigenvalue $\lambda \gt 8$. The matrix is sparse, so $\det\big(8I - A\big)$ comes down to evaluating a 2 x 2 determinant and managing the sign function.

main idea
you know $\lambda \gt 8$, so as increases in magnitude you can 'pinch' the disc around 8 and get a tighter and tighter approximation around that eigenvalue... this tightening stops when your disc around 8 is completely inside the expanding disc around -4. (note: The disc for the top row is invariant to choice of c.) Solving for the $c$ where the gains from pinching stop, gives the result.

user8675309
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