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Assume we have a matrix $A$ with eigenvalues $\lambda_i$ and each eigenvalue has the geoemtrical multiplicity $g_i$ and the algebraic multiplicity $a_i$.

Now assume, that for a specific $i$ we have $g_i=n\cdot a_i$ with $n\in\mathbb N$

E.g. $g_1=2, a_1=4$

Notation: $J_{size,\lambda_i}$

So, for the given example, we would need $2$ blocks, since the geometrical multiplicity is $2$.

The possibilities are:

  1. $J_{2,\lambda_i}\boxplus J_{2,\lambda_i}$
  2. $J_{1,\lambda_i}\boxplus J_{2,\lambda_i}$
  3. $J_{2,\lambda_i}\boxplus J_{1,\lambda_i}$

I'm wondering, how I decide if I should take $(1)$, $(2)$/$(3)$. [I know, that $(2)$ and $(3)$ are equivalent]. But I'm not sure if I can always choose between $(1)$ and $(2)$ or if there are criterias which limit me in my choice.

1 Answers 1

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I think you have written $J_{2,\lambda_i}\boxplus J_{1,\lambda_i}$ and $J_{1,\lambda_i}\boxplus J_{2,\lambda_i}$ instead of $J_{3,\lambda_i}\boxplus J_{1,\lambda_i}$ and $J_{1,\lambda_i}\boxplus J_{3,\lambda_i}$, rispectively.

By the way, the trick consists in calculating $(A-\lambda_i Id)^2$ and looking at its rank, which I'll write as $rank(A-\lambda_i Id)^2:=r(A-\lambda_i Id)^2$.

If $r(A-\lambda_i Id)^2 = r(A-\lambda_i Id)-2$, then you have case $(1)$, otherwise $r(A-\lambda_i Id)^2 = r(A-\lambda_i Id)-1$ and you have case $(2)$ or $(3)$.

Why? Because the rank of the squared matrix decreases by $1$ for every block with size $\ge2$.

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    Guess you were too fast that you catched that typo. :) Thanks for the clarification. Is that the only way or are there others way to decide it? I think you have a typo in you second last line. Guess you wanted to say $r(A-\lambda_i Id) 2=r(A-\lambda_iId)=2$?2017-02-05
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    Of course there are other ways, but I find this one pretty good when $a_i =4$. And no, there shouldn't be any typo2017-02-05
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    Ah, true. I missread. So what are the other ways?2017-02-05
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    I don't really remember, I forgot a bit about linear algebra, but I am sure you can google it2017-02-05
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    of course I'll also google it, just though you might just know it. :) Thank you2017-02-05
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    Don't worry, I usually always check the answers was just in a hurry. ;)2017-02-05