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Let $A$ be a $n\times n$ matrix. If $\lambda$ is an eigenvalue of A with multiplicity $m>1$. Then there could be two cases:

  1. There are $m$ linearly independent eigenvectors of A corresponding to $\lambda$.In this case, the geometric multiplicity is the same as the algebraic multiplicity.
  2. The geometric multiplicity is less than algebraic multiplicity.

This information is quite useful while solving a system of linear ODEs. If we have the situation in case 2, we resort to finding generalized eigenvectors.

For example, suppose A is a $3\times 3$ matrix with repeated eigenvalue $\lambda = 1$ and we found only a single eigenvector $v_1$ by solving $(A-\lambda I)v_1 = 0$. A nontrivial solution is always guaranteed for a homogeneous case $Bx = 0$, where $B$ is singular.

Now, we find the generalized eigenvalue of rank 2 by solving the equation $$(A-\lambda I)v_2 = v_1$$.

See that $(A-\lambda I)$ is singular and it's a non-homogeneous case. Why is it so that the system is not inconsistent? In fact $$(A - \lambda I)^mv = 0$$ for a nonzero $v$ makes it possible to repeat the process of finding generalized eigenvectors up to $m-1$ steps. What makes $(A - \lambda I)^mv = 0$ possible.

Can anyone clear it in simple language? Thanks in advance.

MathGuy
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  • Look at a matrix in Jordan form, https://en.wikipedia.org/wiki/Jordan_normal_form It should then be quite clear what is happening. – Yuval Peres Jul 06 '22 at 15:48
  • Thanks, @YuvalPeres. I will check. – MathGuy Jul 06 '22 at 15:52
  • I like reversing the order. For a Jordan block of size $k$ and eigenvalue $\lambda,$ begin with a vector $v$ with $(A-\lambda I)^k v = 0$ but $(A-\lambda I)^{k-1} v \neq 0.$ Then the basis vectors for that block are $v, (A-\lambda I) v, \ldots (A-\lambda I)^{k-1} v.$ The final one is a genuine eigenvector. This way of doing it, with all integer eigenvalues, is doable by hand. All together, we gate an integer matrix $P,$ so that the only occurrence of fractions occurs in a single number as $P^{-1} = \frac{1}{\det P} P',$ where $P'$ is the adjoint matrix, all integers. – Will Jagy Jul 06 '22 at 23:56
  • Thanks @willJagy – MathGuy Jul 09 '22 at 03:30

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