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Suppose $M$ is a $2 \times 2$ matrix over $\mathbb{C}$ with characteristic polynomial $\chi =(x-\lambda)^2$ and minimal polynomial $m= (x-\lambda)^2$. My text claims that the geometric multiplicity of $\lambda$ is $1$.

I can only show this by "cheating":

$M$ diagonalizable (here) iff $\lambda$ has geometric multiplicity $2$ iff $m$ decomposes into distinct linear factors. Since $m$ does not decompose into distinct linear factors, we know that $\lambda$ can't have geometric multiplicity $2$.

Is there a way to show this without "cheating"? A more direct solution? Many thanks!

  • Why is that cheating? It's a valid argument. – Chrystomath Jun 12 '20 at 08:51
  • Just seems like overkill (essentially referring to the primary decomposition theorem). I thought there might be a more direct and elementary way of showing this. In fact my text hinted at this: It considered some vector $v$ such that $(M-\lambda I) v \neq 0$. Then $(M-\lambda I)v$ must be an eigenvector of $M$ also with eigenvalue $\lambda$. From here my text just claims that this is the only independent eigenvector, but I can't see this ( unless I forget about the previous argument and "cheat"). –  Jun 12 '20 at 08:58
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    Overkill yes, not cheating. If there were another independent vector $w$, then it and $(M-\lambda)v$ would span the whole space. So $M-\lambda = 0$, which is not the case. – Chrystomath Jun 12 '20 at 10:49
  • @Chrystomath Thanks! That's what I was looking for! –  Jun 12 '20 at 13:46
  • @Cheesecake "If there is only one eigenvalue, then its eigenspace equals the whole of $V$"... How about the matrix with rows $(1,0)$ and $(0,0)$? Also this would mean that every $2 \times 2$ matrix is diagonalisable over $\mathbb{C}$, which is false. Since when is $(x−λ)^2$ not monic? –  Jun 12 '20 at 13:50
  • Sorry! I was thinking of the matrix \begin{bmatrix}1&1\0&1\end{bmatrix}! Your first comment is incorrect. –  Jun 12 '20 at 14:24
  • This doesn't make alot of sense to me. $V$ is not necessarily the direct sum of eigenspaces. If it were, then every matrix would be diagonal. Your comments are incorrect. Anyway I'm out of here. –  Jun 12 '20 at 14:35
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    Does this answer your question? Prove that T is diagonalizable if and only if the minimal polynomial of T has no repeated roots. – PinkyWay Jun 12 '20 at 14:59
  • My question was answered by @Chrystomath! –  Jun 12 '20 at 15:08

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