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Let $T: \Bbb C^{10} \to \Bbb C^{10}$ be a linear operator with minimal polynomial $x^3(x-1)^2$ and $\dim Im T=6$. Show $\dim \ker(T-I)\leq 3$.

Since the dimension of the image is $6$, we know that $T$ has $0$ as an eigenvalue of algebraic multiplicity at least $4$. that means $1$ is of multiplicity at most $6$. I know $\dim \ker(T-I)$ corresponds to the multiplicity of the eigenvalue $1$ but I'm not sure how to continue from here.

Kantig Shoter
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  • This is not a rigorous proof but let's bring $T$ in jordan normal form. Then because of $T^3$ appearing in minimal polynomial there should be a $3×3$ jordan block corresponding to $0$ and you are using only one eigenvector of $0$ for this. You still have $3$ more which will give you at least 3 more blocks for $0$. Even if they are all $1×1$ it leaves you with at most $4$ diagonal places that can correspond to eigenvalue $1$. If we had $4$ eigenvectors for $1$ then only $(T-I)$ would appear in minimal polynomial but we have square so it's less than $4$. May be we can make this rigours – Tony Pizza Oct 14 '23 at 15:16
  • With following two results I think this can be made precise. https://math.stackexchange.com/questions/82607/why-does-the-largest-jordan-block-determine-the-degree-for-that-factor-in-the-mi https://math.stackexchange.com/questions/546838/why-is-the-geometric-multiplicity-of-an-eigen-value-equal-to-number-of-jordan-bl – Tony Pizza Oct 14 '23 at 16:39

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