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Let $M_n(\mathbb{C})$ be the ring of $n\times n$ matrices with complex entries. Let $I$ be a minimal left ideal of $M_n(\mathbb{C})$. How can I prove that there is a $i\in\{1,\ldots,n\}$ such that $I$ is the set of matrices that have zero columns, except possibly the $i$th one?

I tried to use this, but did not get that far. Any help?

I am trying to understand the following proof from Fulton-Harris where they seem to be using this fact. enter image description here

user569423
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You can't because this proposition is false.

For example, $\left\{\begin{bmatrix}a& a\\ b&b\end{bmatrix}\middle|\,a,b\in\mathbb C\right\}$ is a minimal left ideal and is not of the form you describe.

Need help seeing why it's minimal? Well I do agree that

$$ L=\left\{\begin{bmatrix}a& 0\\ b&0\end{bmatrix}\middle|\,a,b\in\mathbb C\right\} $$ is also a minimal left ideal, and the ideal I'm proposing is $Lu$ where $u$ is the unit $\begin{bmatrix}1&1\\0&1\end{bmatrix}$, so the two are isomorphic as $M_2(\mathbb C)$ modules.

rschwieb
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  • I'd suggest you take another look at the question you linked, which does not say what you are trying to prove here. Even better, look at the duplicate linked to that question, which is comprehensive. – rschwieb Sep 29 '20 at 13:15
  • I see your point. I was trying to understand a proof from the Fulton-Harris book. There they seem to be claiming that a minimal left ideal is of the form stated in my question. Have I misinterpreted what they are saying in the book? (I have added a picture of the proof in the question) – user569423 Sep 29 '20 at 20:07
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    @user569423 "Any minimal ideal is isomorphic to* one which consists of..." That much is true. A semisimple artinian ring only has one isoclass of simple module. But you cannot show the left ideal is equal* to such a left ideal. – rschwieb Sep 29 '20 at 20:08
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    oh yeah you are right. Need to be more careful next time when reading. Thank you. – user569423 Sep 29 '20 at 20:25