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Let's say we have a linear transformation $f:\mathbb{C}^3\rightarrow \mathbb{C}^3$, so let $F$ be the matrix that represents $f$ in the canonical basis, and we are given that $f$ is not injective.

The question comes here: Since obviously no injectivity imply $\text{det }F=0$, from here... Can we infer that one of its eigenvalues will be 0? Since I've seen in a lot of questions that from $\text{det }F=0$ they directly say that one of its eigenvalues must be 0.

A rural reader
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Acedium 20
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    The determinant is the product of its eigenvalues (considering multiplicity) , so yes. – Peter Jun 24 '23 at 18:52
  • @Peter wait so $\text{det }F=\lambda_1\times \lambda_2\times \lambda_3$? – Acedium 20 Jun 24 '23 at 18:55
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    Yes , where duplicates are possible. – Peter Jun 24 '23 at 18:56
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    Another way to see it is that determinants relate to volumes of parallelepipeds, so the parallelepiped formed does not have any volume. Therefore, the linear map cannot have full range, so it has nontrivial nullspace. – Cameron Williams Jun 24 '23 at 18:57
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    $det F=0$ iff $f$ is not injective, or equivalently $\ker f\neq{0}$, which is the definition of having $0$ as an eigenvalue. – Will Jun 24 '23 at 19:02
  • If there is some non zero $v$ such that $Av = 0 $ then what do you think? – copper.hat Jun 24 '23 at 19:12

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