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I've been struggling all day with this question. I tried to come up with a proof which shows that such an endomorphism does NOT exist, but I'm not sure it is correct.

  • Let $ B = (b_1, b_2, b_3, b_4)$ be a base of $ \mathbb{R}^4$;
  • Since $ \text{dim}(\mathbb{R}^4) = \text{dim}(\text{ker}f) + \text{dim}(\text{im}f)$ , then $\text{dim}(\text{ker}f) = \text{dim}(\text{im}f) = 2$;
  • We begin building an endomorphism such that $f(b_1) = 0_{\mathbb{R}^4}$ and $f(b_2) = 0_{\mathbb{R}^4}$ (I don't think it matters which vectors we choose to span the kernel). Consequently, $\text{mg}(0) = 2$;
  • If $\text{dim}(\text{im}f) = 2$ and $\text{im}(f) = \text{ker}(f)$, then it must be that $f(b_3) = b_1$ and $f(b_4) = b_2$;
  • By doing so, $b_1$ and $b_2$ are the only eigenvectors of $f$, and thus we can't find a base of $\mathbb{R}^4$ made of eigenvectors.

Please bear in mind that my knowledge of linear algebra stops at diagonalization, and that this is my very first attempt at making a proof.

If I'm wrong and building such an endomorphism is actually possible, then what am I missing?

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Here is a very quick proof: show that because $\ker (f) = \operatorname{im}(f)$, it must hold that $f \neq 0$ but $f^2 = 0$. However, the only diagonalizable endomorphism $f$ for which $f^2 = 0$ is the zero endomorphism. So, $f$ cannot be diagonalizable.


Regarding the points that you have written: first of all, you have not proved that $b_1,b_2$ are the only eigenvectors of $f$. Second, showing that the example you tried to make failed to be diagonalizable while satisfying the condition does not prove that there are no such endomorphisms.

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  • $\begingroup$ Thank you for the answer. I realize now that my proof is incomplete, but I'm clueless as to how it can be improved (or if it is a good start at all). About your proof: the linear algebra course I've attended didn't cover the powers of $f$ (if that is the correct name of this topic), so I don't think I can use that. Is there another way to prove it? $\endgroup$ – Remisse Jun 25 '20 at 23:09
  • $\begingroup$ Powers of $f$ correspond to powers of matrices. Are you familiar with those? $\endgroup$ – Osama Ghani Jun 26 '20 at 0:50
  • $\begingroup$ @Omnomnomnom Perhaps he was asking me haha. In my first comment to your answer I said I didn't know what the powers of $f$ are, so I don't really understand why it has to be that $f^2 = 0$. We didn't cover the powers of matrices, either; we stopped right at the diagonalization theorem. $\endgroup$ – Remisse Jun 26 '20 at 7:03
  • $\begingroup$ @Remisse Somehow I missed that you had commented. The "powers of $f$" is not as fancy as you're making it out to be: I'm just saying that $f \circ f = 0$. In other words, for any vector $x$, we have $f(f(x)) = 0$. See if you can prove that this holds using the definitions of the image and kernel. $\endgroup$ – Ben Grossmann Jun 26 '20 at 7:24
  • $\begingroup$ @Remisse From there, show that for any diagonal matrix $D$: if $D^2 = DD = 0$, then $D$ had to be the zero matrix. Using the definition of diagonalizability, use this to conclude that the diagonalizable endomorphisms have the same property: if $g$ is diagonalizable and $g^2 = 0$, then $g = 0$. $\endgroup$ – Ben Grossmann Jun 26 '20 at 7:27

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