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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.