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I got homework to prove some question and after almost 5 hours I gave up. The questions are:

1) operator $T : \Bbb R^n\to \Bbb R^n$, prove that $\operatorname{Im}(T)∩ \operatorname{Ker}(T)={0}.$

2) prove or disprove : Linear operator $T : \Bbb R^n \to\Bbb R^n $, $T$ is diagonalizable if and only if $\operatorname{Im}(T)∩ \operatorname{Ker}(T)=\{0\}$.

3) $T : \Bbb R^n \to \Bbb R^n$ linear operator with all eigenvalues $=0 $, prove that $T$ is diagonalizable only if $T$ is the zero operator.

Thanks

mechanodroid
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Or Yehuda Ben Shimol
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    Have you copied this down correctly? Because the first statement is just wrong. – Aweygan Dec 19 '18 at 14:44
  • Welcome to MSE! It's important to show your work even if it is an incomplete attempt. Especially considering it is a homework question! Also please check the following link on how to properly type in MSE: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Kal S. Dec 19 '18 at 14:48
  • What characterisations of diagonalisability do you know? This could radically change the answer to this question. – Theo Bendit Dec 19 '18 at 14:53
  • @Aweygan Yes that’s the question – Or Yehuda Ben Shimol Dec 19 '18 at 15:43
  • @TheoBendit I know that there is base B with vectors with eigenvalues . – Or Yehuda Ben Shimol Dec 19 '18 at 15:46

2 Answers2

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  1. If $T$ has a basis of eigenvalues, then either $0$ is an eigenvalue or not. If it isn't, then $\operatorname{ker} T = \{ 0 \}$, and the intersection result is trivial. Otherwise, we may express the basis of eigenvectors in the form $$B = (u_1, \ldots, u_m, v_1, \ldots, v_n)$$ where $u_1, \ldots, u_m$ are eigenvectors corresponding to $0$, and $v_i$ is an eigenvector corresponding to $\lambda_i \neq 0$ for all $i = 1, \ldots, n$. I claim that $$\operatorname{Im} T = \operatorname{span} (v_1, \ldots, v_n).$$ To see containment in the $\subseteq$ direction, consider the image of an arbitrary element of the span of $B$. To see containment in the $\supseteq$ direction, show that $v_i = T(\lambda_i^{-1} v_i) \in \operatorname{Im} T$ for all $i$. Then, $$\operatorname{Ker} T \cap \operatorname{Im} T = \operatorname{span} (u_1, \ldots, u_m) \cap \operatorname{span} (v_1, \ldots, v_n) = \{ 0 \},$$ by linear independence of $B$.

  2. Just take an invertible but not diagonalisable operator (e.g. a nilpotent operator with the identity added), and $\operatorname{Ker} T = \{ 0 \}$.

  3. If $0$ is the only eigenvalue of $T$, and there is a full basis of eigenvectors corresponding to $0$, then what does $T$ do to an arbitrary linear combination of these eigenvectors?

Theo Bendit
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  1. This isn't true. Consider $T= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. We have $\ker T = \operatorname{Im} T = \operatorname{span}\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right\}$.

  2. The direction "$T$ diagonalizable $\implies \ker T \cap \operatorname{Im} T = \{0\}$" is true, the other direction isn't. Indeed, let $\{b_1, \ldots, b_n\}$ be a basis of eigenvectors for $T$, and take $y \in \ker T \cap \operatorname{Im} T$. There exists $x = \sum_{i=1}^n \alpha_i b_i$ such that $y = Tx = \sum_{i=1}^n \alpha_i \lambda_ib_i$.

    We have $$0 = Ty = T^2x = \sum_{i=1}^n \alpha_i \lambda^2 b_i$$ so $\alpha_i\lambda_i^2 =0$ for $i=1, \ldots, n$. Hence if $\alpha_i\ne 0$, we get $\lambda_i = 0$ so $y = \sum_{i=1}^n \alpha_i \lambda_ib_i = 0$.

    For the other direction consider $$T = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$$

    $T$ is not diagonalizable but $\ker T = \operatorname{span}\left\{\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}\right\}$ and $\operatorname{Im} T = \operatorname{span}\left\{\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}\right\}$ have trivial intersection.

  3. If $T = 0$ then $T$ is clearly diagonalizable. Conversely, if $T$ is diagonalizable with all zero eigenvalues, then $T$ diagonalizes to the matrix $0$, so $T = 0$.

mechanodroid
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