Csdrhrt

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

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

2) prove or disprove : Linear operator $T : Bbb R^n toBbb 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 🙂

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 $alpha_ilambda_i^2 =0$ for $i=1, ldots, n$. Hence if $alpha_ine 0$, we get $lambda_i = 0$ so $y = sum_{i=1}^n alpha_i lambda_ib_i = 0$.

   
   
   

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

   
   
   

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

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?