Let $V$ be a vectorspace of dimension 6 over a field $F$. There exists $T \in A(V)$ such that $Ker(T) = Range(T)$ (True or false)
I know that Kernel is the subspace of domain and Range is the subspace of Codomain and $x=0$ is the only point in both spaces. My question is can we find a linear operator such that $Ker T = Range T?$
Best Answer
First "decipher" $\DeclareMathOperator \im{im} \ker T = \im T$. For $v \in V$, $Tv = \im T = \ker T$, thus $T^2 v = 0$. Therefore $T^2 = 0$.
Also by the Rank-Nullity theorem, $\dim (\ker T) + \dim (\im T) = \dim V = 6$, thus $\DeclareMathOperator \rank {rank} \dim (\ker T) = 3 $.
For simplicity, we consider the Jordan canonical form of possible $T$. The argument above shows that $T$ is nilpotent, thus the Jordan form exists. Since $\dim (\ker T) = 3$, there should be $3$ Jordan blocks. Easy to see that $X^2$ is the minimal polynomial of $T$, thus the max size of the Jordan blocks is $2$. Since $\dim V = 6$, each Jordan blocks should be of size $2$. Therefore if such operator exists, then under certain basis the matrix of $T$ should be $$\mathrm {diag}(J_2(0),J_2(0), J_2(0)),$$ where $J_2(0)$ denotes the Jordan block of size $2$ with all diagonal entries $0$.
Now we check this $T$. By definition, if the basis is $(e_j)_1^6$, then $$ T e_{2j-1} = 0, Te_{2j} = e_{2j-1}, j=1,2,3. $$ Direct verification shows $\ker T = \mathrm {span} (e_1, e_3, e_5)$ and $\im T = \mathrm {span} (e_1, e_3, e_5)$. In conclusion, such operator exists, whose matrix is described above.