Find If There Are Such Linear Transformations

Question

1. $\def\im{\mathop{\mathrm{Im}}}$$T:\mathbb{R}^3\to \mathbb{R}^3$ s.t $\ker(T)=\im(T)$

2. $T:\mathbb{R}^4\to \mathbb{R}^4$ s.t $\ker(T)=\im(T)$

3. $T:\mathbb{R}^3\to \mathbb{R}^3$ s.t $T\neq 0$ and $T^3=-T$

4. $T:\mathbb{R}^3\to \mathbb{R}^3$ s.t $T\neq 0$ and $T\neq I$ s.t for all $v\in V$ $T(v)=v$ or $T(v)=0$

1. $\dim(v)=\dim(\ker(T))+\dim(\im(T))$ so we have $3=2x\iff x=\frac{2}{3}$ so $\dim(\ker(T))=\dim(\im(T))=\frac{2}{3}$ which can not be and so there is no such linear transformation

2. by $\dim(v)=\dim(\ker(T))+\dim(\im(T))$ we get $\dim(\ker(T))=\dim(\im(T))=2$ but if $\ker(T))=\dim(\im(T)$ then $T^2(v)=0$ and the only matrices that I know that do that are nilpotent matrices which do not work

3. $T^3=-T\iff T^3+T=0\iff T(T^2+I)=0$ So or $T=0$ which can not be or $T^2=-I$ but any number in $\mathbb{R}$ in the power of $2$ is positive so that is not such linear transformation

4. By the definition of linear transformation on basis, there is just one linear transformation that can be set on the basis vectors, so there no two matrices such that $T\neq S=0$ or $T\neq S=I$ or $\mathrm{Hom}(V\to V)$ is a linear space so there is just one identity element for addition and just one identity element for multiplication namely $I$ and $0$

Are the answers valid?

Answer 1

1. What you did is correct.
2. Just take $T(a,b,c,d)=(b,0,d,0)$.
3. Just take $T(x,y,z)=(-y,x,0)$.
4. There is no such $T$, because if $v,w\in\mathbb{R}^2\setminus\{0\}$ are such that $T(v)=v$ and $T(w)=0$, then $T(v+w)=v$, which is neither $v+w$ nor $0$. (I don't understand your answer here.)