If $T\colon \mathbb R^n \to \mathbb R^n $ linear and $T^2 = kT$ [closed]

If $T^2=kT$, then $T$ satisfies the polynomial $t^2-kt = t(t-k)$; hence the minimal polynomial of $T$ is either $t$, $t-k$, or $t(t-k)$. (If $k=0$, then the two possibilities are $t$ and $t^2$).

1. If the minimal polynomial of $T$ is $t$, then $T=0$. In particular, $T$ is singular, so 3 holds (since the consequent is true). 2 is true by vacuity, and 4 is false as written (since $T$ satisfies $T^2=kT$ with $k\neq 0$, and we may not be free to choose $k$; however, if 4 were "$T=\lambda I$ for some $\lambda$", then it would be true with $\lambda = 0$).

2. If the minimal polynomial of $T$ is $t^2$ (when $k=0$), then the Jordan canonical form of $T$ is a block diagonal matrix, with each block either a $1\times 1$ block of $0$, or a $2\times 2$ block $$\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right),$$ and at least one such block. Again, the matrix is singular, so 3 holds because the consequent it true. It is the only of the options that is true.

3. If $k\neq 0$ and the minimal polynomial of $T$ is $t-k$, then $T=kI$. Then 1, 2, and 4 are all true, and 3 is true if and only if $|k|\leq 1$ (in which case it is true by vacuity).

4. If $k\neq 0$ and the minimal polynomial of $T$ is $t(t-k)$, then $T$ is diagonalizable and every diagonal entry of the diagonal form is either $0$ or $k$. In this case, 1 is false, 2 is true, 3 is true (because the consequent is true), and 4 is false.

Unless I'm mistaken, all statements are wrong. The first one is wrong because it implies $\left|\lambda\right|=1$ if $T$ is nonzero, which is not true because you can let $T=2\mathrm{Id}$. (Perhaps you formulated this statment wrongly? Maybe it should be $\left\Vert T(x)\right\Vert =\left|\lambda\right|\left\Vert x\right\Vert$ ?).

The second statement is also wrong: definte $T$ to be $(x,y)\mapsto(\sqrt{2}x,0)$; then $\left\Vert (1,1)\right\Vert =\sqrt{2}=\left\Vert (\sqrt{2},0)\right\Vert =\left\Vert T((1,1))\right\Vert$ but in this case $\lambda=\sqrt{2}$ (because $T^{2}=\sqrt{2}T$).

The third one is not correct because you can take $T$ to be $(x,y)\mapsto x$.

The fourth statement is wrong because you can let $T=2\mathrm{Id}$, which is not singular. This is essentially the only possibility, though: if we assume $T$ is invertible, then clearly $\lambda\neq0$, so we can set $S=\frac{1}{\lambda}T$, which is also invertible (with inverse $\lambda T^{-1}$). We have $S^{2}=\frac{1}{\lambda^{2}}T^{2}=\frac{1}{\lambda^{2}}\lambda T=\frac{1}{\lambda}T=S$, so $S=\mathrm{Id}$, i.e. $T=\lambda\mathrm{Id}$.