If $T\colon \mathbb R^n \to \mathbb R^n $ linear and $T^2 = kT$   It is given that $T$ is a linear transformation from $\mathbb R^n$ to $\mathbb R^n$ such that $T^2 = k T $  for some $k\in \mathbb R$. 
Then, one or more of the options are true 


*

*$\|T(x)\| = |k| \|x\|$ for all $x\in \mathbb R^n$. 

*If $\|T(x)\| = \|x\|$ , for some non- zero vector $x\in \mathbb R^n$ then $ k =\pm 1 $. 

*$\|T(x)\|  >  \|x\|$ for some non- zero vector $x\in \mathbb R^n$, then T is singular. 

*$ T = k I $ where I is an identity transformation. 


Please suggest how to proceed.
 A: 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 $t^2$).


*

*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 $\lambda = 0$).

*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.

*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). 

*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.
A: 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. $T=\lambda\mathrm{Id}$.
A: I love to give silly answers to questions like this because they really make you wonder "What was the professor thinking"?
If you define $T = O$ (the zero matrix), then it satisfies your requirement T^2=kT (in particular, for ANY k).
1) True, for $k=0$ and False for any other $k$.
2) Not applicable!  It's only equal to to the norm of the zero vector!
3) Not applicable!   But kinda interesting, in the sense that its never satisfied (i.e. the norm of the zero matrix is always less than the norm of a non-zero vector.... BUT the (zero) matrix is clearly non-singular.
4) False, FOR ANY K.
Clearly this problem is "ill-posed", right?  But my "solution" is still satisfactory because it satisfied $T^2=kT$ and it also found "at least one true" solution.
