If T^2 = T, where T is a linear operator on V, T=I or T=0?

[JJ__]

TL;DR

If T^2 = T, where T is a linear operator on a nonzero vector space V, does this imply that either T equals the identity operator on V or that T is the zero operator on V?

I can't think of a counterexample.

 

[fresh_42]

$T^2=T$ means that $T$ is a projection. So any projection will do, not just $T=1$ or $T=0$.

Example: $T=\begin{bmatrix}1&1\\0&0\end{bmatrix}$

 

[Eclair_de_XII]

Why not start with the equality $A^2=A$ where $A$ is the matrix for the transformation $T$, and prove the claim using the matrix algebra of transformations?

 

[HallsofIvy]

Why not start with the equality $A^2=A$ where $A$ is the matrix for the transformation $T$, and prove the claim using the matrix algebra of transformations?

Because, as fresh_42 pointed out, the claim is NOT TRUE!

 

[Eclair_de_XII]

My mistake. I really should learn to read better.

 

[StoneTemplePython]

Summary: If T^2 = T, where T is a linear operator on a nonzero vector space V, does this imply that either T equals the identity operator on V or that T is the zero operator on V?

I can't think of a counterexample.

In terms of basic problem solving technique:

whenever you can imagine only two possibilities, (a) or (b), and they seem to be mutually exclusive... you should ask, why can't it be 'both'? And blocked matrices work perfect for this, e.g.:

$T^{(k)} = \begin{bmatrix} \mathbf I_k & \mathbf 0\mathbf 0^T \\ \mathbf 0\mathbf 0^T & \big(\mathbf 0\mathbf 0\big)_{n-k}^T \end{bmatrix}$

for finite dimensions, all idempotetent matrices are similar to this.

 

[fresh_42]

... all idempotetent matrices are similar to this.

And how does my idempotent example in post #2 fit into this scheme?

 

[StoneTemplePython]

And how does my idempotent example in post #2 fit into this scheme?

in reals, with
$S := \left[\begin{matrix}1 & -1\\0 & 1\end{matrix}\right]$

consider
$S^{-1} \begin{bmatrix}1&1\\0&0\end{bmatrix}S$

and of course $k=1$ and $n-k = 1$

 

[mathwonk]

by the formula T^2 = T, it follows that T is the identity on its actual range, a subspace of V. So choose a basis of this range, and then throw in also a basis of the kernel of T. By the rank nullity theorem, this gives a basis of V, and in this basis the matrix of T is the block matrix in post #6.

 

[WWGD]

Why not start with the equality $A^2=A$ where $A$ is the matrix for the transformation $T$, and prove the claim using the matrix algebra of transformations?

Because the ring of matrices is not an integral domain, so that $A^2-A=0 \rightarrow A(A-I)=0$ does not imply A=0 or A=I.

 

[Eclair_de_XII]

Right, because $A$ can also represent the matrix of the projection transformation for some proper subspace $U$ of $V$, and $(A-I)$ can represent the projection transformation for the complement of $U$. In turn, the composition of these two linear transformations is just the zero transformation, I think?

 

[fresh_42]

Right, because $A$ can also represent the matrix of the projection transformation for some proper subspace $U$ of $V$, and $(A-I)$ can represent the projection transformation for the complement of $U$. In turn, the composition of these two linear transformations is just the zero transformation, I think?

$A^2=A$ is the definition of a projection.