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

#### JJ__

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.

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#### fresh_42

Mentor
2018 Award
$T^2=T$ means that $T$ is a projection. So any projection will do, not just $T=1$ or $T=0$.

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

• JJ__

#### 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?

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#### HallsofIvy

Science Advisor
Homework Helper
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.

• HallsofIvy

#### StoneTemplePython

Science Advisor
Gold Member
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

Mentor
2018 Award
... all idempotetent matrices are similar to this.
And how does my idempotent example in post #2 fit into this scheme?

#### StoneTemplePython

Science Advisor
Gold Member
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

Science Advisor
Homework Helper
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.

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