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

• I
• JJ__
In summary: So any projection will do, not just ##T=1## or ##T=0##.In summary, the claim that T^2 equals T is not always true.
JJ__
TL;DR 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.

##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__
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?

Last edited:
Eclair_de_XII said:
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!

My mistake. I really should learn to read better.

HallsofIvy
JJ__ said:
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.

mathwonk
StoneTemplePython said:
... all idempotetent matrices are similar to this.
And how does my idempotent example in post #2 fit into this scheme?

fresh_42 said:
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##

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.

Eclair_de_XII said:
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.

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?

Eclair_de_XII said:
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.

## 1. What is a linear operator?

A linear operator is a mathematical function that maps one vector space to another, while preserving the operations of addition and scalar multiplication.

## 2. What does T^2 = T mean?

This notation means that when the linear operator T is applied twice, it results in the same output as when it is applied once. In other words, the operator T is idempotent.

## 3. What is the significance of T=I or T=0 in this equation?

If T=I, it means that the linear operator T is the identity operator, which means it does not change the input vector. If T=0, it means that the linear operator T is the zero operator, which means it always outputs the zero vector.

## 4. Can there be other solutions besides T=I and T=0?

Yes, there can be other solutions depending on the vector space V and the specific linear operator T. For example, if T is a projection operator, then T^2 = T for any nonzero projection operator.

## 5. How can this equation be useful in studying linear operators?

This equation can be useful in simplifying calculations and understanding the behavior of the linear operator T. It can also help identify special properties of the operator, such as being an identity or zero operator.

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