# I Confused on definition of projection

1. Mar 12, 2016

### JonnyG

My textbook says: "if $V = W_1 \oplus W_2$,, then a linear operator $T$ on $V$ is the projection on $W_1$ along $W_2$ if, whenever $x = x_1 + x_2$, with $x_1 \in W_1$ and $x_2 \in W_2$, we have $T(x) = x_1$"

It then goes on to say that "$T$ is a projection if and only $T^2 = T$.

But what if $T = I$ (the identity operator)? Then suppose $V$ is finite dimensional and $W$ is a subspace of $V$. Then $V = W \oplus W^{\perp}$ so that any $x \in V$ has the form $x = x_1 + x_2$. So by the first definition, $I$ is not a projection because $I(x) = x = x_1 + x_2 \neq x_1$. But by the second definition, $I$ is a projection, because $I^2 = I$.

What's going on here?

2. Mar 12, 2016

### Simon Bridge

Are you sure the second statement is a definition ... could it be a consequence of applying the definition in a particular situation.

3. Mar 12, 2016

### JonnyG

The exact wording in the book is "In fact, it can be shown (see Exercise 17 of Section 2.3) that $T$ is a projection if and only $T^2 = T$. The article on Wikipedia also gives this definition.

4. Mar 13, 2016

### mathwonk

Id is a trivial projection, based on the trivial decomposition of V into the sum of V and the zero subspace. note: Vperp = {0}. I.e. x1+x2 = x1 precisely when x2 =0.