1. Jul 8, 2009

### evilpostingmong

Say we have a transformation T$$\in$$L(V). Now suppose a subspace of V (U) is in the rangespace of T. Now suppose PUv=u with u=a1u1+...+amvm.
Now apply T to u to get T(u)=b1u1+...+bmum=/=a1u1+...+amvm. What would happen
if we apply PUto T(u)? In other words, what would we end up with after computing PUT(u)?
I'm just wondering whether or not applying a transformation (a non-projection in this case) after a projection would result
in a different output after applying the projection to the image of T? In other words, would
PUT(u) map to a1u1+..+amum or would it map to b1u1+..+bmum?

2. Jul 8, 2009

### kof9595995

Em..It's a bit hard to understand your post, could you explain what do these mean:
T $$\in$$ L(V),
V (U) (what does U mean)
PU (projection?)
T(u)=b1u1+...+bmum=/=a1u1+...+amvm
It would be helpful if you can clarify these

3. Jul 9, 2009

### evilpostingmong

T is a transformation that sends a vector from V to V.
U is a subspace of V. PU is a projection onto U. And T(u)=b1u1+..+bmum=/=a1u1+...+amum means that T is not
an identity operator, since u=a1u1+...+amum.

Last edited: Jul 9, 2009
4. Jul 9, 2009

### kof9595995

I think in general cases they wouldn't be same, just take an exmple
projection
$$P = \left( {\begin{array}{*{20}c} 1 & 0 \\ 0 & 0 \\ \end{array}} \right)$$
transformation(I take a rotation)
$$T = \left( {\begin{array}{*{20}c} 0 & { - 1} \\ 1 & 0 \\ \end{array}} \right)$$
take a vector
$$\overrightarrow v = \left[ \begin{array}{l} 1 \\ 1 \\ \end{array} \right]$$
then compute as what you said
$$P\overrightarrow v = \left[ \begin{array}{l} 1 \\ 0 \\ \end{array} \right]$$
$$PT\overrightarrow v = \left[ \begin{array}{l} - 1 \\ 0 \\ \end{array} \right]$$

5. Jul 9, 2009