Another question about projections.

[evilpostingmong]

Say we have a transformation T\inL(V). Now suppose a subspace of V (U) is in the rangespace of T. Now suppose P_Uv=u with u=a1u1+...+amvm.
Now apply T to u to get T(u)=b1u1+...+bmum=/=a1u1+...+amvm. What would happen
if we apply P_Uto T(u)? In other words, what would we end up with after computing P_UT(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
P_UT(u) map to a1u1+..+amum or would it map to b1u1+..+bmum?
Thank you for your response!

 

[kof9595995]

Em..It's a bit hard to understand your post, could you explain what do these mean:
T <br /> \in<br /> 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

 

[evilpostingmong]

Em..It's a bit hard to understand your post, could you explain what do these mean:
T <br /> \in<br /> 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

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.

 

[kof9595995]

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

 

[evilpostingmong]

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

That answers it! Thank you!