# Linear Transfrm. Projection Problem

• April 6th 2007, 11:28 PM
ruprotein
Linear Transfrm. Projection Problem
Let T: R^2->R^2. Include figures for each of the following parts.
a) Find a formula T(a,b) where T represents the projectionon the y-axis along x-axis

My Solution, (CAN SOMEONE SEE IF THIS IS RIGHT):

R(T) = Y-AXIS
N(T) = X-AXIS
T(x,y) = (0,y)

b) find a formula T(a,b) where T represents the projectin on the y-axis along the line L = {(s,s): s is in R}.
solution
T(s,y) = ( s, s)
R(T) = R^2
N(T) = {0}

is this right?
• August 24th 2007, 05:32 AM
Rebesques
For the second, R(T)=L.
• August 24th 2007, 10:26 AM
JakeD
Quote:

Originally Posted by ruprotein
Let T: R^2->R^2. Include figures for each of the following parts.

b) find a formula T(a,b) where T represents the projectin on the y-axis along the line L = {(s,s): s is in R}.
solution
T(s,y) = ( s, s)
R(T) = R^2
N(T) = {0}

is this right?

Quote:

Originally Posted by Rebesques
For the second, R(T)=L.

A projection along L onto the y-axis means L is the null space and the y-axis is the range. See this.
So

T(a,b) = (0,b-a)
R(T) = y-axis
N(T) = L

$
\setlength{\unitlength}{.7cm}
\begin{picture}(4,4)

\qbezier(-1,-1)(-1,-1)(4,4)
\qbezier(2,3)(2,3)(0,1)
\put(1.8,3.3){(a,b)}
\put(-2,1){T(a,b)}
\put(1.3,.8){L}

\qbezier(0,0)(0,0)(4,0)
\qbezier(0,-2)(0,-2)(0,4)

\end{picture}
$

PS: I think the answer to problem (a) is correct.
• August 24th 2007, 01:51 PM
Rebesques
Yeah I guess you 're right. The way I read it I thought it was the other way round.