- #1

nateHI

- 146

- 4

## Homework Statement

We have seen that the linear transformation ##T(x_1,x_2)=(x_1,0)## on ##\mathcal{R}^2## has the matrix ##A = \left( \begin{smallmatrix} 1&0\\ 0&0 \end{smallmatrix} \right)## with respect to the standard basis. This operator satisfies ##T^2=T##. Prove that if ##S:\mathcal{R}^2 \to \mathcal{R}^2##, is a linear transformation satisfying ##S^2=S##, then ##S=0## or ##S=I## or there exists a basis ##\mathcal{B}## for ##R^2## such that the matrix of ##S## with respect to ##\mathcal{B}## is ##A##.

## Homework Equations

The professor would prefer if we used basic principals to solve this since it is a undergraduate course.

This homework is past due and I will not get credit for this but it might be on the test so I would appreciate the typical, "make me think" type of replies.

## The Attempt at a Solution

The fact that ##S^2=S## when ##S=0## or ##S=I## is obvious. However, looking at the dimension of the kernel and the range of ##S## in those cases gives a hint on how to solve the remaining case.

1st Case: ##S=0\implies## the ##dim(n(S))=2## and ##dim(range(S))=0##

2nd Case: ##S=I\implies## the ##dim(n(S))=0## and ##dim(range(S))=2## (I has an inverse and therefore is a bijection so the kernel is {0})

The final case must have that ##dim(n(S))=1=dim(range(S))## since that is the only option left to us.

...This is as far as I got.