Show that T - I is a projection.

Question

Came across a question where:

I could solve (a) and (b), but have no clue what (c) means.

Please advice.

Answer 1

Hint: Show that $$T=\begin{bmatrix} 1 & 0 & 0\\ 1 & 1 & 1 \\ 1 & 0 & 2\\\end{bmatrix}$$ by looking at the images of $T$ under the canonical basis $\{1,x, x^2\}$.

Answer 2

By definition:

A projection is a linear transformation $P$ from a vector space to itself such that $P^2 = P$.

You have found that $$ T=\begin{bmatrix} 1&0&0\\ 1&1&1\\ 1&0&2 \end{bmatrix} $$ now you have to prove that: $$ P=(T-I)= \begin{bmatrix} 1&0&0\\ 1&1&1\\ 1&0&2 \end{bmatrix}- \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} =\begin{bmatrix} 0&0&0\\ 1&0&1\\ 1&0&1 \end{bmatrix} $$ is such that $P^2=P$ (that is easy)

Answer 3

Check whether $(T-I)^2=(T-I)$. If yes, what are those operators called?

Answer 4

You can show that $(T-I)(a+bx+cx^2)=(a+c)x+(a+c)x^2 \Rightarrow(T-I)^2(a+bx+cx^2)=(T-I)((a+c)x+(a+c)x^2)=(a+c)x+(a+c)x^2\Rightarrow(T-I)^2=(T-I)$