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I'm trying to solve the following exercise:

Consider the vectorspace $V$ of polynomials of the form $ax + b$, $a, b \in \mathbb{R}$. Find two linear transformations $f_1, f_2 : V \rightarrow V$ for which $f_1 \circ f_2 = \operatorname{id}$ and $f_2 \circ f_1 \neq \operatorname{id}$.

Surely this must be impossible, right? Let $B = \{1, x\}$ be a basis of $V$ and thus:

$$ M_B^B(f_1 \circ f_2) = M_B^B(f_1) M_B^B(f_2) = I $$

such that:

$$ M_B^B(f_2) = M_B^B(f_1)^{-1} $$

and thus:

$$ M_B^B(f_2 \circ f_1) = M_B^B(f_1)^{-1} M_B^B(f_1) = I \implies f_2 \circ f_1 = \operatorname{id} $$

Where am I wrong here?

Peter
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1 Answers1

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I think your argument is correct.

But it involves a bit of a leap ... you have two matrices, which I'll call $P$ and $Q$, for short, and you show that $PQ=I$. From this, you immediately conclude that $Q = P^{-1}$. That is true, in fact, but it's not at all obvious. See here.

In fact, the given problem is essentially equivalent to the problem of finding two $2 \times 2$ matrices $P$ and $Q$ with $PQ=I$ and $QP \ne I$, and that's not possible.

bubba
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