Infinite Dimensional Vector Space.

**joe909** · February 15th 2010, 09:13 AM

There is a 3 part question that I have been working on.

The first 2 parts are show that for a finite dimensional vector space:

$S \circ T$ is invertible if and only if $S$ and $T$ are invertible.
$S \circ T = I$ if and only if $T \circ S = I$

Theese two I have successfully proven. It is the last part I am having trouble with which is:

Give an example for each that shows that the statements are false for infinite dimensional vector spaces.

Any help would be appreciated, Thanks!

**Bruno J.** · February 15th 2010, 12:07 PM

Hint : consider the vector space $V$ of sequences of real numbers with the map $S: (a_1,\dots,a_n,\dots) \mapsto (0,a_1,\dots,a_n,\dots)$ . Show that $S$ has a left inverse which is not a right inverse.

**joe909** · February 15th 2010, 01:09 PM

Thanks for the hint!

Here is what I got

Let

and $T: (b_1,...,b_n,...) \mapsto (b_2,...,b_{n+1},...)$

$S \circ T$ will give us the original sequence however
$T \circ S$ wont. This proves the second
"

if and only if

" does not hold for infinite dimensional matrices.

However Im still not sure how to prove the first part "

is invertible if and only if

and

are invertible." does not hold for infinite dimensional matrices.

**Bruno J.** · February 15th 2010, 05:59 PM

Well you have to be careful; $T\circ S$ is the identity map, not $S\circ T$ . You are right that it proves the second part of the problem.

For the first part, notice that $T\circ S$ is invertible (it's the identity!) but that $T$ is not. So the part of the statement which says "If $T\circ S$ is invertible, then both $T$ and $S$ are invertible" is false. The part which says "If both $T$ and $S$ are invertible, then $T\circ S$ is invertible" is always true; the inverse of $T\circ S$ is then given by $S^{-1}\circ T^{-1}$ .

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