Infinite Dimensional Vector Space.

• Feb 15th 2010, 09:13 AM
joe909
Infinite Dimensional Vector Space.
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!
• Feb 15th 2010, 12:07 PM
Bruno J.
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.
• Feb 15th 2010, 01:09 PM
joe909
Thanks for the hint!

Here is what I got

Let http://www.mathhelpforum.com/math-he...1f6dd197-1.gif
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
"http://www.mathhelpforum.com/math-he...47f907c1-1.gif if and only if http://www.mathhelpforum.com/math-he...ccc97ce5-1.gif" does not hold for infinite dimensional matrices.

However Im still not sure how to prove the first part "http://www.mathhelpforum.com/math-he...a7e5c5f6-1.gif is invertible if and only if http://www.mathhelpforum.com/math-he...1a47546e-1.gif and http://www.mathhelpforum.com/math-he...4ff731d3-1.gif are invertible." does not hold for infinite dimensional matrices.
• Feb 15th 2010, 05:59 PM
Bruno J.
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 $S^{-1}\circ T^{-1}$.