# 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:

\$\displaystyle S \circ T\$ is invertible if and only if \$\displaystyle S\$ and \$\displaystyle T\$ are invertible.
\$\displaystyle S \circ T = I\$ if and only if \$\displaystyle 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 \$\displaystyle V\$ of sequences of real numbers with the map \$\displaystyle S: (a_1,\dots,a_n,\dots) \mapsto (0,a_1,\dots,a_n,\dots)\$. Show that \$\displaystyle 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 \$\displaystyle T: (b_1,...,b_n,...) \mapsto (b_2,...,b_{n+1},...)\$

\$\displaystyle S \circ T\$ will give us the original sequence however
\$\displaystyle 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; \$\displaystyle T\circ S\$ is the identity map, not \$\displaystyle S\circ T\$. You are right that it proves the second part of the problem.

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