Linear operators mapping finite dimensional vector spaces are bounded,

Posted byayushkhaitan3437Posted inUncategorized

Theorem: Every linear operator T:V\to W, where V is finite dimensional, is bounded.

Proof \frac{\|Tx\|}{\|x\|}=\frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|a_1e_1+a_1e_2+\dots+a_ne_n\|}\leq \frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{c(|a_1|+|a_2|+\dots+|a_n|)}\leq \frac{\|T(e_i)\|}{c}

where \|T(e_i)\|=\max\{\|T(e_1)\|,\|T(e_2)\|,\dots\}.

What we learn from here is

\|e_i\|(|a_1|+|a_2|+\dots+|a_n|)\geq\|a_1e_1+a_1e_2+\dots+a_ne_n\|\geq c(|a_1|+|a_2|+\dots+|a_n|)

where

\|e_i\|=\max\{\|e_1\|,\|e_2\|,\dots,\|e_n\|\}.

Similarly,

\|e_i\|(|a_1|+|a_2|+\dots+|a_n|)\geq\|a_1e_1+a_1e_2+\dots+a_ne_n\|\geq \|e_k\|(|a_1|+|a_2|+\dots+|a_n|)

where

\|e_k\|=\min\{\|e_1\|,\|e_2\|,\dots,\|e_n\|\}

Another proof of the assertion is

\frac{\|Tx\|}{\|x\|}=\frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|a_1e_1+a_1e_2+\dots+a_ne_n\|}\leq \frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|e_k\|(|a_1|+|a_2|+\dots+|a_n|)}\leq \frac{\|T(e_i)\|}{\|e_k\|}

which is a constant.

Note: why does this not work in infinite dimensional spaces? Because the max and min of \|e_r\| and \|Te_r\| might not exist.

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. I am always excited about talking to people about their research. Please please set up a meeting with me if you feel that I might have an interesting perspective to offer- https://calendly.com/ayushkhaitan/meeting-with-ayush

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