Norm of operator vs. norm of its inverse

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SUMMARY

The discussion centers on the relationship between the norm of an invertible, bounded linear operator T and the norm of its inverse T^{-1}. It is established that the equation \| T^{-1} \| = \frac{1}{\| T \|} does not hold universally, even in finite-dimensional spaces, as demonstrated with the matrix \(\left(\begin{array}{cc} 2 & 0\\ 0 & 1\end{array}\right)\) and its inverse \(\left(\begin{array}{cc} 1/2 & 0\\ 0 & 1\end{array}\right)\). Instead, the inequality \(\frac{1}{\|T\|} \leq \|T^{-1}\|\) is valid for bounded operators, emphasizing the importance of testing functional analysis statements in finite dimensions first.

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Are there any circumstances under which we can conclude that, for an invertible, bounded linear operator T,

<br /> \| T^{-1} \| = \frac{1}{\| T \|} ?<br />

E.g., does this always hold if we know the inverse is bounded?
 
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No, this doesn't even hold for finite-dimensional spaces! (i.e. for matrices).

Consider the matrix

\left(\begin{array}{cc} 2 &amp; 0\\ 0 &amp; 1\end{array}\right).

The norm of this operator is 2. However, the inverse operator is

\left(\begin{array}{cc} 1/2 &amp; 0\\ 0 &amp; 1\end{array}\right)

and this has norm 1.

However, you do have an inequality (for bounded operators of course): Since 1=\|id\|=\|TT^{-1}\|\leq \|T\|\|T^{-1}\|, it follows that \frac{1}{\|T\|}\leq \|T^{-1}\|.
 
Or simpler, the 1x1-matrix (a) has inverse (1/a), and these have norms a and 1/a, respectively :p

In general, it's good advice to test statements in functional analysis in the easy case of finite dimensions first.
 
Landau said:
Or simpler, the 1x1-matrix (a) has inverse (1/a), and these have norms a and 1/a, respectively :p

In general, it's good advice to test statements in functional analysis in the easy case of finite dimensions first.

Good advice. Thanks to all of you :biggrin:
 

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