# Operator norm --- Remarks by Browder After Lemma 8.4 ....

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In summary, Peter is explaining to the reader that the sup and inf in (8.2) and (8.3) are actually max and min, i.e. they are actually attained. He also mentions that there is another approach to (8.2) that can be found in an algebra textbook. Lastly, he mentions that the sigmas are the singular values of $\mathbf A$.
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I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.1 Linear Algebra ...

I need some help in fully understanding some remarks by Browder after Lemma 8.4 pertaining to the "operator norm" Lemma 8.4 ... The relevant text including Lemma 8.4 reads as follows:

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Near the end of the above text we read the following:

" ... ... We leave it to the reader to show that the sup and inf in (8.2) and (8.3) are actually max and min, i.e. that they are actually attained ... ... "
Could someone please demonstrate rigorously that the sup and inf in (8.2) and (8.3) are actually max and min, i.e. that they are actually attained ... ...

Help will be much appreciated ...

Peter

#### Attachments

• Browder - 1 - Lemma 8.4 ... PART 1 .. .png
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• Browder - 2 - Lemma 8.4 ... PART 2 ... .png
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Peter said:
Could someone please demonstrate rigorously that the sup and inf in (8.2) and (8.3) are actually max and min, i.e. that they are actually attained ... ...
I think that you are going to have to quote a theorem from earlier in the Analysis course in order to prove these results.

The function given by $f(\mathbf{v}) = |T\mathbf{v}|$ is a continuous function from $\Bbb{R}^n$ to $\Bbb{R}$. The set $\{\mathbf{v}\in \Bbb{R}^n:|\mathbf{v}|\leqslant1\}$ is a closed, bounded subset of $\Bbb{R}^n$. There is a theorem that a continuous function on a closed, bounded set is bounded and attains its bounds. That explains why the sup in (8.2) is attained.

As for (8.3), you can check that the inf in (8.3) is equal to the sup in (8.2). So if the sup in (8.2) is attained at some vector $\mathbf{v}$, then the inf in (8.3) is attained at that same vector..

the above approach using Compactness probably is the right answer for an analysis text

- - - - -
However, there is an alternative algebraic approach to (8.2) that I hope is of considerable interest to OP. First focus on $\big \Vert \mathbf v \big \Vert_2 = 1$ and consider
$0 \leq \sigma_n \leq \sigma_{n-1} \leq ... \leq \sigma_2 \leq \sigma_1$
then for $j\in\{1,2,..., n\}$ we have the point-wise bound

$\sigma_j^2 \leq \sigma_1^2$
which is preserved under rescaling by real non-negative numbers $w_j$ (in particular for convex combinations, where $\sum_{j=1}^n w_j = 1$)

so we have
$w_j \sigma_j^2 \leq w_j \sigma_1^2$
and summing over the bound
$\sum_{j=1}^n w_j \sigma_j^2 \leq \sum_{j=1}^n w_j \sigma_1^2 =\sigma_1^2 \cdot \sum_{j=1}^n w_j = \sigma_1^2$ (now to accommodate the 'more general' case of $\big \Vert \mathbf v \big \Vert_2 \leq 1$, insert a slack parameter $\sigma_{n+1} = 0$ and $w_{n+1}\geq 0$, where again $\sum_{j}w_j = 1$ and re-run the above argument)

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what are these sigmas? They are the singular values of $\mathbf A$ and recall that one way of constructing them is to look at the eigenvalues of $\big(\mathbf A^* \mathbf A\big)$ (and then take square roots). Why should the eigenvalues of $\mathbf A^* \mathbf A$ come up? Because $\big \Vert \mathbf A\mathbf x \big \Vert_2^2 = \mathbf x^* \mathbf A^*\mathbf A \mathbf x$ and hopefully one recalls how to examine quadratic form problems...

There are a few more details that need filled in here, and I'd like to leave them as an exercise for the reader Peter

incidentally the trace norm is better known as a Frobenius norm (or Schatten 2 norm, in which case the operator norm is the Schatten $\infty$ norm)

## 1. What is the operator norm?

The operator norm is a mathematical concept used to measure the size or magnitude of a linear operator on a vector space. It is defined as the maximum possible length of the operator applied to a vector of unit length.

## 2. How is the operator norm calculated?

The operator norm is calculated by taking the maximum value of the operator applied to a vector of unit length. This can be done using various methods, such as finding the supremum of the operator's values or using the singular value decomposition.

## 3. Why is the operator norm important?

The operator norm is important because it allows us to quantify the size or magnitude of a linear operator. This is useful in many areas of mathematics, such as functional analysis, where the properties of linear operators are studied.

## 4. What is the relationship between the operator norm and the eigenvalues of a matrix?

The operator norm is related to the eigenvalues of a matrix through the spectral theorem, which states that the operator norm is equal to the maximum absolute value of the eigenvalues. In other words, the operator norm is the maximum possible magnitude of the eigenvalues.

## 5. How is the operator norm used in applications?

The operator norm is used in many applications, such as in the analysis of differential equations, optimization problems, and in the study of linear transformations in geometry. It is also used in the development of numerical methods for solving linear systems of equations.

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