# Norm of a Linear Transformation .... Junnheng Proposition 9.2.3 .... ....

• MHB
• Math Amateur
In summary, the conversation discusses Proposition 9.2.3 and Definition 9.2.2 from Chapter 9 of Hugo D. Junghenn's book "A Course in Real Analysis." The conversation focuses on the proof of Proposition 9.2.3 and the substitution of $c = \| \mathbf{x} \|^{-1}$ in the equation $\| c \mathbf{y}\| = c\| \mathbf{y}\|$. The conversation also addresses a mistake in the equation $\| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1$ and clarifies that it should be $\| T ( \| \mathbf{x} Math Amateur Gold Member MHB I am reading Hugo D. Junghenn's book: "A Course in Real Analysis" ... I am currently focused on Chapter 9: "Differentiation on $$\displaystyle \mathbb{R}^n$$" I need some help with the proof of Proposition 9.2.3 ... Proposition 9.2.3 and the preceding relevant Definition 9.2.2 read as follows: https://www.physicsforums.com/attachments/7889 View attachment 7890 In the above proof we read the following: " ... ... If $$\displaystyle \mathbf{x} \neq \mathbf{0}$$ then $$\displaystyle \| \mathbf{x} \|^{-1} \mathbf{x}$$ has a norm $$\displaystyle 1$$, hence $$\displaystyle \| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1$$ ... ... " Now I know that $$\displaystyle T( c \mathbf{x} ) = c T( \mathbf{x} )$$... BUT ...... how do we know that this works "under the norm sign" ...... that is, how do we know ...$$\displaystyle \| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \|$$ ... and further ... how do we know that ... $$\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1$$Help will be appreciated ... Peter Peter said: I know that $$\displaystyle T( c \mathbf{x} ) = c T( \mathbf{x} )$$ ... BUT ... ... how do we know that this works "under the norm sign" ... ... that is, how do we know ... $$\displaystyle \| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \|$$ If you make the substitution$c = \| \mathbf{x} \|^{-1}$in the equation$\| c \mathbf{y}\| = c\| \mathbf{y}\|$(where$c$is a positive constant), then it becomes$\| \| \mathbf{x} \|^{-1} \mathbf{y}\| = \| \mathbf{x} \|^{-1}\| \mathbf{y}\|$. Do that when$\mathbf{y} = T\mathbf{x}$, to get$\| \|T( \mathbf{x} \|^{-1} \mathbf{x})\| = \| \| \mathbf{x} \|^{-1}(T \mathbf{x})\| = \| \mathbf{x} \|^{-1}\|T \mathbf{x}\|$. Peter said: ... and further ... how do we know that ... $$\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1$$ That is a mistake. It should read $$\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le \|T\|$$. Opalg said: If you make the substitution$c = \| \mathbf{x} \|^{-1}$in the equation$\| c \mathbf{y}\| = c\| \mathbf{y}\|$(where$c$is a positive constant), then it becomes$\| \| \mathbf{x} \|^{-1} \mathbf{y}\| = \| \mathbf{x} \|^{-1}\| \mathbf{y}\|$. Do that when$\mathbf{y} = T\mathbf{x}$, to get$\| \|T( \mathbf{x} \|^{-1} \mathbf{x})\| = \| \| \mathbf{x} \|^{-1}(T \mathbf{x})\| = \| \mathbf{x} \|^{-1}\|T \mathbf{x}\|$.That is a mistake. It should read $$\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le \|T\|$$. Thanks Opalg ... Appreciate your help ... Peter Opalg said: If you make the substitution$c = \| \mathbf{x} \|^{-1}$in the equation$\| c \mathbf{y}\| = c\| \mathbf{y}\|$(where$c$is a positive constant), then it becomes$\| \| \mathbf{x} \|^{-1} \mathbf{y}\| = \| \mathbf{x} \|^{-1}\| \mathbf{y}\|$. Do that when$\mathbf{y} = T\mathbf{x}$, to get$\| \|T( \mathbf{x} \|^{-1} \mathbf{x})\| = \| \| \mathbf{x} \|^{-1}(T \mathbf{x})\| = \| \mathbf{x} \|^{-1}\|T \mathbf{x}\|$.That is a mistake. It should read $$\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le \|T\|$$. Hi Opalg ... Just realized that I don't fully understand why $$\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le \|T\|$$ ... Can you please help further and demonstrate why this is the case ... Peter Peter said: Hi Opalg ... Just realized that I don't fully understand why $$\displaystyle \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le \|T\|$$ ... Can you please help further and demonstrate why this is the case ... Peter The definition$\|T\| = \sup\{\|T \mathbf{y} \| : \| \mathbf{y} \| = 1\}$says that if$\| \mathbf{y} \| = 1$then$\|T \mathbf{y} \| \leqslant \|T\|$. But$ \| \mathbf{x} \|^{-1} \mathbf{x} $has norm$1$, so you can substitute that vector for$ \mathbf{y}$, to get$\bigl\|T( \| \mathbf{x} \|^{-1} \mathbf{x} )\bigr\| \leqslant \|T\|$. Opalg said: The definition$\|T\| = \sup\{\|T \mathbf{y} \| : \| \mathbf{y} \| = 1\}$says that if$\| \mathbf{y} \| = 1$then$\|T \mathbf{y} \| \leqslant \|T\|$. But$ \| \mathbf{x} \|^{-1} \mathbf{x} $has norm$1$, so you can substitute that vector for$ \mathbf{y}$, to get$\bigl\|T( \| \mathbf{x} \|^{-1} \mathbf{x} )\bigr\| \leqslant \|T\|\$.
Hi Opalg ... thanks again for the help ...

Peter

## 1. What is the norm of a linear transformation?

The norm of a linear transformation is a measure of the size or magnitude of the transformation. It is a non-negative real number that represents the maximum amount by which the transformation can change the length of a vector.

## 2. How is the norm of a linear transformation calculated?

The norm of a linear transformation is calculated using the formula ||T|| = sup{||T(x)|| : x ∈ V and ||x|| = 1}, where sup stands for the supremum or the least upper bound. Essentially, it is the maximum value of ||T(x)|| as x ranges over all possible unit vectors in the vector space V.

## 3. What is the significance of the norm of a linear transformation?

The norm of a linear transformation is important because it tells us how much the transformation can stretch or compress vectors in the vector space. It is also useful in determining the convergence of a sequence of linear transformations and in solving optimization problems.

## 4. How is the norm of a linear transformation related to the eigenvalues of the transformation?

The norm of a linear transformation is equal to the absolute value of the largest eigenvalue of the transformation. This means that the norm can also be used to find the eigenvalues of a linear transformation.

## 5. Can the norm of a linear transformation ever be negative?

No, the norm of a linear transformation is always a non-negative real number. This is because it represents the maximum amount by which the transformation can change the length of a vector, and a negative length is not possible.

• Topology and Analysis
Replies
3
Views
2K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
7
Views
2K
• Topology and Analysis
Replies
24
Views
2K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
3
Views
2K
• Special and General Relativity
Replies
1
Views
480
• Topology and Analysis
Replies
4
Views
2K
• Linear and Abstract Algebra
Replies
11
Views
2K