📂Linear Algebra

Norm of Linear Transformations

Definition¹

A linear transformation $T \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$’s norm is defined as follows.

$$\begin{equation} \| T \| := \sup \limits_{| \mathbf{x} | = 1} | T ( \mathbf{x} ) | \end{equation}$$

Explanation

If we look at (a), the following equation holds, so $\| T \|$ is the rate of change in size when $T$ maps the elements of $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$. In other words, no matter how much the size changes, it means to the extent of $\| T \|$.

$$\dfrac{|T(\mathbf{x})|}{|\mathbf{x}|} \le \| T \|$$

Also, by definition, $\| T \|$ is known to be the smallest value among $\lambda$ that satisfies the following.

$$| T (\mathbf{x}) | \le \lambda | \mathbf{x} | , \quad \forall \mathbf{x} \in \mathbb{R}^{n}$$

It can be easily verified that $\| T \|$ satisfies the definition of a norm.

• $\| T \| \ge 0$
• $\| T \| = 0 \iff T = 0$
• $\| c T \| = | c | \| T \|$
• $\| T_{1} + T_{2} \| \le \| T_{1} \| + \| T_{2} \|$

Then, since the distance of $L(\mathbb{R}^{n}, \mathbb{R}^{m})$ can be given as follows, $L(\mathbb{R}^{n}, \mathbb{R}^{m})$ becomes a metric space.

$$d(T_{1}, T_{2}) = \| T_{1} - T_{2} \|,\quad T_{1},T_{2} \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$$

Definition $(1)$ and theorem (a) are necessary and sufficient conditions.

$$\| T \| := \sup \limits_{| \mathbf{x} | = 1} | T ( \mathbf{x} ) | \implies | T(\mathbf{x}) | \le \| T \| | \mathbf{x} |,\quad \forall \mathbf{x} \in \mathbb{R}^{n}$$

$$\| T \| := \min \left\{ K : | T(\mathbf{x}) | \le K | \mathbf{x} |,\quad \forall \mathbf{x} \in \mathbb{R}^{n} \right\} \implies \| T \| = \sup \limits_{| \mathbf{x} | = 1} | T ( \mathbf{x} ) |$$

Theorem

• (a) If $T \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, the following holds. $$| T(\mathbf{x}) | \le \| T \| | \mathbf{x} |,\quad \forall \mathbf{x} \in \mathbb{R}^{n}$$

• (b) If $T \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, $\| T \| < \infty$ and $T$ are uniformly continuous.

• (c) If $T_{1} \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$ and $T_{2} \in L(\mathbb{R}^{m}, \mathbb{R}^{k})$, the following holds. $$\|T_{2}\circ T_{1} \| \le \| T_{2} \| \| T_{1} \|$$

Proof

(a)

Let’s say $\mathbf{x} \ne \mathbf{0}$. Then, since $T$ is a linear transformation, the following holds.

$$\begin{align*} \dfrac{ | T(\mathbf{x}) |}{|\mathbf{x}|} &= \dfrac{1}{|\mathbf{x}|}|T(\mathbf{x})| \\ &= \left| \dfrac{1}{|\mathbf{x}|} T(\mathbf{x}) \right| \\ &= \left| T\left( \dfrac{\mathbf{x}}{|\mathbf{x}|} \right) \right| \end{align*}$$

Then, since $\left| \dfrac{\mathbf{x}}{|\mathbf{x}|} \right| = 1$, by the definition of $\| T \|$, the following holds.

$$\begin{align*} && \dfrac{ | T(\mathbf{x}) |}{|\mathbf{x}|} &\le \| T \| \\ \implies && | T(\mathbf{x}) | &\le \| T \| |\mathbf{x}|,\quad \forall \mathbf{x} \in \mathbb{R}^{n} \end{align*}$$

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(b)

Let’s consider $\left\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \right\}$ as the standard basis of $\mathbb{R}^{n}$. Then, for $\mathbf{x} \in \mathbb{R}^{n}$ which is $| \mathbf{x} | \le 1$, the following holds.

$$\mathbf{x} = \sum c_{i}\mathbf{e}_{i} \quad \text{and} \quad |c_{i}| \le 1$$

Then, since $T$ is a linear transformation, the following holds.

$$| T (\mathbf{x}) | = \left| T \left( \sum _{i=1}^{n} c_{i} \mathbf{e}_{i} \right) \right| = \left| \sum _{i=1}^{n} c_{i} T \left(\mathbf{e}_{i} \right) \right| \le \sum _{i=1}^{n} | c_{i} | | T (\mathbf{e}_{i}) | \le \sum _{i=1}^{n} | T (\mathbf{e}_{i}) |$$

Therefore, we obtain the following.

$$\| T \| \le \sum _{i=1}^{n} | T (\mathbf{e}_{i}) | < \infty$$

By (a), for $\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}$, the following holds.

$$|T(\mathbf{x}) - T(\mathbf{y})| \le \| T \| | \mathbf{x} - \mathbf{y} |$$

Let’s assume $\varepsilon > 0$. Let’s say $\delta = \dfrac{\varepsilon}{\| T \|}$. Then, since the following holds, $T$ is uniformly continuous.

$$| \mathbf{x} - \mathbf{y} | < \delta \implies |T(\mathbf{x}) - T(\mathbf{y})| \le \| T \| | \mathbf{x} - \mathbf{y} | = \| T \| \dfrac{\varepsilon}{\| T \|} = \varepsilon$$

(c)

By (a), the following holds.

$$| (T_{2}\circ T_{1}) (\mathbf{x}) | = | T_{2} (T_{1} (\mathbf{x})) | \le \| T_{2} \| |T_{1} (\mathbf{x}) | \le \| T_{2} \| \|T_{1}\| | \mathbf{x} |$$

Therefore, we obtain the following.

$$\| T_{2} \circ T_{1} \| \le \| T_{2} \| \|T_{1}\|$$