📂Linear Algebra

Properties of the Space of Invertible Linear Transformations

Theorem¹

Let’s call the set of all invertible linear transformations on $\Omega$.

$$\Omega = \left\{ \text{all invertible linear operator on } \mathbb{R}^{n} \right\}$$

• (a) If the following holds for $T_{1} \in \Omega$ and $T_{2} \in L(\mathbb{R}^{n})$, then $T_{2} \in \Omega$ is true.

  $$\| T_{2} - T_{1} \| \| T_{1}^{-1} \| < 1$$

  Here, $\| T \|$ is the norm of the linear transformation.

• (b) The following holds for $h \gt 0$: $$\| T^{-1} \| \le \dfrac{1}{h} \iff \| T \mathbf{x} \| \ge h | \mathbf{x} |\quad \forall \mathbf{x}$$

• (c) $\Omega$ is an open set.

• (d) The given $f : \Omega \to \Omega$ is continuous as follows:

  $$f( T ) = T^{-1}$$

Proof

(a)

Suppose the following holds for $T_{1} \in \Omega$ and $T_{2} \in L(\mathbb{R}^{n})$.

$$\| T_{2} - T_{1} \| \| T_{1}^{-1} \| < 1$$

Let’s denote $\| T_{1}^{-1} \| = \dfrac{1}{\alpha}, \| T_{2} - T_{1} \| = \beta$. Then, by the assumption, $\beta < \alpha$ is true. By the property of the norm of a linear transformation, the following holds for all $\mathbf{x} \in \mathbb{R}^{n}$.

$$\begin{align*} \alpha | \mathbf{x} | &= \alpha | T_{1}^{-1}( T_{1} ( \mathbf{x})) | \\ &\le \alpha \| T_{1}^{-1} \| | T_{1} (\mathbf{x}) | \\ &= \alpha \dfrac{1}{\alpha} | T_{1} (\mathbf{x}) | = | T_{1} (\mathbf{x}) | \\ &= | T_{1} (\mathbf{x}) - T_{2}(\mathbf{x}) + T_{2}(\mathbf{x}) | \\ &\le | T_{1} (\mathbf{x}) - T_{2}(\mathbf{x}) | + | T_{2}(\mathbf{x}) | \\ &\le \| T_{1} - T_{2} \| |\mathbf{x}| + | T_{2}(\mathbf{x}) | \\ &= \beta |\mathbf{x}| + | T_{2}(\mathbf{x}) | \end{align*}$$

Therefore, we obtain the following:

$$\begin{equation} (\alpha - \beta) | \mathbf{x} | \le | T_{2} ( \mathbf{x} ) |,\quad \forall \mathbf{x}\in \mathbb{R}^{n} \end{equation}$$

At this time, because $\alpha - \beta > 0$, the following is true.

$$\mathbf{x} \ne \mathbf{0} \implies T_{2}(\mathbf{x}) \ne \mathbf{0}$$

Since this is an equivalence condition for a linear transformation to be injective, $T_{2}$ is injective, thus it is an invertible transformation.

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$$\| T^{-1} \| \le \dfrac{1}{h} \iff \| T \mathbf{x} \| \ge h | \mathbf{x} |\quad \forall \mathbf{x}$$

(b)

• $(\Longrightarrow)$

  Let’s denote $\| T^{-1} \| \le \dfrac{1}{h}$. Then,

  $$\begin{align*} && \| T^{-1} \| &\le \dfrac{1}{h} \\ \implies && h \| T^{-1} \| &\le 1 \\ \implies && h \| T^{-1} \| | T \mathbf{x} | &\le | T \mathbf{x} | \\ \implies && h |T^{-1} T \mathbf{x} | \le h \| T^{-1} \| | T \mathbf{x} | &\le | T \mathbf{x} | \\ \implies && h | \mathbf{x} | &\le | T \mathbf{x} | \\ \end{align*}$$

• $(\Longleftarrow)$

  $$\begin{align*} && | T \mathbf{x} | &\ge h | \mathbf{x} | \\ && \| T \| | T \mathbf{x} | &\ge h | \mathbf{x} | \\ \implies && \| T^{-1} \| | T \mathbf{x} | &\ge h \| T^{-1} \| | \mathbf{x} | \\ \implies && \| T^{-1} \| | T \mathbf{x} | &\ge h \| T^{-1} \| | \mathbf{x} | \\ \end{align*}$$

(c)

According to (a), all $T_{2} \in L(\mathbb{R}^{n})$ that satisfy the following are $T_{2} \in \Omega$.

$$d(T_{1}, T_{2}) = \| T_{1} - T_{2} \| < \alpha$$

Then, all $T_{1} \in \Omega$ have a neighborhood that is a subset of $\Omega$, thus they are interior points. Since all elements of $\Omega$ are interior points, $\Omega$ is an open set.

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(d)

Let’s denote composition simply as follows.

$$T_{2} \circ T_{1} = T_{2}T_{1}$$

Substitute $(1)$ for $\mathbf{x} = T_{2}^{-1}(\mathbf{y})$.

$$\begin{align*} && (\alpha - \beta) | T_{2}^{-1}(\mathbf{y})| &\le | T_{2} ( T_{2}^{-1}(\mathbf{y}) ) | = | \mathbf{y} |,\quad \forall \mathbf{y}\in \mathbb{R}^{n} \\ \implies && | T_{2}^{-1}(\mathbf{y})| &\le \dfrac{1}{\alpha - \beta}|\mathbf{y}| \end{align*}$$

Then, by the property of the norm of a linear transformation, the following holds.

$$\| T_{2} ^{-1} \| \le \dfrac{1}{\alpha - \beta}$$

Moreover, the following is true.

$$T_{2}^{-1} - T_{1}^{-1} = T_{2}^{-1}( T_{1} - T_{2} ) T_{1}^{-1}$$

Then, since the norm of a product is greater than the product of norms, the following holds.

$$\| T_{2}^{-1} - T_{1}^{-1} \| \le \| T_{2}^{-1} \| \| T_{1} - T_{2} \| \| T_{1}^{-1}\| \le \dfrac{\beta }{\alpha (\alpha-\beta)}$$

Therefore, when $d(T_{1}, T_{2}) = \|T_{2} - T_{1} \| =\beta \to 0$, since $d(T_{1}^{-1}, T_{2}^{-1})=\|T_{2}^{-1} - T_{1}^{-1}\| \to 0$, the mapping of $T \mapsto T^{-1}$ is continuous.

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