Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous

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Theorem

Linear transformations between finite-dimensional normed vector spaces are continuous.


Proof

We have that Norms on Finite-Dimensional Real Vector Space are Equivalent.

Choose the Euclidean norm.

Let $X = \struct {\R^n, \norm {\, \cdot \,}_2}$ and $Y = \struct {\R^m, \norm {\, \cdot \,}_2}$ be normed vector spaces.

Let the matrix $A \in \R^{m \times n}$ be given by:

$A = \begin {bmatrix} a_{1 1} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots \\ a_{m 1} & \cdots & a_{m n} \\ \end{bmatrix}$


We have that Set of Linear Transformations is Isomorphic to Matrix Space.

Let $T_A : \R^n \to \R^m$ be the linear transformation such that:

$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$


Then:

\(\ds \norm {T_A \mathbf x}^2_2\) \(=\) \(\ds \norm {A \mathbf x}^2_2\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^m \paren {\sum_{j \mathop = 1}^n a_{ij} x_j}^2\) Definition of Euclidean Norm
\(\ds \) \(\le\) \(\ds \sum_{i \mathop = 1}^m \paren {\sum_{j \mathop = 1}^n a_{ij}^2} \paren{\sum_{j \mathop = 1}^n x_j^2}\) Cauchy's Inequality
\(\ds \) \(=\) \(\ds \norm {\mathbf x}_2^2 \paren {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2}\) Definition of Euclidean Norm

Hence:

$\ds \norm{T_A \mathbf x}_2 \le \norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 }$

By Continuity of Linear Transformation between Normed Vector Spaces $T_A$ is continuous.

Hence, all linear transformations between finite-dimensional normed vector spaces are continuous.

$\blacksquare$


Sources