description/proof of that matrices multiplications map is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% matrices space.
- The reader knows a definition of Euclidean topology.
- The reader knows a definition of product topology.
- The reader knows a definition of continuous map.
Target Context
- The reader will have a description and a proof of the proposition that any matrices multiplications map is continuous.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(M (d_1 x d_2, \mathbb{R})\): \(= \{\text{ the } d_1 x d_2 \text{ real matrices } \}\) with the topology of the Euclidean topological space, \(\mathbb{R}^{d_1 d_2}\)
\(M (d_2 x d_3, \mathbb{R})\): \(= \{\text{ the } d_2 x d_3 \text{ real matrices } \}\) with the topology of the Euclidean topological space, \(\mathbb{R}^{d_2 d_3}\)
\(M (d_1 x d_3, \mathbb{R})\): \(= \{\text{ the } d_1 x d_3 \text{ real matrices } \}\) with the topology of the Euclidean topological space, \(\mathbb{R}^{d_1 d_3}\)
\(f\): \(: M (d_1 x d_2, \mathbb{R}) \times M (d_2 x d_3, \mathbb{R}) \to M (d_1 x d_3, \mathbb{R}), (M, N) \mapsto M N\)
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Statements:
\(f \in \{\text{ the continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: take each \((M, N) \in M (d_1 x d_2, \mathbb{R}) \times M (d_2 x d_3, \mathbb{R})\) and each open cube around \(M N\), \(C_{M N, \epsilon}\); Step 2: take an open neighborhood of \((M, N)\), \(C_{M, \delta} \times C_{N, \delta}\), and choose \(\delta\) to satisfy \(f (C_{M, \delta} \times C_{N, \delta}) \subseteq C_{M N, \epsilon}\).
Step 1:
Let \((M, N) \in M (d_1 x d_2, \mathbb{R}) \times M (d_2 x d_3, \mathbb{R})\) be any.
Let \(C_{M N, \epsilon}\) be any open cube around \(M N\).
Step 2:
Let us take an open neighborhood of \((M, N)\), \(C_{M, \delta} \times C_{N, \delta}\), where \(C_{M, \delta}\) and \(C_{N, \delta}\) are the open cubes: we are going to choose \(\delta\) to satisfy \(f (C_{M, \delta} \times C_{N, \delta}) \subseteq C_{M N, \epsilon}\).
\((M N)^j_k = M^j_l N^l_k\).
Let \(\overline{M}\) be the maximum absolute component of \(M\); let \(\overline{N}\) be the maximum absolute component of \(N\).
As \(M^j_l\) becomes \(M^j_l + \lambda^j_l\) and \(N^l_k\) becomes \(N^l_k + \tau^l_k\), \(M^j_l N^l_k\) becomes \((M^j_l + \lambda^j_l) (N^l_k + \tau^l_k) = M^j_l N^l_k + M^j_l \tau^l_k + \lambda^j_l N^l_k + \lambda^j_l \tau^l_k\).
Its absolute difference from \(M^j_l N^l_k\) is \(\vert M^j_l \tau^l_k + \lambda^j_l N^l_k + \lambda^j_l \tau^l_k \vert \le \vert M^j_l \vert \vert \tau^l_k \vert + \vert \lambda^j_l \vert \vert N^l_k \vert + \vert \lambda^j_l \vert \vert \tau^l_k \vert \le \sum_{l} \overline{M} \delta + \delta \sum_{l} \vert \overline{N} \vert + \sum_{l} \delta^2 = d_2 \overline{M} \delta + \delta d_2 \vert \overline{N} \vert + d_2 \delta^2\).
It is obvious that \(\delta\) can be chosen small enough such that \(d_2 \overline{M} \delta + \delta d_2 \vert \overline{N} \vert + d_2 \delta^2 \lt \epsilon\).
That means that \(f (C_{M, \delta} \times C_{N, \delta}) \subseteq C_{M N, \epsilon}\).
