Linear Map Comes from a Matrix: Explanation

Every linear transformation \(L\from \RNrSpc{n}\to \RNrSpc{m}\) may be represented by a matrix, ... which matrix? – Here we explain first how the representing matrix is built. Then we explain which fundamental property of a linear map which makes this result possible.

How to build the matrix representing a linear map   Here we explain the procedure by which we build the matrix \(\Mtrx{A}\) representing a linear transformation \(L\from \RNrSpc{n}\to \RNrSpc{m}\):

  1. First apply \(L\) to the vector \(\StdBssVec{1}=(1,0,\dots ,0)\); then use the resulting vector of \(\RNrSpc{m}\) as the first column of \(\Mtrx{A}\).
  2. Next apply \(L\) to the vector \(\StdBssVec{2}=(0,1,0,\dots ,0)\), then use the resulting vector of \(\RNrSpc{m}\) as the second column of \(\Mtrx{A}\).
  3. etc. until
  4. Finally apply \(L\) to the vector \(\StdBssVec{n}=(0,\dots ,0,1)\), then use the resulting vector of \(\RNrSpc{m}\) as the \(n\)-th and last column of \(\Mtrx{A}\).

Why does the representing matrix exist?   Recall that for \(n\geq 1\), \(\RNrSpc{n}\) contains infinitely many points. Therefore the function \(L\) must make infinitely many assignments, one assignment of a point \(L(\Vect{x})\) in \(\RNrSpc{m}\) for each \(\Vect{x}\) in \(\RNrSpc{n}\).

What we learn here is that, because \(L\) is linear, all of these infinitely many assignments are determined by the finite collection of assignments

  1. \(L\) sends \(\StdBssVec{1}\) to \(L(\StdBssVec{1})\).
  2. \(L\) sends \(\StdBssVec{2}\) to \(L(\StdBssVec{2})\).
  3. etc.
  4. \(L\) sends \(\StdBssVec{n}\) to \(L(\StdBssVec{n})\).

So this is it: finite base information implies information about infinitely many transformation situations. – What we describe here is a generalization of familiar linear functions like