Explanation: Meaning of the Linear Transformation Properties

Given a linear transformation \(L\from \RNrSpc{n}\to \RNrSpc{m}\), let's analyze what is happening here. First of all, \(L\) is a function from \(\RNrSpc{n}\to \RNrSpc{m}\), and

• \(\RNrSpc{n}\) is the domain of \(L\),
• \(\RNrSpc{m}\) is the codomain or target of \(L\).
• So \(L\) receives as input points \(\Vect{x}\) in \(\RNrSpc{n}\), and it transforms such a point \(\Vect{x}\) into a point \(L(\Vect{x})\) in \(\RNrSpc{m}\).

Now, let's turn to the question: what does it mean that this function \(L\) is linear? In intuitive geometric terms, this means that \(L\) transforms any linear motion in \(\RNrSpc{n}\) into a linear motion in \(\RNrSpc{m}\). In symbolic terms, this is equivalent to the requirement that \(L\) satisfies the identities

• \(L(\Vect{x}+\Vect{y}) = L(\Vect{x}) + L(\Vect{y})\) for all \(\Vect{x},\Vect{y}\in\RNrSpc{n}\)
• \(L(t\cdot \Vect{x}) = t\cdot L(\Vect{x})\) for all \(t\in\RNr\), and all \(\Vect{x}\in\RNrSpc{n}\).

What do these identities mean? – The short answer is that a linear \(L\) commutes with the operations ‘vector addition’ and ‘scalar multiplication’. For a more detailed description, read on:

What does the identity \(L(\Vect{x}+\Vect{y}) = L(\Vect{x}) + L(\Vect{y})\) mean?

This identity begins with two vectors \(\Vect{x}\) and \(\Vect{y}\) in \(\RNrSpc{n}\), and then demands the following:

• Form the sum in \(\RNrSpc{n}\): \(\Vect{x} + \Vect{y}\)
• Transform the resulting vector \((\Vect{x} +\Vect{y})\) of \(\RNrSpc{n}\) with \(L\).
• The result is the vector \(L(\Vect{x}+\Vect{y})\) of \(\RNrSpc{m}\). It must be equal to the outcome of the process:
• Using \(L\), transform \(\Vect{x}\) into the vector \(L(\Vect{x})\) of \(\RNrSpc{m}\).
• Using \(L\), transform \(\Vect{y}\) into the vector \(L(\Vect{y})\) of \(\RNrSpc{m}\).
• In \(\RNrSpc{m}\), form the sum \(L(\Vect{x}) + L(\Vect{y})\)

In other words, we get the same result along each of these two computational paths:

1. First add \(\Vect{x}\) and \(\Vect{y}\) in \(\RNrSpc{n}\), then transform the sum vector \(\Vect{x}+\Vect{y}\) using \(L\).
2. First transform \(\Vect{x}\) and \(\Vect{y}\) individually, and then add the transformed vectors \(L(\Vect{x})\) and \(L(\Vect{y})\) in \(\RNrSpc{m}\).

So, it doesn’t matter if we first add and then transform or if we first transform and then add. This is what it means when we say: \(L\) commutes with vector addition.

What does the identity \(L(t\cdot \Vect{x}) = t\cdot L(\Vect{x})\) mean?

This identity begins with a number \(t\) and a vector \(\Vect{x}\) in \(\RNrSpc{n}\), and then demands the following:

• Form the scalar product in \(\RNrSpc{n}\): \(t\cdot \Vect{x}\)
• Transform the resulting vector \(t\cdot \Vect{x}\) of \(\RNrSpc{n}\) with \(L\).
• The result must be equal to the outcome of the process:
• Using \(L\), transform \(\Vect{x}\) into the vector \(L(\Vect{x})\) of \(\RNrSpc{m}\).
• In \(\RNrSpc{m}\), form the scalar product \(t\cdot L(\Vect{x})\)

In other words, we get the same result along each of these two computational paths:

1. First scalar multiply \(\Vect{x}\) by the scalar \(t\), then transform the resulting vector \(t\cdot \Vect{x}\).
2. First transform \(\Vect{x}\) using \(L\), and then scalar multiply the transformed vector by \(t\).

So, it doesn’t matter if we first scalar multiply and then transform or if we first transform and then scalar multiply. This is what it means when we say: \(L\) commutes with scalar multiplication.

Examples of two processes which do not commute include

• Putting socks and shoes on.
• Washing the salad and eating it (a cynical recommendation offered by a computing scientist to a fellow worker who just refused to pay attention to the order in which he invoked certain functions in his code).