Advanced Linear Algebra: Are you ready?

Solution

• \(L: \mathbb R^n \rightarrow \mathbb R^m \) is a linear transformation if and only if it satisfies two conditions.

  1. If \(\alpha \in \mathbb R \) and \(x \in \mathbb R^n \) then \(L( \alpha x ) = \alpha L( x ) \text{.}\)

     In other words: scaling a vector first and then transforming yields the same result as transforming the vector first and then scaling.

  2. If \(x,y \in \mathbb R^n \) then \(L( x + y ) = L( x + y ) \text{.}\)

     In other words: adding two vectors first and then transforming yields the same result as transforming the vectors first and then adding.

  We can prove that these conditions are met by

  \begin{equation*} f( \left( \begin{array}{c} \chi_0 \\ \chi_1 \\ \chi_2 \end{array} \right) ) = \left( \begin{array}{c} \chi_0 + \chi_1 \\ \chi_0 - \chi_2 \end{array} \right) . \end{equation*}

  1. Let \(\alpha \in \mathbb R \) and \(x \in \mathbb R^n \) then

     \begin{equation*} \begin{array}{l} f( \alpha x ) \\ ~~~ = ~~~~ \lt \mbox{ instantiate } \gt \\ f( \alpha \left( \begin{array}{c} \chi_0 \\ \chi_1 \\ \chi_2 \end{array} \right) ) \\ ~~~ = ~~~~ \lt \mbox{ scale the vector } \gt \\ f( \left( \begin{array}{c} \alpha \chi_0 \\ \alpha \chi_1 \\ \alpha \chi_2 \end{array} \right) ) \\ ~~~ = ~~~~ \lt \mbox{ definition of } f \gt \\ \left( \begin{array}{c} \alpha \chi_0 + \alpha \chi_1 \\ \alpha \chi_0 - \alpha \chi_2 \end{array} \right) \\ ~~~ = ~~~~ \lt \mbox{ factor out } \alpha \gt \\ \left( \begin{array}{c} \alpha ( \chi_0 + \chi_1 )\\ \alpha ( \chi_0 - \chi_2 ) \end{array} \right) \\ ~~~ = ~~~~ \lt \mbox{ scale the vector } \gt \\ \alpha \left( \begin{array}{c} \chi_0 + \chi_1 \\ \chi_0 - \chi_2 \end{array} \right)\\ ~~~ = ~~~~ \lt \mbox{ definition of } f \gt \\ \alpha f( \left( \begin{array}{c} \chi_0 \\ \chi_1 \\ \chi_2 \end{array} \right) \\ ~~~ = ~~~~ \lt \mbox{ instantiate } \gt \\ \alpha f( x ). \end{array} \end{equation*}

     Thus, scaling a vector first and then transforming yields the same result as transforming the vector first and then scaling.

  2. Let \(x,y \in \mathbb R^n \) then

     \begin{equation*} \begin{array}{l} f( x + y ) \\ ~~~ = ~~~~ \lt \mbox{ instantiate } \gt \\ f( \left( \begin{array}{c} \chi_0 \\ \chi_1 \\ \chi_2 \end{array} \right) + \left( \begin{array}{c} \psi_0 \\ \psi_1 \\ \psi_2 \end{array} \right) ) \\ ~~~ = ~~~~ \lt \mbox{ vector addition } \gt \\ f( \left( \begin{array}{c} \chi_0 + \psi_0 \\ \chi_1 + \psi_1 \\ \chi_2 + \psi_2 \end{array} \right) ) \\ ~~~ = ~~~~ \lt \mbox{ definition of } f \gt \\ \left( \begin{array}{c} ( \chi_0 + \psi_0 ) + ( \chi_1 + \psi_1 )\\ ( \chi_0 + \psi_0 ) - ( \chi_2 + \psi_2 ) \end{array} \right) \\ ~~~ = ~~~~ \lt \mbox{ associativity; communitivity } \gt \\ \left( \begin{array}{c} ( \chi_0 + \chi_1 ) + ( \psi_0 + \psi_1 )\\ ( \chi_0 - \chi_2 ) + ( \psi_0 - \psi_2 ) \end{array} \right) \\ ~~~ = ~~~~ \lt \mbox{ matrix addition } \gt \\ \left( \begin{array}{c} \chi_0 + \chi_1 \\ \chi_0 - \chi_2 \end{array} \right) + \left( \begin{array}{c} \psi_0 + \psi_1 \\ \psi_0 - \psi_2 \end{array} \right) \\ ~~~ = ~~~~ \lt \mbox{ definition of } f \gt \\ f( \left( \begin{array}{c} \chi_0 \\ \chi_1 \\ \chi_2 \end{array} \right) ) + f ( \left( \begin{array}{c} \psi_0 \\ \psi_1 \\ \psi_2 \end{array} \right) ) \\ ~~~ = ~~~~ \lt \mbox{ instantiation } \gt \\ f( x ) + f ( y ) \end{array} \end{equation*}

     Hence, adding two vectors first and then transforming yields the same result as transforming the vectors first and then adding.

• Some may be able to discover matrix \(A \) by examination. However, what we want you to remember is that for our linear transformation

  \begin{equation*} \begin{array}{l} f( x ) \\ ~~~ = ~~~~ \lt \mbox{ instantiate } \gt \\ f( \left( \begin{array}{c} \chi_0 \\ \chi_1 \\ \chi_2 \end{array} \right) ) \\ ~~~ = ~~~~ \lt \mbox{ write vector as linear combination of standard basis vectors } \gt \\ f( \chi_0 \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right) + \chi_1 \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right) + \chi_2 \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) ) \\ ~~~ = ~~~~ \lt f \mbox{ is a linear transformation } \gt \\ \chi_0 f( \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right) ) + \chi_1 f ( \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right) ) + \chi_2 f ( \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) ) \\ ~~~ = ~~~~ \lt \mbox{ definition of } f \gt \\ \chi_0 \left( \begin{array}{c} 1 + 0 \\ 1 - 0 \end{array} \right) + \chi_1 \left( \begin{array}{c} 0 + 1 \\ 0 - 0 \end{array} \right) ) + \chi_2 \left( \begin{array}{c} 0 + 0 \\ 0 - 1 \\ \end{array} \right) \\ ~~~ = ~~~~ \lt \mbox{ arithmetic } \gt \\ \chi_0 \left( \begin{array}{r} 1 \\ 1 \end{array} \right) + \chi_1 \left( \begin{array}{r} 1 \\ 0 \end{array} \right) ) + \chi_2 \left( \begin{array}{r} 0 \\ - 1 \\ \end{array} \right) \\ ~~~ = ~~~~ \lt \mbox{ definition of matrix-vector multiplication } \gt \\ \left( \begin{array}{r r r} 1 \amp 1 \amp 0 \\ 1 \amp 0 \amp -1 \end{array} \right) \left( \begin{array}{c} \chi_0 \\ \chi_1 \\ \chi_2 \end{array} \right) \end{array} \end{equation*}

  More directly, if we partition \(A \) by columns

  \begin{equation*} A = \left( \begin{array}{ c | c | c } a_0 \amp a_1 \amp a_2 \end{array} \right) \end{equation*}

  then

  \begin{equation*} a_0 = f( \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right) ), a_1 = f( \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right) ), a_2 = f( \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) ). \end{equation*}

  Hence

  \begin{equation*} A = \left( \begin{array}{r r r} 1 \amp 1 \amp 0 \\ 1 \amp 0 \amp -1 \end{array} \right) . \end{equation*}

  This captures the very important insight that the matrix captures the linear transformation by storing how the standard basis vectors are transformed by that linear transformation.

Relevance

The relation between linear transformations and matrices is central to linear algebra. It allows the theory of linear algebra to be linked to practical matrix computations, because matrices capture the action of llinear transformations.

The exercise is relevant to ALAFF in a number of ways by

• Ilustrating a minimal exposure to proofs that will be expected from learners at the start of the course.

• Highlighting that matrices first and foremost are objects that capture the "action" of a linear transformation.

• Emphasizing some of the notation that we will use in the notes for ALAFF:

  • Greek lower case letters are generally reserved for scalars.

  • Roman lower case letters are used to denote vectors.

  • Roman upper case letters are used for matrices.

• Showing the level of detail that will be found in the solutions in the notes for ALAFF.

Resources

A learner who deduces from this exercise that they need a refresher may wish to spend some time with Week 1 and Week 2 of LAFF.