Linear Algebra II

October 30, 2014

Lecture 19

Filed under: 2014 Fall — Y.K. Lau @ 7:26 PM

The main object today is the change of basis theorem for matrix representation (see p.1 of Lect19-20.pdf). Both the statement and its proof are not difficult as we have an illustrative diagram for it. Then we looked at some simple and direct application (in p.2) and also some more sophisticated application (in p.3). Let me repeat the key-points (for the proof in p.3):

  • Apply Example 7 in p.4 to {T}. We get a pair of bases {B} and {D} in {V} such that {M_{DB}(T)=\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}}.
  • Apply Example 7 in p.4 to {T_A}. We get a pair of bases {H} and {K} in {{\mathbb R}^n} such that {M_{KH}(T_A)=\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}}.
  • Apply the basis change Theorem (p.1) to {T_A}. Note that {M_{StSt}(T_A)=A} and {M_{KH}(T_A)=\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}}. We find a pair of invertible matrices {P,Q} such that {A= Q\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}P}.
  • Apply the 2nd Example in p.2 to find a pair of bases {E} and {F} in {V} such that {P_{BE}= P} and {P_{DF}=Q^{-1}}.

The pair of bases {E} and {F} are desired bases because {M_{FE}(T)=A} (which can be seen by using the basis change theorem in p.1).

Example 4 in p.5 is another a bit difficult but important example.

There is one very important point underlying these examples, that is, viewing a linear operator through its matrix representations and viewing matrices via a linear operator. (That’s what you have to master!) Recall that at the end of lecture, we mentioned that

{(*)} if {A} is similar to {B}, then {A} and {B} are actually the matrix representations of the same linear transformation with respect to some bases.

This provides a way to explain some properties of two similar matrices. For example, if {A} is similar to {B}, then {{\rm rank}(A)={\rm rank}(B)}. By {(*)}, both {{\rm rank}(A)} and {{\rm rank}(B)} equal {{\rm rank}(T)}; thus {{\rm rank}(A)={\rm rank}(B)}.

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