Using coordinates, Algebra Transformations, Change of Basis, change of basis for transformation

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23 Jun 2011
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Tuesday25/01/2011, Lecture notes byY. Burda
1Using coordinates
Weare going to expand on the idea that while one should think about vectors
in vector spaces and linear transformations, computations should bedone
with coordinates and transformation matrices1.
Example:Let T:P2P1bethe differentiation mapping T(p)=p.
Let A=(1 x, 1+x, 1+x+x2), B=(1 x, x)bebases of P2and P1
respectively.
Find [T]B,Aand use it to find abasis for ker Tand ImT.
Solution:Tofind [T]B,Aweshould apply Tto basis vectors from Aand
express the results as linear combinations of vectors from B.
For the vector 1xwehave[T(1 x)]B=[1]B
Tofind [1]Bweshould express 1as λ1·(1 x)+λ2·xfor some
λ1,λ2R.If 1=λ1·(1 x)+λ2·x,then λ1=1,λ2=1.
Thus
[T(1 x)]B=[1]B=1
1
Similarly wefind
[T(1 +x)]B=[1]B=(1
1)
T(1 +x+x2)B=[1 +2x]B=(1
3)
Hence
[T]B,A=1 1 1
1 1 3
Tofind ker Twerecall that vker Tif and only if its coordinates vector
x=[v]Asolves [T]B,A·x=0.
Tosolvethe system [T]B,A·x=0werow-reduce the matrix
1 1 1
1 1 3
1And remember:hours of calculations can often spareyou tens of minutes of thinking!
1
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to its reduced row-echelon form
11 0
001
So xsolves [T]B,A·x=0if and only if it is amultiple of 1
1
0.
This translates to the fact that [v]Asatisfies [Tv]B=0if and only if [v]A
is amultiple of 1
1
0.
Equivalently vker Tif and only if vis amultiple of 1·(1 x)+1·x,
whichis equal to 2.
In fact wecould haveanticipated this result: the only polynomials whose
derivativeis zero are constantpolynomials. However the use of an unconve-
nientbasis made the calculation somewhat lengthyand non-transparent.
Tofind abasis of ImTwerecal that wImTif and only if [w]Bis in the
column space of [T]B,A.
So all wehaveto do is to find abasis for colum space of
1 1 1
1 1 3
Todo so weshould choose asubset from the column space whichis linearly
independent, but is still spanning the whole column space.
One staightforward wayto do so is to use column operations to reduce
the matrix to its column-reduced form
1 0 0
0 0 1
and then wesee that the vectors (1
0)and (0
1)form abasis for the column
space. Thus the vectors 1xand xare abasis of ImT.
Alternatively wecould use row-operations to reduce the matrix to its
row-echelon form 11 0
001
and notice that the leading 1’s appear inthe first and the third columns.
Thus if wetakethe first and third columns of the original matrix, they
2
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