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

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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:P2→P1bethe 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 1−xwehave[T(1 −x)]B=[−1]B

Toﬁnd [−1]Bweshould express −1as λ1·(1 −x)+λ2·xfor some

λ1,λ2∈R.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 v∈ker 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!

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to its reduced row-echelon form

1−1 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 v∈ker 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 w∈ImTif 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 1−xand xare abasis of ImT.

Alternatively wecould use row-operations to reduce the matrix to its

row-echelon form 1−1 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

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