[<< wikibooks] Linear Algebra/Change of Basis
Representations, whether of vectors or of maps, vary with the bases.
For instance, with respect to the two bases

E

2

{\displaystyle {\mathcal {E}}_{2}}
and

B
=
⟨

(

1

1

)

,

(

1

−
1

)

⟩

{\displaystyle B=\langle {\begin{pmatrix}1\\1\end{pmatrix}},{\begin{pmatrix}1\\-1\end{pmatrix}}\rangle }
for

R

2

{\displaystyle \mathbb {R} ^{2}}
, the vector

e
→

1

{\displaystyle {\vec {e}}_{1}}
has two different representations.

R
e
p

E

2

(

e
→

1

)
=

(

1

0

)

R
e
p

B

(

e
→

1

)
=

(

1

/

2

1

/

2

)

{\displaystyle {\rm {Rep}}_{{\mathcal {E}}_{2}}({\vec {e}}_{1})={\begin{pmatrix}1\\0\end{pmatrix}}\qquad {\rm {Rep}}_{B}({\vec {e}}_{1})={\begin{pmatrix}1/2\\1/2\end{pmatrix}}}
Similarly, with respect to

E

2

,

E

2

{\displaystyle {\mathcal {E}}_{2},{\mathcal {E}}_{2}}
and

E

2

,
B

{\displaystyle {\mathcal {E}}_{2},B}
, the identity map
has two different representations.

R
e
p

E

2

,

E

2

(

id

)
=

(

1

0

0

1

)

R
e
p

E

2

,
B

(

id

)
=

(

1

/

2

1

/

2

1

/

2

−
1

/

2

)

{\displaystyle {\rm {Rep}}_{{\mathcal {E}}_{2},{\mathcal {E}}_{2}}({\text{id}})={\begin{pmatrix}1&0\\0&1\end{pmatrix}}\qquad {\rm {Rep}}_{{\mathcal {E}}_{2},B}({\text{id}})={\begin{pmatrix}1/2&1/2\\1/2&-1/2\end{pmatrix}}}
With our point of view that the objects of our studies are vectors and
maps, in fixing bases
we are adopting a scheme of tags or names for these objects, that are
convienent for computation.
We will now see how to translate among these names— we will
see exactly how representations vary as the bases vary.