Grassmann Algebra

TheRegressiveProduct.nb 43
Now let �Βi � Α� � bi�Β, where the bi are scalars, and substitute into the above equation to
n�m m n
give the required expression for Α. m
Ν
Α � � �bi�Βi
m
Example: Expressing a 3-element in terms of another basis
i�1
Consider a 4-space with basis �e1, e2, e3, e4� and let:
X � 2e1 � e2 � e3 � 3e1 � e3 � e4 ��2e2 � e3 � e4
be a 3-element expressed in terms of this basis.
Suppose now that we have another basis related to the e basis by:
Ε1 � 2�e1 � 3�e3
Ε2 � 5�e3 � e4
Ε3 � e1 � e3
Ε4 � e1 � e2 � e3 � e4
m
3.50
We wish to express X in terms of the new basis. Instead of inverting the transformation to find
the ei in terms of the Εi , substituting for the ei in X, and simplifying, we can almost write the
result required by inspection.
X � �Ε1 � X���Ε2 � Ε3 � Ε4� � �Ε2 � X���Ε1 � Ε3 � Ε4� �
�Ε3 � X���Ε1 � Ε2 � Ε4� � �Ε4 � X���Ε1 � Ε2 � Ε3�
� �5�e1 � e2 � e3 � e4���Ε2 � Ε3 � Ε4� �
�2�e1 � e2 � e3 � e4���Ε1 � Ε3 � Ε4� �
��2�e1 � e2 � e3 � e4���Ε1 � Ε2 � Ε4� �
��e1 � e2 � e3 � e4���Ε1 � Ε2 � Ε3�
But from the transformation we find that Ε1 � Ε2 � Ε3 � Ε4 � 19�e1 � e2 � e3 � e4 whence X
is:
X � 5
�������
19 �Ε2 � Ε3 � Ε4 � 2
�������
19 �Ε1 � Ε3 � Ε4 � 2
�������
19 �Ε1 � Ε2 � Ε4 � 1
�������
19 �Ε1 � Ε2 � Ε3
Expressing a 1-element in terms of another basis
For the special case of a 1-element, the scalar coefficients bi in formula 3.49 can be written
more explicitly, and without undue complexity, resulting in a decomposition of Α in terms of n
independent elements Βi equivalent to expressing Α in the basis Βi .
2001 4 5
n
�
Α�� ��
�
Β1 � Β2 � � � Α � � � Βn �
�������������������������������� ���������������������� ���Βi
Β1 � Β2 � � � Βi � � � Βn �
i�1
3.51