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Grassmann Algebra

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Explor<strong>Grassmann</strong>Matrix<strong>Algebra</strong>.nb 20<br />

Let A be the 2�2 matrix which differs from the matrix discussed above by having distinct<br />

eigenvalues 1 and 2. We will discuss eigenvalues in Section 13.9 below.<br />

A � ��1, x�, �y, 2��; MatrixForm�A�<br />

1 x<br />

�<br />

y 2 �<br />

We compute the square root of A.<br />

sqrtA � <strong>Grassmann</strong>MatrixPower�A, 1<br />

����<br />

2 �<br />

��1 � �� 3<br />

���� �<br />

2 �����<br />

2 � x � y, ��1 � �����<br />

2 � x�,<br />

���1 � �����<br />

2 � y, �����<br />

2 � 1 ���� ��4 � 3<br />

4 �����<br />

2 � x � y��<br />

To verify that this is indeed the square root of A we 'square' it.<br />

����sqrtA � sqrtA��<br />

More generally, we can extract a symbolic pth root. But to have the simplest result we need<br />

to ensure that p has been declared a scalar first,<br />

DeclareExtraScalars��p��<br />

�a, b, c, d, e, f, g, h, p, �, �_ � _�? InnerProductQ, _� 0<br />

pthRootA � <strong>Grassmann</strong>MatrixPower�A, 1 ����<br />

p �<br />

��1 � ��1 � 2 1<br />

����<br />

p � 1<br />

����<br />

���1 � 2 1<br />

����<br />

p � x � y, ��1 � 2 1<br />

p � y, 2 1<br />

���� p � �<br />

1 �����<br />

����<br />

�1 � 2<br />

�<br />

p �<br />

����<br />

p � x�,<br />

�<br />

��������������<br />

����� x � y��<br />

p<br />

�<br />

1<br />

����<br />

2�1� p<br />

We can verify that this gives us the previous result when p = 2.<br />

2001 4 26<br />

sqrtA � pthRootA �. p� 2 �� Simplify<br />

True

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