B.1. BASIS VECTORS 341

Conversely, suppose A has nonzero determinant. Why are the ek a basis? Supposexkek = 0. Is each xk = 0? Then xka j

ki j = 0 and so for each j, a jkxk = 0 and since A has

nonzero determinant, xk = 0. ■Summarizing what has been shown so far, we know that {ei}p

i=1 is a basis for Rp if andonly if when ei = a j

i i j,

det(

a ji

)̸= 0. (2.5)

If {ei}pi=1 is a basis, then there exists a unique dual basis,

{e j}p

j=1 satisfying

e j ·ei = δji , (2.6)

and that if v is any vector,v = v je

j, v = v je j. (2.7)

The components of v which have the index on the top are called the contravariant compo-nents of the vector while the components which have the index on the bottom are called thecovariant components. In general vi ̸= v j! We also have formulae for these components interms of the dot product.

v j = v ·e j, v j = v ·e j. (2.8)

As indicated above, define gi j ≡ ei ·e j and gi j ≡ ei ·e j. The next theorem describes theprocess of raising or lowering an index.

Theorem B.1.6 The following hold.

gi je j = ei, gi jej = ei, (2.9)

gi jv j = vi, gi jv j = vi, (2.10)

gi jg jk = δik, (2.11)

det(gi j)> 0, det(gi j)> 0. (2.12)

Proof: First,ei = ei ·e je j = gi je j

by 2.7 and 2.8. Similarly, by 2.7 and 2.8,

ei = ei ·e jej = gi je

j.

This verifies 2.9. To verify 2.10,

vi = ei ·v = gi je j ·v = gi jv j.

The proof of the remaining formula in 2.10 is similar.To verify 2.11,

gi jg jk = ei ·e je j ·ek =((ei ·e j)e j

)·ek = ei ·ek = δ

ik.

This shows the two determinants in 2.12 are non zero because the two matrices are inversesof each other. It only remains to verify that one of these is greater than zero. Lettingei = a j

i i j = biji

j, we see that since i j = i j,a ji = bi

j. Therefore,

ei ·e j = ari ir ·b j

kik = ar

i bjkδ

kr = ak

i b jk = ak

i akj.