31.1. BASIS VECTORS 615

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. (31.5)

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

{e j}p

j=1 satisfying

e j ·ei = δji , (31.6)

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

j, v = v je j. (31.7)

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. (31.8)

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 31.1.6 The following hold.

gi je j = ei, gi jej = ei, (31.9)

gi jv j = vi, gi jv j = vi, (31.10)

gi jg jk = δik, (31.11)

det(gi j)> 0, det(gi j)> 0. (31.12)

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

by 31.7 and 31.8. Similarly, by 31.7 and 31.8,

ei = ei ·e jej = gi je

j.

This verifies 31.9. To verify 31.10,

vi = ei ·v = gi je j ·v = gi jv j.

The proof of the remaining formula in 31.10 is similar.To verify 31.11,

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

)·ek = ei ·ek = δ

ik.

This shows the two determinants in 31.12 are non zero because the two matrices are in-verses of 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.