Suppose that:

(*) for every vector .

Let i, j be any two particular coordinate indices.
(i.e. these are NOT abstract indices, they're constants, in the range 0..3 when working in four dimensions).

Define the vectors:

p =
q =
r =

(e.g. for i=1 and j=3, p = [0,1,0,0], q = [0,0,0,1], r = [0,1,0,1]; for i=j=0, p,q = [1,0,0,0], r = [2,0,0,0]; ...etc.)

Then:



.

Combining the above gives:

(1)

And similarly:

(2)


By (*), the RHS of (1) and of (2) must be equal; so we have proven that for every choice of i, j.
Hence .


If and are symmetric tensors, and , and so in that case (*) implies that .

Another way of looking at this is to note that any (arbitrary) symmetric tensor can be constructed as a sum of vector (outer-)products:





So (*) implies that for any .

This is equivalent to for any (even non-symmetric).

...So once again we get .