Verifying Tensor Math: Is Step 0 True for Any Tensor A?

[redstone]

Looking for a check on my tensor math to make sure I've done this correctly...
Where D equals the dimension of the metric -
Step 0: {{A}^{ab}}=\frac{1}{D}{{g}^{ab}}{{g}_{cd}}{{A}^{cd}}
Step 1: {{g}_{ab}}{{A}^{ab}}={{g}_{ab}}\frac{1}{D}{{g}^{ab}}{{g}_{cd}}{{A}^{cd}}
Step 2: {{g}_{ab}}{{A}^{ab}}=\frac{1}{D}{{g}_{ab}}{{g}^{ab}}{{g}_{cd}}{{A}^{cd}}
Step 3: {{g}_{ab}}{{A}^{ab}}=\frac{1}{D}g_{a}^{a}{{g}_{cd}}{{A}^{cd}}
Step 4: {{g}_{ab}}{{A}^{ab}}=\frac{1}{D}D{{g}_{cd}}{{A}^{cd}}
Step 5: {{g}_{ab}}{{A}^{ab}}={{g}_{cd}}{{A}^{cd}}
Step 6: {{g}_{ab}}{{A}^{ab}}={{g}_{ab}}{{A}^{ab}}

So I know that the equation in step 0 is true for any tensor A, is that correct?

 

[Dick]

No. Step 0 says A is equal to g*trace(A)/D. If that were true, then it would say any rank two tensor is a multiple of the metric tensor. That doesn't sound right, does it? It isn't true for any tensor. You can't go backward from Step 1 to Step 0. Just because the contractions of two tensors are equal, you can't say the original two tensors are equal.