A well-known homework problem from General Relativity and Tensor Analysis textbooks asks the student to prove that:
$A^{\alpha \beta }B_{\alpha \beta } = 0$,
if A is an antisymmetric $\left ( \frac{2}{0} \right )$ tensor and B is a symmetric $\left ( \frac{0}{2} \right )$ tensor.
Now, this problem will do your head in if you treat the above equation as a matrix product. Remember that we're looking at Einstein Summation Notation here and all will be fine. If we sum on repeated indices, we can work things out in the following way:
$A^{\alpha \beta }B_{\alpha \beta } =-A^{\beta\alpha}B_{\beta\alpha}$ (antisymmetry of A and symmetry of B) (1)
$=-A^{\mu \nu }B_{\mu \nu }$ (relabeling dummy indices) (2)
$=-A^{\alpha \beta }B_{\alpha \beta }$ (relabeling dummy indices again) (3)
Now, if you're not comfortable at this point with just relabeling indices, it's instructive to work out the expressions on both the L.H.S. and the R.H.S. of the equation:
$A^{\alpha \beta }B_{\alpha \beta } = -A^{\beta\alpha}B_{\beta\alpha}$ (1)
L.H.S.: $A^{\alpha \beta }B_{\alpha \beta } = A^{11}B_{11}+A^{12}B_{12}+A^{21}B_{21}+A^{22}B_{22}+\cdot \cdot \cdot$
R.H.S.: $-A^{\beta\alpha}B_{\beta\alpha} = -A^{11}B_{11}-A^{21}B_{21}-A^{12}B_{12}-A^{22}B_{22}+\cdot \cdot \cdot$
Rearranging terms and factoring out a negative one on the R.H.S. gives us:
$-A^{\beta\alpha}B_{\beta\alpha} = -1\ast \left [ A^{11}B_{11}+A^{12}B_{12}+A^{21}B_{21}+A^{22}B_{22}+\cdot \cdot \cdot \right ]$
Which shows that:
$-A^{\beta\alpha}B_{\beta\alpha}=-A^{\alpha \beta }B_{\alpha \beta }$
So, now we go back to (3) above:
$A^{\alpha \beta }B_{\alpha \beta }=-A^{\alpha \beta }B_{\alpha \beta }$
$\Rightarrow 2A^{\alpha \beta }B_{\alpha \beta }=0$
$\Rightarrow A^{\alpha \beta }B_{\alpha \beta } =0$ Q.E.D.
