# Tensor multiplication

1. Nov 20, 2011

### PhyPsy

This book says that if $W^aX_a=0$ and $X_a$ is arbitrary, then I should be able to prove that $W^a=0$. I don't see how this is possible. This is the equivalent of the vector dot product, so if, say, $X_a=(1,0,0,0)$, then $W^a$ could be (0,1,1,1), and the dot product would be $1*0+0*1+0*1+0*1=0$. Why would $W^a$ have to be 0?

2. Nov 20, 2011

### lurflurf

It means for any Xa not for some Xa. So in your vector example the only vector orthogonal to all vectors is the zero vector.

3. Nov 21, 2011

### HallsofIvy

Staff Emeritus
If $X_\alpha$ can be any vector, it can be $W_\alpha$. If $W^\alpha X_\alpha= 0$ for $X+_\alpha$ any vector then $W^\alpha W_\alpha= 0$ which immediately gives $W^\alpha= 0$