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I've encounter many proves of Tensorail identity that begin with assuming our tensor can be written in form of: $T^{\alpha\beta}=u^{\alpha}v^{\beta}$ . As helpful is it might be, I'm not sure if its a valid assumption, and if there are $\alpha\times\beta$ Independent components.

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In general the tensor will be a linear combination of such products, so if it is clear from the context that if the statement is true for some tensors then it is true for any linear combination of them, then it is enough to prove it for these products. – MBN Jun 30 '13 at 13:23
@MBN thanks! just to make sure , please look at my comment for the answer of vibert – Franz Unberlaude Jun 30 '13 at 13:48
Yes, the covariant derivative is linear. This of course will generalize it only for rank two tensors. – MBN Jun 30 '13 at 13:53
up vote 1 down vote accepted

It's not a good asssumption: a general rank-2 tensor cannot be written in this way. Suppose the vectors live in $D$-dimensional Euclidean space for definiteness. Then the tensor

$$T_{ab} = u_a v_b$$ projects almost all vectors to zero. To be precise, if $x^a v_a = 0,$ then $(T \cdot x)^a = 0$ and similarly if $y^a u_a = 0$ then $(y \cdot T)_b = 0$. Yet the subspace $\{ x^a \, | \, x \cdot v = 0 \}$ is $(D-1)$-dimensional, and the same goes for $\{ y^a \, | \, y \cdot u = 0 \}$.

In general, a tensor can have much smaller null spaces (kernels) - $\delta_{ab}$ is a perfectly fine tensor but it has an empty null space - it doesn't project anything to zero.

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ok. thanks. a specific example I have is using it for developing to covariant derivative for contravariant tensor (2). Is there anything special in that example or in the context of general relativity that justify the assumption? – Franz Unberlaude Jun 30 '13 at 12:53
I don't fully understand what you are trying to do. The covariant derivative acting on tensors (both covariant, contravariant and mixed tensors) already exists, see the big formula here:…. Or are trying to do something else? – Vibert Jun 30 '13 at 13:23
I meant that I had it as an exercise of generalizing the covariant derivative of a vector field to a tensor. the solution assumed that its suffice to to show for $u^{\alpha}v^{\beta}$ – Franz Unberlaude Jun 30 '13 at 13:46
Oh, I see. What is true is that any tensor can be written as a linear combination of things like $u^a v^b$, and you will probably impose linearity on your covariant derivative. So together these two facts might simplify some proofs (but you'd have to open a separate question if you want to get into details). – Vibert Jun 30 '13 at 14:21
thanks again. what do mean by details? if it's about calculating the covariant derivative then I'm now settled. – Franz Unberlaude Jun 30 '13 at 16:27

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