How do you decompose a general tensor into a sum of outer products? Dirac's book "General Theory of Relativity" says on p. 2 that a general rank-2 tensor can be written as a sum of outer products:
$$ T^{\mu\nu} = A^\mu B^\nu + A'^\mu B'^\nu + A''^\mu B''^\nu + \cdots $$
Importantly, he repeats this on p. 18, in developing the covariant derivative, where he mentions that a tensor $T_{\mu\nu}$ is "expressible as a sum of terms like $A_\mu B_\nu$".
Is this obvious? Can someone show or explain this?
 A: Sticking to the 2-by-2 case for brevity: if $T^{\mu\nu}$ in components looks like $$T^{\mu\nu}=\begin{bmatrix}a&b\\c&d\end{bmatrix},$$ then you can write $$T^{\mu\nu}=\begin{bmatrix}a&0\\0&0\end{bmatrix}+\begin{bmatrix}0&b\\0&0\end{bmatrix}+\begin{bmatrix}0&0\\c&0\end{bmatrix}+\begin{bmatrix}0&0\\0&d\end{bmatrix}.$$ Each term can be found by multiplying two vectors with zeroes in all positions except for one, for example $$A^\mu=\begin{bmatrix}a\\0\end{bmatrix},B^\mu=\begin{bmatrix}1\\0\end{bmatrix},A^\mu B^\nu=\begin{bmatrix}a&0\\0&0\end{bmatrix}$$ and $$A'^\mu=\begin{bmatrix}b\\0\end{bmatrix},B'^\mu=\begin{bmatrix}0\\1\end{bmatrix},A'^\mu B'^\nu=\begin{bmatrix}0&b\\0&0\end{bmatrix},$$ etc.
In index notation, we have $$T^{\mu\nu}=\sum_{\lambda,\kappa}T^{\lambda\kappa}{\delta_\lambda}^\mu{\delta_\kappa}^\nu,$$ where I've kept the summation explicit. Each term (in which $\lambda,\kappa$ are fixed) is a product of the two vectors $T^{\lambda\kappa}{\delta_\lambda}^\mu$ and ${\delta_\kappa}^\nu$ (again, where $\lambda,\kappa$ are fixed and $\mu,\nu$ vary).
A tensor can more-or-less be defined as a sum of products of vectors.
