Undergrad Is Second rank tensor always tensor product of two vectors?

[arpon]

Suppose a second rank tensor $T_{ij}$ is given. Can we always express it as the tensor product of two vectors, i.e., $T_{ij}=A_{i}B_{j}$ ? If so, then I have a few more questions:
1. Are those two vectors $A_i$ and $B_j$ unique?
2. How to find out $A_i$ and $B_j$
3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?

 

[fresh_42]

Suppose a second rank tensor $T_{ij}$ is given. Can we always express it as the tensor product of two vectors, i.e., $T_{ij}=A_{i}B_{j}$ ?

No, it is a sum of such products.

If so, then I have a few more questions:
1. Are those two vectors $A_i$ and $B_j$ unique?

No. Even for dyadics $A_i \otimes B_j$ you always have $A_i \otimes B_j = c \cdot A_i \otimes \frac{1}{c}B_j$ for any scalar $c \neq 0$.

2. How to find out $A_i$ and $B_j$

$T_{ij}$ is basically any matrix and $A_i \otimes B_j$ a matrix of rank $1$. So write your matrix as a sum of rank-$1$ matrices and you have a presentation.

3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?

No. See the rank explanation above.

 

[haushofer]

3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?

To add to Fresh's answer: this last remark already should make you suspicious. An arbitrary (!) second-rank tensor in three dimensions has 3*3=9 components, but two vectors have 2*3=6 components. What you can do, is to decompose a second rank tensor like T_{ij} as

$$T_{ij} = T_{[ij]} + T_{(ij)} $$

where [ij] stands for antisymmetrization, whereas (ij) stands for symmetrization. Both parts transform independently under coordinate transfo's. The antisymmetric part has 3 independent components, whereas the symmetric part has 6 components. You can even go further in this decomposition, because the trace of the tensor components also does not change under a coordinate transformation.