Prove that tensor is second order

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robc

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I need to prove that a tensor X[sub:2nqew2lh]ij[/sub:2nqew2lh] is second order, where X[sub:2nqew2lh]ij[/sub:2nqew2lh] satisfies:

X[sub:2nqew2lh]ij[/sub:2nqew2lh]a[sub:2nqew2lh]j[/sub:2nqew2lh] = b[sub:2nqew2lh]i[/sub:2nqew2lh] (1)

where a[sub:2nqew2lh]j[/sub:2nqew2lh] is an arbitrary vector, and b[sub:2nqew2lh]i[/sub:2nqew2lh] is a vector.

I would have my proof using the transformation laws if I could arrive at:

X'[sub:2nqew2lh]ij[/sub:2nqew2lh] = l[sub:2nqew2lh]im[/sub:2nqew2lh]l[sub:2nqew2lh]jn[/sub:2nqew2lh]X[sub:2nqew2lh]mn[/sub:2nqew2lh] (2)

But this is where it gets difficult. My first move is the transformation of (1):

X'[sub:2nqew2lh]ij[/sub:2nqew2lh]a'[sub:2nqew2lh]j[/sub:2nqew2lh] = b'[sub:2nqew2lh]i[/sub:2nqew2lh] => X'[sub:2nqew2lh]ij[/sub:2nqew2lh]l[sub:2nqew2lh]jk[/sub:2nqew2lh]a[sub:2nqew2lh]k[/sub:2nqew2lh] = l[sub:2nqew2lh]im[/sub:2nqew2lh]b[sub:2nqew2lh]m[/sub:2nqew2lh] = l[sub:2nqew2lh]im[/sub:2nqew2lh]X[sub:2nqew2lh]mk[/sub:2nqew2lh]a[sub:2nqew2lh]k[/sub:2nqew2lh]

I'm trying to factor out a and arrive at a relationship between X'[sub:2nqew2lh]ij[/sub:2nqew2lh] and X[sub:2nqew2lh]mk[/sub:2nqew2lh], with suitable index swapping so that I end up with (2), but the steps in between elude me. Any ideas?
 
No further help required with this.

Problem : prove the tensor X is second order, given that a is an arbitrary vector, and b is a vector, satisfying:

X[sub:3afft8v9]ij[/sub:3afft8v9]a[sub:3afft8v9]j[/sub:3afft8v9] = b[sub:3afft8v9]i[/sub:3afft8v9]

Solution:

Transforming with orthogonal matrix [ l[sub:3afft8v9]ij[/sub:3afft8v9] ]:

X'[sub:3afft8v9]ij[/sub:3afft8v9]a'[sub:3afft8v9]j[/sub:3afft8v9] = b'[sub:3afft8v9]i[/sub:3afft8v9]

=> X'[sub:3afft8v9]ij[/sub:3afft8v9]l[sub:3afft8v9]jk[/sub:3afft8v9]a[sub:3afft8v9]k[/sub:3afft8v9] = l[sub:3afft8v9]im[/sub:3afft8v9]b[sub:3afft8v9]m[/sub:3afft8v9] = l[sub:3afft8v9]im[/sub:3afft8v9]X[sub:3afft8v9]mk[/sub:3afft8v9]a[sub:3afft8v9]k[/sub:3afft8v9]
=> a[sub:3afft8v9]k[/sub:3afft8v9](X'[sub:3afft8v9]ij[/sub:3afft8v9]l[sub:3afft8v9]jk[/sub:3afft8v9] - l[sub:3afft8v9]im[/sub:3afft8v9]X[sub:3afft8v9]mk[/sub:3afft8v9]) = 0
=> X'[sub:3afft8v9]ij[/sub:3afft8v9]l[sub:3afft8v9]jk[/sub:3afft8v9] - l[sub:3afft8v9]im[/sub:3afft8v9]X[sub:3afft8v9]mk[/sub:3afft8v9] = 0 (a[sub:3afft8v9]k[/sub:3afft8v9] an arbitrary vector)
=> X'[sub:3afft8v9]ij[/sub:3afft8v9]l[sub:3afft8v9]jk[/sub:3afft8v9]l[sub:3afft8v9]jk[/sub:3afft8v9] = l[sub:3afft8v9]im[/sub:3afft8v9]l[sub:3afft8v9]jk[/sub:3afft8v9]X[sub:3afft8v9]mk[/sub:3afft8v9]
=> X'[sub:3afft8v9]ij[/sub:3afft8v9]l[sup:3afft8v9]t[/sup:3afft8v9][sub:3afft8v9]kj[/sub:3afft8v9]l[sub:3afft8v9]jk[/sub:3afft8v9] = l[sub:3afft8v9]im[/sub:3afft8v9]l[sub:3afft8v9]jk[/sub:3afft8v9]X[sub:3afft8v9]mk[/sub:3afft8v9]
=> X'[sub:3afft8v9]ij[/sub:3afft8v9] = l[sub:3afft8v9]im[/sub:3afft8v9]l[sub:3afft8v9]jk[/sub:3afft8v9]X[sub:3afft8v9]mk[/sub:3afft8v9]

which is the form of a second order tensor.
 
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