AES Affine Transformation Polynom Representation I have been reading on the polynomial representation of the AES Sbox in the PDF “Essential Algebraic Structure Within the AES” by Murphy and Robshaw (www.isg.rhul.ac.uk/~sean/crypto.pdf) on page 7, that the affine transformation can be represented as $f(a) = \sum^7_{k=0} \lambda_ka^{2^k}$ where $(\lambda_0, \lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_5, \lambda_6, \lambda_7)$ = (0x05, 0x09, 0xf9, 0x25, 0xf4, 0x01, 0xb5, 0x8f). Can me anybody tell how to compute this values $\lambda_0,\ldots,\lambda_7$ from the Matrix $L_A$ on page 7?

Best regards
One way to derive the values $\lambda_0, \lambda_1, ..., \lambda_7$ would be to formulate the eight simultaneous equations in GF(256):

$$\lambda_7 01^{128} + \lambda_6 01^{64} + \lambda_5 01^{32} + \lambda_4 01^{16} + \lambda_3 01^{8} + \lambda_2 01^{4} + \lambda_1 01^{2}  + \lambda_0 01^{1}  = Aff(01)$$
$$\lambda_7 02^{128} + \lambda_6 02^{64} + \lambda_5 02^{32} + \lambda_4 02^{16} + \lambda_3 02^{8} + \lambda_2 02^{4} + \lambda_1 02^{2}  + \lambda_0 02^{1}  = Aff(02)$$
$$\lambda_7 04^{128} + \lambda_6 04^{64} + \lambda_5 04^{32} + \lambda_4 04^{16} + \lambda_3 04^{8} + \lambda_2 04^{4} + \lambda_1 04^{2}  + \lambda_0 04^{1}  = Aff(04)$$
$$...$$
$$\lambda_7 80^{128} + \lambda_6 80^{64} + \lambda_5 80^{32} + \lambda_4 80^{16} + \lambda_3 80^{8} + \lambda_2 80^{4} + \lambda_1 80^{2}  + \lambda_0 80^{1}  = Aff(80)$$

(where $Aff$ is the affine transform, not counting the addition of the constant), and 80 is the GF(256) element represented by the bit pattern '10000000')

and solve this set of 8 linear equations in $\lambda_0, \lambda_1, ..., \lambda_7$

Both sides of these equations are linear; that is $(x^{2^k} + y^{2^k}) = (x+y)^{2^k}$ and $Aff(x) + Aff(y) = Aff(x+y)$, and so if find such a set $\lambda_0, \lambda_1, ..., \lambda_7$, that set will hold for a sum of any subset of (01, 02, 04, ..., 80); since that's the entire field $GF(256)$, we have:

$$\lambda_7 x^{128} + \lambda_6 x^{64} + \lambda_5 x^{32} + \lambda_4 x^{16} + \lambda_3 x^{8} + \lambda_2 x^{4} + \lambda_1 x^{2}  + \lambda_0 x^{1}  = Aff(x)$$

for all $x$.