Exercise 8.C.9

Answers

Proof. Since

4 + 5T 6T2 7T3 + 2T4 + T5 = 0,

multiplying both sides by T5 we get

4T5 + 5T4 6T3 7T2 + 2T1 + I = 0.

Hence 4z5 + 5z4 6z3 7z2 + 2z + 1 is a polynomial multiple of the minimal polynomial of T1 (by 8.46). As it turns out, this is actually the minimal polynomial of T1 (with the coefficients multiplied by 4). To see this, suppose by contradiction that it is not. Hence, the minimal polynomial of T has degree at most 4. This means that

a0I + a1T1 + a 2T2 + a 3T3 + a 4T4 = 0

for some a0,a1,a2,a3,a4 𝔽, not all equal 0. Multiplying both sides of the equation above by T4, we get

a0T4 + a 1T3 + a 2T2 + a 3T + a4I = 0,

which is a contradiction, because the minimal polynomial of T has degree 5. □

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2017-10-06 00:00
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