Exercise 8.C.4

Answers

Proof.

A = (0000 0 0 0 0 0000 0 0 0 0 )

Let

A = (1100 0 5 1 0 0050 0 0 0 5 ).

Then

(A 5I)2(A I) = (4100 0 0 1 0 0 000 0 0 0 0 )2 ( 0100 0 4 1 0 0040 0 0 0 4 ) = (4100 0 0 1 0 0 000 0 0 0 0 ) (0010 0 0 4 0 0000 0 0 0 0 ) = (0000 0 0 0 0 0000 0 0 0 0 ).

Define T L(4) by

T(z1,z2,z3,z4) = (z1 + z2,5z2 + z3,5z3,5z4).

Then M(T) = A and the eigenvalues of T are thus 1 and 5 (the entries on the diagonal). Now 8.36 implies that the minimal polynomial of T is a polynomial multiple of (z 5)(z 1). The previous work shows that (T 5I)(T I)0 and (T 5I)2(T I) = 0. Hence (z 5)2(z 1) is the minimal polynomial of T. By Exercise 11 in section 8B, the multiplicity of 1 is 1 and of 5 is 3. Thereby the characteristic polynomial of T is (z 1)(z 5)3. □

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2017-10-06 00:00
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