Module with matrix multiplication Define the rational 3×3-matrix A:=(141010012) We then define a Q[X]-module on V=Q3 such that Q[X]×V→V,(P,v)→P(A)⋅v where P(A)∈Q3×3 is achieved by plugging in matrix A into polynomial P and P(A)⋅v is therefore the matrix-vector-multiplication. Call this module VA. Argue if VA=Q[X](101)⊕Q[X](01−1).

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Module with matrix multiplication  Define the rational 3×3-matrix  A:=(141010012)  We then define a Q[X]-module on V=Q3 such that  Q[X]×V→V,(P,v)→P(A)⋅v  where P(A)∈Q3×3 is achieved by plugging in matrix A into polynomial P and P(A)⋅v is therefore the matrix-vector-multiplication. Call this module VA.  Argue if VA=Q[X](101)⊕Q[X](01−1).

utripljigmp · Accepted Answer

Your attempt is wrong. In fact, VA equals the direct sum.  The elementary divisors of A (that is, of XI3−A) are (X−1)2,X−2, and then VA is isomorphic to QX(X−1)2⊕QXX−2.  On the other side, one can check that (X−2)(101)=0,(X−1)2(01−1)=0 and (X−1)(01−1)≠0, so the annihilator of the first vector is (X−2), and the annihilator of the second vector is ((X−1)2).  Knowing all these and using the uniqueness of the elementary divisors one can conclude that  VA=Q[X](101)⊕Q[X](01−1)

Module with matrix multiplication Define the rational 3 \times 3-matrix A:=\left(\b

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