Exercise 6.3.1
Solution: The minimal polynomial for the identity operator is . It annihilates the identity operator and the monic zero degree polynomial does not, so it must be the minimal polynomial. The minimal polynomial for the zero operator is . It is a monic polynomial that annihilates the zero operator and again the monic zero degree polynomial does not, so it must be the minimal polynomial.
Exercise 6.3.2
Solution: The characteristic polynomial is
Now for any
Thus for all such that . Thus the minimum polynomial cannot have degree two, it must therefore have degree three. Since it divides it must equal .
Exercise 6.3.3
Solution: The characteristic polynomial equals
Here we used (5-20) page 158. The minimum polynomial is clearly not linear, thus the minimal polynomial is one of , , or . We will plug in to the first three and show it is not zero. It will follow that the minimum polynomial must be .
and
Thus
and
Thus the minimal polynomial must be .
Exercise 6.3.4
Solution: Not diagonalizable, because for characteristic value the matrix and is row equivalent to
which has rank three. So the null space has dimension one. So if is the null space for then has dimension one, which is less than the power of in the characteristic polynomial. So by Theorem 2, page 187, is not diagonalizable.
Exercise 6.3.5
Solution: the only characteristic value is zero. We know the minimal polynomial divides this so the minimal polynomial is of the form for some . Thus by Theorem 3, page 193, the characteristic polynomial’s only root is zero, and the characteristic polynomial has degree . So the characteristic polynomial equals . By Theorem 4 (Caley-Hamilton) .
Exercise 6.3.6
Solution: If and then the minimal polynomial is or . So any such that has minimal polynomial . E.g.
Exercise 6.3.7
Solution: is a basis.
The matrix for is therefore
Suppose is a matrix such that except when . Then has except when . has except when . Etc., where finally . Thus if then for and . Thus the minimum polynomial divides and cannot be for . Thus the minimum polynomial is .
Exercise 6.3.8
Solution: can be given in the standard basis by left multiplication by . Since is given by left multiplication by a matrix, is clearly linear. Since is diagonal, the characteristic values are the diagonal values. Thus the characteristic values of are and . The characteristic polynomial is a degree two monic polynomial for which both and are roots. Therefore the characteristic polynomial is . If the characteristic polynomial is a product of distinct linear terms then it must equal the minimal polynomial. Thus the minimal polynomial is also .
Exercise 6.3.9
Solution: Suppose is . Claim: . Proof by induction: case . . . The trace of is so we have established the claim for the case . Suppose true for up to . Let . Then
Now expanding by minors using the first column, and using induction, we get that this equals
Now if then the coefficient of is so it must be that .
Exercise 6.3.10
Solution: If we represent a matrix as a column vector by stacking the columns of the matrix on top of each other, with the first column on the top, then the transformation is represented in the standard basis by the matrix
And since
it is evident that .
Exercise 6.3.11
Solution: We have
Similarly,So and have the same characteristic polynomials.
But the minimum polynomials need not be equal. To see this let and . Then and so the minimal polynomial of is and the minimal polynomial of is clearly not (it is in fact ).