Problem Set #11 Part B Official Solutions Total Points: 44

Inna Zakharevich

January 18, 2004

(7) 1.

(5) (1) We construct a basis for V by taking pairs of vectors v, fv, so that every vector v in a basis has a fv also in the basis (or has a multiple in the basis). Then the vectors v such that fv = v are eigenvectors with eigenvalue 1. Consider the vectors v such that fv ≠ v. If fv = −v then v is an eigenvector with eigenvalue −1. Now suppose that v is not proportional to fv. Then consider the vectors fv + v, fv − v. These will also form a basis of the space spanned by v, fv. f(fv + v) = f^{2}v + fv = v + fv, so it is an eigenvector with eigenvalue 1. f(fv − v) = f^{2}v − fv = v − fv, so it is an eigenvector with eigenvalue −1. Thus we have a basis of eigenvectors, so f is diagonalizable, with eigenvalues ±1.

(2) (2) Such a map is called a symmetry because the subspace V_{−} is reflected over the subspace V_{+}.

(5) 2. First note that if V is a bijection then ker V = {0} so V = ker V ⊕ im V. Thus we will ignore this case.

Suppose that 0 is not a root of the characteristic polynomial. Then we know that Π_{f} = fP (f) + c for c ≠ 0. Thus we see that v = −\frac{1}{c}fP (f)v, so v ∈ im f. Thus f is injective, and therefore it is a bijection, which is the case we are not considering. Now suppose that 0 is a root with multiplicity k > 1. Then Π_{f} = f^{k}P (f) with P (0) ≠ 0. Then we know that 0 = f^{k}P (f)v for all v. Thus f^{k−1}P (f)v is both in the kernel and the image, since f(f^{k−1}P (f)v) = 0, and f(f^{k−2}P (f)v) = f^{k−1}P (f)v. Thus ker f ∩ im f ≠ {0}, so V ≠ ker f ⊕ im f. Thus if V = ker f ⊕ im f then 0 is a simple root of Π_{f}.

Now suppose that 0 is a simple root of Π_{f}. Then we know that Π_{f} = fP (f), with P (0) ≠ 0. Notice that if P (f)v = 0 then v = −\frac{1}{P (0)}(P (f) − P (0)), which means that v ∈ im f. On the other hand, if v ∈ ker f then P (f)v ≠ 0, since P (f)v = P (0)v. Thus ker f ∩ im f = {0}. Since we already knew that dim V = dim ker f + dim im f, we see that V = ker f ⊕ im f.

(10) 3.

(4) (1) Consider the minimal polynomial for f. Suppose that it is linear; this means that f is a multilple of the identity. In this case ker f = {0} and im f = V, so V = ker f ⊕ im f and we are done. Now suppose that the minimal polynomial is not linear. We know that the polynomial f^{2} − f works, so we see that it is the minimal polynomial. Since 0 is a simple root of x^{2} − x we see, by the previous problem, that V = im f ⊕ ker f.

(4) (2) Notice that, for all v such that v = fw (for some w), fv = v. Thus on the image of f, f is the identity and so has eigenvalues 1. On the kernel it has eigenvalues 0. From the first part we see that we can have a basis which consists only of vectors in the image or the kernel, so we see that f is diagonalizable, and has eigenvalues 1 and 0.

(2) (3) f is called a projection because it kills takes a splitting of V as V_{1} ⊕ V_{2} and kills the vectors that are in V_{1}, “projecting” all vectors onto their component in V_{2}.

(5) 4. Consider a polynomial of degree m. We will show that for any m ≥ 0 we can find a polynomial P (x) such that f(P (x)) = m(m − 1)P (x).

For m = 0, 1 this is obvious since f kills all polynomials of degrees 0, 1, so the polynomials 1, x work. Suppose m > 1. Then, if P (x) = ∑ a_{n}x^{n} we know that f(P (x)) = (x^{2} − 1) ∑ n(n − 1)a_{n}x^{n−2} = ∑((n − 2)(n − 3)a_{n−2} − n(n − 1)a_{n})x^{n−2}. Notice that for n < m − 1 this means that we can express a_{n} in terms of a_{n+2}. If we set a_{m} = 1 and a_{m−1} = 0 and then let a_{n} = \frac{(n+1)(n+1)}{n(n−1)−m(m−1)}a_{n+2} we a polynomial with the desired characteristics. Since this worked for every degree we see that the map is diagonalizable, since we can take an eigenvector of every degree and this will form a basis.

(5) 5. Notice that

M^{k} = ( \begin{matrix}A^{k} & 0 \\ 0 \\ B^{k} \end{matrix} ) .

Therefore for any polynomial P

P (M) = ( \begin{matrix}P (A) & 0 \\ 0 \\ P (B) \end{matrix} ) .

We see that P (M) = 0 if and only if P (A) = 0 and P (B) = 0. Therefore Π_{A}|Π_{M} and Π_{B}|Π_{M}. The polynomial of least degree such that this holds is lcm(Π_{A}, Π_{B}), so we see that Π_{M} = lcm(Π_{A}, Π_{B}).

(7) 6.

(2) (1) We will prove this by induction. The base case is true by definition. Suppose that M^{k}N − NM^{k} = kλM^{k}. Then

kλM^{k+1} = M(M^{k}N − NM^{k}) = M^{k+1}N + (MN)M^{k}
= M^{k+1}N − (MN − λM)M^{k} = M^{k+1}N − NM^{k+1} − λM

which implies the desired result.

(5) (2) Notice that if π_{M} = ∑_{i=0}^{n} a_{i}x^{i} we have

0 = Π_{M}N − NΠ_{M} = λ ∑_{i=0}^{n} a_{i}ix^{i}.

The polynomial on the right has the same degree as Π_{M}, so it must be an integer multiple of it. It is not unless all a_{i} = 0 for i < n. Thus we see that Π_{M} = x^{n}, which means that M is nilpotent.