Solution Set 11: Part A

Benjamin Bakker Mathematics 25a Prof. Berger

December 15, 2003

A.1. Matrices of trace 0 I. Let M ∈ M_{n}(K) be a (nonzero) matrix whose trace is 0. (1) Prove that there exists an invertible matrix P such that the first column of P^{−1}MP is

(0 1 0 ⋮ 0)

Solution. Let e_{1}, ⋯, e_{n} be the standard basis for K^{n} (i.e., e_{i} has a 1 in the ith position and zeroes everywhere else) and let f : K^{n} → K^{n} be the map represented by M in the basis e_{i}. We must find another basis v_{i} such that v_{2} = f(v_{1}); it will be sufficient to find a v that isn’t an eigenvector (i.e., f(v), v is free). If every v ∈ K^{n} is an eigenvector, then for every i, j, if λ_{i} is the eigenvalue of e_{i}, there must exist λ ∈ K with

λ(e_{i} + e_{j}) = f(e_{i} + e_{j}) = λ_{i}e_{i} + λ_{j}e_{j} ⇒ λ = λ_{i} = λ_{j}

since e_{i} is a basis; thus, f = λId. But 0 = Tr(f) = nλ ⇒ λ = 0, and f = 0, contrary to hypothesis. Thus, such a v must exist.

(2) Prove that there exists an invertible matrix Q such that all the diagonal entries of Q^{−1}MQ are 0.

Solution. This follows from applying (1) n times. After applying (1) once, we have a matrix of the form

( \begin{matrix}0 & ∗ ∗ ∗ ∗ \\ 1 0 ⋮ 0 & A \end{matrix} )

where A is an n − 1 × n − 1 matrix; The trace is independent of basis, so because the first diagonal entry is 0, Tr(A) = 0. Now apply (1) repeatedly until we obtain the desired form.

A.2. Matrices of trace 0 II. Let

D = (\begin{matrix}1 \\ & 2 \\ & & . & . & . \\ & & & & & n \end{matrix})

and let ϕ : M_{n}(K) → M_{N}(K) be the map defined by ϕ(M) = MD − DM.

(1) Prove that ker ϕ is the set of all diagonal matrices. What is the rank of ϕ?

Solution. Multiplying some M ∈ M_{n}(K) on the right by D will multiply the j the column by j, while multiplying by D on the left will multiply the ith row by i. Thus, ϕ(M)_{ij} = (j − i)M_{ij}, and ϕ(M) = 0 if and only if M_{ij} = 0 for all i ≠ j, i.e., M is diagonal.

M_{n}(K) is a K-vector space of dimension n^{2} (with basis the matrices E_{i,j} in A.4.); the space of diagonal matrices has dimension n (with basis E_{i,i}). Thus, rk(ϕ) = dim M_{n}(K)− dim ker(ϕ) = n^{2} − n = n(n − 1).

(2) Prove that im(ϕ) is a subspace of the space of matrices whose diagonal entries are 0.

Solution. This follows immediately from the above formula for ϕ(M).

(3) Prove that im(ϕ) is the space of matrices whose diagonal entries are 0.

Solution. The space of matrices with zeroes along the diagonal has dimension n^{2} − n = n(n − 1) = rk(ϕ), so the two must in fact be equal.

A.3. Matrices of trace 0 III (Shoda’s theorem). Using the previous two problems, prove that a matrix M satisfies Tr(M) = 0 if and only if there exist A, B ∈ M_{n}(K) such that AB − BA.

Solution. Clearly 0 ∈ im(ϕ). If M ≠ 0 has trace 0, then by A.1.1 there exists an invertible matrix Q such that Q^{−1}MQ ∈ im(ϕ), by A.2.3, so there is matrix C such that Q^{−1}MQ = CD − DC. Thus,

M = QCDQ^{−1} − QDCQ^{−1} = (QCQ^{−1})(QDQ^{−1}) − (QDQ^{−1})(QCQ^{−1})

Conversely, every matrix in im(ϕ) is traceless.

A.4. Matrices of trace 0 IV. Prove that the subspace of M_{n}(K) generated by all nilpotent matrices is the space of matrices whose trace is 0.

Solution. Let T be the space of traceless matrices. We know that a matrix M is nilpotent if and only if there is an invertible matrix P such that P^{−1}MP is upper triangular with zeroes along the diagonal. The trace is independent of basis, so M ∈ T. Conversely, for any M ∈ T, there exists an invertible P such that Q^{−1}MQ has 0’s along the diagonal. Take A to be the lower-triangular part of Q^{−1}MQ, and B the upper triangular part, both with zeroes along the diagonal. A, B are thus both nilpotent, and therefore so are QAQ^{−1}, QBQ^{−1}; we then have M = QAQ^{−1} + QBQ^{−1}.

A.5. Skolem-Noether’s theorem. Let ϕ : M_{n}(K) → M_{n}(K) be a bijective map such that ϕ(A + λB) = ϕ(A) + λϕ(B), ϕ(Id) = Id, and ϕ(AB) = ϕ(A)ϕ(B). Such a map is called an automorphism of M_{n}(K). The goal of this problem is to find all automorphisms of M_{n}(K).

(1) Check that if P is an invertible matrix, then the map ϕ_{P} defined by ϕ_{P} (M) = P MP^{−1} is an automorphism of M_{n}(K).

Solution. The inverse of ϕ_{P} is ϕ_{P^{−1}}, so ϕ_{P} is bijective. Since multiplication is distributive, ϕ_{P} (A + λB) = P (A + λB)P^{−1} = P AP^{−1} + λP BP^{−1} = ϕ_{P} (A) + λϕ_{P} (B). Clearly ϕ_{P} (Id) = P IdP^{−1} = Id, and ϕ_{P} (AB) = P ABP^{−1} = (P AP^{−1})(P BP^{−1}) = ϕ_{P} (A)ϕ_{P} (B).

(2) Let E_{i,j} be the matrix all of whose entries are 0 except the (i, j)-th which is 1. Compute E_{i,j}E_{k,ℓ} (the answer depends on whether j = k or not).

Solution. After a little thought, its pretty obvious that E_{i,j}E_{k,ℓ} = E_{i,ℓ} if j = k and 0 otherwise.

(3) Now let ϕ be an unspecified automorphism and let U_{i,j} = ϕ(E_{i,j}). Prove that ker(U_{1,1} −Id) is non-trivial.

Solution. Note that ϕ(0) = 0, by linearity. We have E_{1,1}E_{1,1} = E_{1,1} by (2), so (U_{1,1} − Id)U_{1,1} = 0. ϕ is bijective, so U_{1,1} ≠ 0, and im(U_{1,1}) ≠ 0, implying that ker(U_{1,1} − Id) ⊃ im(U_{1,1}) is nontrivial.

(4) Let f_{1} ∈ ker(U_{1,1} − Id) be a nontrivial element and for k ≥ 2 set f_{k} = U_{k,1}(f_{1}). Prove that {f_{1}, …, f_{n}} is a basis of K^{n}.

Solution. Consider the matrix

T = U_{1,n} + U_{n,n−1} + ⋯ + U_{2,1} = ϕ(A)

where

A = E_{1,n} + E_{n,n−1} + ⋯ + E_{2,1} = (\begin{matrix}& & & & & 1 \\ 1 \\ & . & . & . \\ & & & & 1 \end{matrix})

A is invertible, so T is as well, and therefore f_{k} = T^{k−1}f_{1} is nonzero for each k. Note that U_{k,k}f_{i} is 0 for i ≠ k and f_{i} for i = k. Suppose we have

0 = λ_{1}f_{1} + ⋯ + λ_{n}f_{n}

Applying U_{k,k}, we find λ_{k}f_{k} = 0, and so by the above λ_{k} = 0. Thus, the f_{1}, …, f_{n} are free. There are n of them, so they form a basis.

(5) Let P be the matrix of f_{i} in the usual basis of K^{n}. prove that ϕ and ϕ_{P} coincide on the E_{k,1}’s.

Solution. Consider E_{k,1} as representing a map in the f_{i} basis. Thus, E_{k,1}f_{i} is 0 for k ≠ i and f_{i} for i = k. As we saw above, U_{k,1} represents the same map but in the usual basis. The relations between them is

ϕ_{P} (E_{k,1}) = P E_{k,1}P^{−1} = U_{k,1} = ϕ(E_{k,1})

(6) Prove by recurrence on n that if ϕ is an automorphism of M_{n}(K), then there exists an invertible Q such that ϕ(M) = QMQ^{−1}.

Solution. We don’t really need induction. Let e_{i} be the standard basis of K^{n}. With the same notation as in (5), we have

U_{i,j}f_{k} = U_{i,j}U_{k,1}f_{1} = \{ \begin{matrix}e_{i} & j = k \\ 0 & i ≠ j \end{matrix}

and similarly

E_{i,j}e_{k} = \{ \begin{matrix}f_{i} & j = k \\ 0 & i ≠ j \end{matrix}

Thus, E_{i,j} is the matrix of U_{i,j} in the f_{i} basis. Therefore, P E_{i,j}P^{−1} = U_{i,j} for all i, j, and ϕ = ϕ_{P}, by linearity.