As always, M_n(\mathbb{C}) is the ring of n\times n matrices with complex entries.

Problem. Let

Z:=\{A \in M_n(\mathbb{C}): \ \ \text{tr}(A)=0\}, \ \ \ \ \ N:=\{A \in M_n(\mathbb{C}): \ \ A \ \text{is nilpotent}\}.

i) Show that Z is a vector space but N is not.

ii) Show that Z=\text{span}(N).

SolutionLet \{e_{ij}: \ 1 \le i,j \le n\} \subset M_n(\mathbb{C}) be the standard basis of M_n(\mathbb{C}).

i) Let A_1,A_2 \in Z and a_1,a_2 \in \mathbb{C}. Then


So a_1A_1+a_2A_2 \in Z and thus Z is a vector space. To show that N is not a vector space, see that, for example, e_{12}, e_{21} \in N, because e_{12}^2=e_{21}^2=0, but e_{12}+e_{21} \notin N because (e_{12}+e_{21})^3=e_{12}+e_{21}.

ii) Since all the eigenvalues of a nilpotent matrix are zero, the trace of a nilpotent matrix zero and thus N \subseteq Z implying that \text{span}(N) \subseteq Z, because Z is a vector space.
So, to complete the proof of ii), we need to show that Z \subseteq \text{span}(N). Let A \in Z. Let U be an upper triangular element of M_n(\mathbb{C}) similar to A (such U exists because the field of complex numbers is algebraically closed). So


for some invertible element P \in M_n(\mathbb{C}). Clearly we can write U=D+V, where D \in M_n(\mathbb{C}) is diagonal and V \in M_n(\mathbb{C}) is strictly upper triangular. Notice that V is nilpotent because all of its eigenvalues are zero (because its eigenvalues are the entries on the main diagonal and those entries are all zero). So


and PVP^{-1} \in N. Thus, in order to complete the solution, we only need to show that D \in \text{span}(N). Clearly D \in Z because A,V \in Z. Let d_i be the (i,i)-entry of D. Since D \in Z, we have d_n=-\sum_{i=1}^{n-1}d_i and thus

\displaystyle D=\sum_{i=1}^{n-1}d_i(e_{ii}-e_{nn}).

So, in order to show that D \in \text{span}(N), we only need to show that e_{ii}-e_{nn} \in \text{span}(N) for all 1 \le i < n and to do that, just write 

\displaystyle e_{ii}-e_{nn}=\frac{1}{2}(e_{ii}+e_{in}-e_{ni}-e_{nn})+\frac{1}{2}(e_{ii}-e_{in}+e_{ni}-e_{nn}). \ \ \ \ \ \ \ \ \ \ \ (*) 

and see that 


So both e_{ii}+e_{in}-e_{ni}-e_{nn} and e_{ii}-e_{in}+e_{ni}-e_{nn} are in N and therefore e_{ii}-e_{nn} \in \text{span}(N), by (*). \ \Box


Problem. Let H_1, H_2, \cdots , H_k, \ \ k \ge 2, be some normal subgroups of a group G and suppose that H_i \cap H_j=\{1\} for all i \ne j. Show that G contains a subgroup isomorphic to H_1 \times H_2 \times \cdots \times H_k if k=2, but not necessarily if k \ge 3.

Solution. Suppose first that k=2. Since H_1,H_2 are normal, H_1H_2 is a subgroup of G and

h_1h_2h_1^{-1}h_2^{-1} \in H_1 \cap H_2=\{1\},

for all h_1 \in H_1, h_2 \in H_2 implying that h_1h_2=h_2h_1. Thus the map

f: H_1 \times H_2 \to H_1H_2

defined by f(h_1,h_2)=h_!h_2 is an onto group homomorphism. Since H_1 \cap H_2=\{1\}, \ f is also one-to-one, hence an isomorphism.

A counter-example for k \ge 3 is G=\mathbb{Z}_2^{k-1}, the direct product of k-1 copies of \mathbb{Z}_2. Clearly every 1 \ne g \in G has order two. Now choose k distinct elements

g_1, \cdots , g_k \in G\setminus \{1\}

(we can do that because k < 2^{k-1}=|G|) and let H_i be the subgroup of G generated by g_i. Since G is abelian, each H_i is normal in G and clearly H_i \cap H_j=\{1\} for all i \ne j because H_i,H_j are distinct subgroups of order two. Now H_1 \times H_2 \times \cdots \times H_k has 2^k > 2^{k-1}=|G| elements and so it can’t be isomorphic to a subgroup of G. \ \Box

As usual, M_n(\mathbb{C}) is the ring of n\times n matrices with entries from \mathbb{C}, the field of complex numbers.

Problem. Let p be a prime number and A \in M_{p-1}(\mathbb{C}). Show that if A^{p+1}=A and \text{tr}(A)=0, then A=0.

Solution. Since A^{p+1}=A, the possible eigenvalues of A are the roots of the polynomial x^{p+1}-x, which are

0, \ \ \alpha^k, \ \ k=0, \cdots , p-1,

where \alpha=e^{2\pi i/p}, the p-th primitive root of unity. Let n_k \ge 0 be the number of eigenvalues of A equal to \alpha^k. Then, since \text{tr}(A)=0, we must have

\displaystyle \sum_{k=0}^{p-1}n_k\alpha^k=0. \ \ \ \ \ \ \ \ \ \ \ (*)

Now consider the polynomials

\displaystyle q_1(x)=\sum_{k=0}^{p-1}x^k, \ \ \ \ q_2(x):=\sum_{k=0}^{p-1}n_kx^k

in \mathbb{Q}[x]. We know that q_1(x) is irreducible over \mathbb{Q} and so, since q_1(\alpha)=0, it is the minimal polynomial of \alpha. Hence q_1(x) divides q_2(x) because we also have q_2(\alpha)=0, by (*). So q_2(x) is a constant multiple of q_1(x) and hence

n_0=n_1= \cdots = n_{p-1}=n.

If n \ge 1, then A would have at least np > p-1 eigenvalues, which is impossible because A \in M_{p-1}(\mathbb{C}).

Hence n=0, i.e. n_0=n_1= \cdots =n_{p-1}=0, which means that \alpha^k, \ \ k=0, 1, \cdots , p-1, is not an eigenvalue of A. Thus all the eigenvalues of A are zero and therefore A is nilpotent.

Let m \ge 1 be the smallest integer such that A^m=0. We have m=s(p+1)+r for some integers s,r with s \ge 0 and 0 \le r \le p. Then, since A^{p+1}=A, we get


So, by the minimality of m, we must have s+r \ge m=s(p+1)+r, which gives s=0. So m=r \le p and thus

0=A^{p+1-m}A^m=A^{p+1}=A. \ \Box

Let M_n(\mathbb{C}) be the ring of n\times n matrices with entries from \mathbb{C}, the field of complex numbers.
Let A,N \in M_n(\mathbb{C}) and suppose that N is nilpotent, i.e. N^k=0 for some integer k \ge 1. Then \det(N)=0. But what can we say about \det(A+N) ? Is it equal to \det(A) ? Not necessarily, of course! For example, consider

A =\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}, \ \ \  N=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}.

Then N^2=0 but \det(A+N)=-1 \ne 0=\det(A). See that in this example, AN \ne NA.

Problem. Let A,N \in M_n(\mathbb{C}) and suppose that N is nilpotent. Show that if A,N commute, i.e. AN=NA, then \det(A+N)=\det(A).

Solution. since \mathbb{C} is algebraically closed and A,N commute, A,N are simultaneously triangularizable, i.e. there exists an invertible element P \in M_n(\mathbb{C}) such that both PNP^{-1} and PAP^{-1} are triangular. Since PNP^{-1} is both nilpotent and triangular, all its diagonal entries are zero and so the diagonal entries of P(A+N)P^{-1}=PAP^{-1}+PNP^{-1} are the same as the diagonal entries of PAP^{-1}. Thus


because P(A+N)P^{-1}, \ PAP^{-1} are both triangular and the determinant of a triangular matrix is the product of its diagonal entries. So

\det(A+N)=\det(P(A+N)P^{-1})=\det(PAP^{-1})=\det(A). \ \Box

Problem. Let R be a finite ring with 1 and let n be the number of elements of R. Show that if n is square-free, then R \cong \mathbb{Z}/n\mathbb{Z}.

Solution. Since (R,+) is an abelian group of order n and n is square-free, (R,+) is cyclic. Let a \in R be a generator of (R,+). Then

R=\{0,a,2a, \cdots , (n-1)a\}.


pa=1, \ \ \ qa=a^2,

for some integers p,q and so


Thus n \mid 1-pq because n is the order of a. Hence q,n are coprime. Now consider the map

f: R \to \mathbb{Z}/n\mathbb{Z}, \ \ \ f(ka)=kq+n\mathbb{Z}, \ \ \ \ \ k \in \mathbb{Z}.

We show that f is a ring isomorphism, which is what we need.

i) f is well-defined. Because if ka=0 for some integer k, then n \mid k and so f(ka)=kq+n\mathbb{Z}=0.

ii) f is additive. That’s clear.

iii) f is bijective. Suppose that f(ka)=kq+n\mathbb{Z}=0 for some integer k. Then n \mid kq and so n \mid k, because q,n are coprime, implying that ka=0. So f is injective and hence bijective because R and \mathbb{Z}/n\mathbb{Z} have the same number of elements.

iv) f is multiplicative. Because for integers k,k' we have

\begin{aligned} f(kak'a)=f(kk'a^2)=f(kk'qa)=kk'q^2+n\mathbb{Z}=(kq+n\mathbb{Z})(k'q+n\mathbb{Z})=f(ka)f(k'a).\end{aligned} \ \Box

Problem. For integer n \ge 1, let

p_n(x):=x^{2^n}+x^{2^{n-1}}+1 \in \mathbb{Z}[x].

Find all irreducible factors of p_n(x).

Solution. Let


and see that

p_n(x)=p_{n-1}(x)q_{n-1}(x), \ \ \ \ \ n \ge 2.


p_n(x)=p_1(x)q_1(x)q_2(x) \cdots q_{n-1}(x). \ \ \ \ \ \ \ \ \ \ \ (*)

It’s clear that p_1(x)=x^2+x+1 is irreducible over \mathbb{Z}. Now, for n \ge 1, let \Phi_n(x) be the n-th cyclotomic polynomial. Using well-known properties of \Phi_n, we have

\displaystyle \Phi_{3 \cdot 2^n}(x)=\frac{\Phi_{2^n}(x^3)}{\Phi_{2^n}(x)}=\frac{x^{3 \cdot 2^{n-1}}+1}{x^{2^{n-1}}+1}=x^{2^n}-x^{2^{n-1}}+1=q_n(x).

Thus q_n is irreducible over \mathbb{Z} because cyclotomic polynomials are irreducible over \mathbb{Z}. Hence, by (*), \ p_n has exactly n irreducible factors and they are p_1, q_1, q_2, \cdots , q_{n-1}. \ \Box

Recall that a matrix A \in M_n(\mathbb{C}) is said to be idempotent if A^2=A.

Problem. Let A,B \in M_n(\mathbb{C}) be two idempotent matrices such that A-B is invertible and let \alpha, \beta \in \mathbb{C}. Let I \in M_n(\mathbb{C}) be the identity matrix. Show that

i) if \alpha \notin \{0,-1\}, then I+\alpha AB is not necessarily invertible

ii) if \alpha \in \{0,-1\}, then I+\alpha AB is invertible

iii) A+B-AB is invertible

iv) if \alpha \beta \ne 0, then \alpha A+\beta B is invertible.

Solution. i) Choose

\displaystyle A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \ \ \ \ \ B=\begin{pmatrix} -\frac{1}{\alpha} & -\frac{1}{\alpha} \\ 1+\frac{1}{\alpha} & 1+\frac{1}{\alpha}\end{pmatrix}.

See that A^2=A, \ B^2=B and A-B is invertible but \alpha AB+I is not invertible.

ii) It’s clear for \alpha =0. For \alpha =-1, suppose that ABv=v for some v \in \mathbb{C}^n. We need to show that v=0. Well, we have ABv=A^2Bv=Av and thus Bv-v \in \ker A. But since B^2=B, we also have Bv -v \in \ker B and hence

Bv -v \in \ker A \cap \ker B \subseteq \ker(A-B)=(0),

because A-B is invertible. So Bv=v and therefore Av=ABv=v. So v \in \ker(A-B) =(0).

iii) Let A'=I-A, \ B'=I-B. See that A',B' are idempotents and so, since A'-B'=-(A-B) is invertible, I-A'B' is invertible, by ii). The result now follows because


iv) Let \displaystyle \gamma:=-\frac{\beta}{\alpha} \ne 0 and suppose that Av=\gamma Bv for some v \in \mathbb{C}^n. We are done if we show that v=0. Well, we have BAv=\gamma Bv=Av=A^2v and thus (A-B)Av=0 implying that Av=0 because A-B is invertible. Hence \gamma Bv=Av=0, which gives Bv=0 because \gamma \ne 0. Thus (A-B)v=0 and therefore v=0. \ \Box