## Space of trace zero matrices = span of nilpotent matrices

Posted: August 14, 2019 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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

$\text{tr}(a_1A_1+a_2A_2)=a_1\text{tr}(A_1)+a_2\text{tr}(A_2)=0.$

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

$A=PUP^{-1}$

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

$A=PDP^{-1}+PVP^{-1}$

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

$(e_{ii}+e_{in}-e_{ni}-e_{nn})^2=(e_{ii}-e_{in}+e_{ni}-e_{nn})^2=0.$

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 $(*). \ \Box$

## Finite direct product of normal subgroups with trivial intersections

Posted: August 6, 2019 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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 $G. \ \Box$

## For p prime & A in M_(p-1)(C), A^(p+1)=A & tr(A)=0 imply A=0

Posted: August 6, 2019 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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

$0=A^m=A^{s(p+1)+r}=A^{s+r}.$

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

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

## Determinant of sum of a matrix and a nilpotent matrix

Posted: August 3, 2019 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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

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

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

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

## Finite rings of square-free order

Posted: July 24, 2019 in finite rings, Noncommutative Ring Theory Notes, Uncategorized

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

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

Thus

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

for some integers $p,q$ and so

$(1-pq)a=a-pqa=a-pa^2=0.$

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

## Irreducible factors of x^2^n + x^2^(n-1) + 1

Posted: May 27, 2019 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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

$q_n(x):=x^{2^n}-x^{2^{n-1}}+1$

and see that

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

Thus

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

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 $p_1, q_1, q_2, \cdots , q_{n-1}. \ \Box$

## On idempotent matrices A, B such that A – B is invertible

Posted: May 27, 2019 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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

$I-A'B'=I-(I-A)(I-B)=A+B-AB.$

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 $v=0. \ \Box$