Let $V$ be the vector space of all $n\times n$ real matrices.
Let us fix a matrix $A\in V$.
Define a map $T: V\to V$ by
\[ T(X)=AX-XA\] for each $X\in V$.

(a) Prove that $T:V\to V$ is a linear transformation.

(b) Let $B$ be a basis of $V$. Let $P$ be the matrix representation of $T$ with respect to $B$. Find the determinant of $P$.

Proof.

(a) Prove that $T:V\to V$ is a linear transformation.

To prove $T$ is a linear transformation, we need to show the following properties.

1. For any $X, Y\in V$, we have $T(X+Y)=T(X)+T(Y)$.
2. For any $X\in V, r\in \R$, we have $T(rX)=rT(X)$.

To check condition 1, let $X, Y \in V$. Then we have
\begin{align*}
T(X+Y)&=A(X+Y)-(X+Y)A && \text{by definition of $T$}\\
&=AX+AY-XA-YA\\
&=AX-XA+AY-YA\\
&=T(X)+T(Y) && \text{by definition of $T$}.
\end{align*}
Hence condition 1 is met.

To verify condition 2, let $X\in V, r\in \R$.
Then we have
\begin{align*}
T(rX)&=A(rX)-(rX)A && \text{by definition of $T$}\\
&=rAX-rXA && \text{$r$ is a scalar}\\
&=r(AX-XA)\\
&=rT(X) && \text{by definition of $T$}.
\end{align*}
So condition 2 is also met, hence $T$ is a linear transformation.

(b) Find the determinant of the matrix representation of $T$.

Let $B$ be a basis of the vector space $V$ and let $P$ be the matrix of the linear transformation $T$ with respect to $B$. We prove that the determinant of $P$ is zero.

Let $I$ be the $n\times n$ identity matrix. Then we have
\begin{align*}
T(I)=AI-IA=A-A=O,
\end{align*}
where $O$ is the $n\times n$ zero matrix.
Since $T(O)=O$, this implies that the linear transformation $T$ is not injective, hence $P$ is a singular matrix.

Let us explain the details.
Let $v=[I]_B \in \R^{n^2}$ be the coordinate vector of $I$ with respect to the basis $B$.
Then since $I\neq O$, the vector $v$ is not zero.
Then $T(I)=O$ implies that
\[Pv=0\in \R^{n^2}.\] As the nonzero vector $v$ is a solution of the matrix equation $Px=0$, the matrix $P$ is singular.

Since $P$ is singular, the determinant of $P$ is zero.

The post Linear Transformation $T(X)=AX-XA$ and Determinant of Matrix Representation first appeared on Problems in Mathematics.

Let $f:G\to G’$ be a group homomorphism. We say that $f$ is monic whenever we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$, we have $g_1=g_2$.

Then prove that a group homomorphism $f: G \to G’$ is injective if and only if it is monic.

Proof.

$(\implies)$ Injective implies monic

Suppose that $f: G \to G’$ is an injective group homomorphism.
We show that $f$ is monic.

So suppose that we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$.
Then for any $x\in K$, we have
\[f(g_1(x))=f(g_2(x)).\] Since $f$ is injective, it follows that
\[g_1(x)=g_2(x)\] for any $x\in K$, and thus we obtain $g_1=g_2$. Thus $f$ is monic.

$(\impliedby)$ Monic implies injective

For the opposite implication, we prove the contrapositive statement. Namely, we prove that if $f$ is not injective, then $f$ is not monic.

Suppose that $f$ is not injective. Then the kernel $\ker(f)$ is a non-trivial subgroup of $G$.
We define the group homomorphism $g_1: \ker(f)\to G$ to be the identity map on $\ker(f)$. That is $g_1(x)=x$ for all $x\in \ker(f)$.
Also we define the group homomorphism $g_2:\ker(f)\to G$ by the formula $g_2(x)=e$ for all $x\in \ker(f)$, where $e$ is the identity element of $G$.
Since $\ker(f)$ is a nontrivial group, these two homomorphisms are distinct: $g_1\neq g_2$.

However, we have
\[fg_1=fg_2.\] In fact, we have for $x\in \ker(f)$
\[fg_1(x)=f(x)=e’,\] where $e’$ is the identity element of $G’$,
and
\[fg_2(x)=f(e)=e’.\] Thus, by definition, the homomorphism $f$ is not monic as required to complete the proof.

The post A Group Homomorphism is Injective if and only if Monic first appeared on Problems in Mathematics.

Show that any finite integral domain $R$ is a field.
 

Definition.

A commutative ring $R$ with $1\neq 0$ is called an integral domain if it has no zero divisors.
That is, if $ab=0$ for $a, b \in R$, then either $a=0$ or $b=0$.

Proof.

We give two proofs.

Proof 1.

Let $r \in R$ be a nonzero element in $R$.
We show that $r$ is a unit.

Consider the map $f: R\to R$ sending $x\in R$ to $f(x)=rx$.
We claim that the map $f$ is injective.
Suppose that we have $f(x)=f(y)$ for $x, y \in R$. Then we have
\[rx=ry\] or equivalently, we have
\[r(x-y)=0.\]

Since $R$ is an integral domain and $r\neq 0$, we must have $x-y=0$, and thus $x=y$.
Hence $f$ is injective. Since $R$ is a finite set, the map is also surjective.

Then it follows that there exists $s\in R$ such that $rs=f(s)=1$, and thus $r$ is a unit.
Since any nonzero element of a commutative ring $R$ is a unit, $R$ is a field.

Proof 2.

Let $r\in R$ be a nonzero element.
We show that the inverse element of $r$ exists in $R$ as follows.
Consider the powers of $r$:
\[r, r^2, r^3,\dots.\] Since $R$ is a finite ring, not all of the powers cannot be distinct.
Thus, there exist positive integers $m > n$ such that
\[r^m=r^n.\]

Equivalently we have
\[r^n(r^{m-n}-1)=0.\] Since $R$ is an integral domain, this yields either $r^n=0$ or $r^{m-n}-1=0$.
But the former gives $r=0$, and this is a contradiction since $r\neq 0$.

Hence we have $r^{m-n}=1$, and thus
\[r\cdot r^{m-n-1}=1.\] Since $r-m-1 \geq 0$, we have $r^{m-n-1}\in R$ and it is the inverse element of $r$.

Therefore, any nonzero element of $R$ has the inverse element in $R$, hence $R$ is a field.

Related Question.

For a solution, check out the post “Every Prime Ideal of a Finite Commutative Ring is Maximal“.

The post Finite Integral Domain is a Field first appeared on Problems in Mathematics.

Let $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that
\[b^m=a.\]  

We give two proofs.

Proof 1.

Since $m$ and $n$ are relatively prime integers, there exists integers $s, t$ such that
\[sm+tn=1.\]

Then we have
\begin{align*}
a&=a^1=a^{sm+tn}=a^{sm}a^{tn} \tag{*}
\end{align*}
Note that since the order of the group $G$ is $n$, any element of $G$ raised by the power of $n$ is the identity element $e$ of $G$.
Thus we have
\[a^{tn}=(a^n)^t=e^t=e.\] Putting $b:=a^s$, we have from (*) that
\[a=b^me=b^m.\]

Now we show the uniqueness of such $b$. Suppose there is another $g’\in G$ such that
\[a=b’^m.\] Then we have
\begin{align*}
&\quad b^m=a=b’^m\\
&\Rightarrow b^{sm}=b’^{sm} \quad \text{ by taking $s$-th power}\\
&\Rightarrow b^{1-tn}=b’^{1-tn}\\
&\Rightarrow b(b^n)^t=b'(b’^n)^t\\
&\Rightarrow b=b’ \quad \text{ since } b^n=e=b’^n.
\end{align*}
Therefore, we have $b=b’$ and the element $b$ satisfying $a=b^m$ is unique.

Proof 2.

Consider a map $f$ from $G$ to $G$ itself defined by sending $g$ to $f(g)=g^m$.
We show that this map is injective.
Suppose that $f(g)=f(g’)$.

Since $m$ and $n$ are relatively prime integers, there exists integers $s, t$ such that
\[sm+tn=1.\]

We have
\begin{align*}
f(g)&=f(g’)\\
&\Rightarrow g^m=g’^m \\
&\Rightarrow g^{sm}=g’^{sm} \quad \text{ by taking $s$-th power}\\
&\Rightarrow g^{1-tn}=g’^{1-tn}\\
&\Rightarrow g(g^n)^t=g'(g’^n)^t\\
&\Rightarrow g=g’ \quad \text{ since } g^n=e=g’^n, n=|G|.
\end{align*}

Therefore the map $f$ is injective. Since $G$ is a finite set, it also follows that the map is bijective.
Thus for any $a \in G$, there is a unique $b \in G$ such that $f(b)=a$, namely, $b^m=a$.

The post Finite Group and a Unique Solution of an Equation first appeared on Problems in Mathematics.

Let $G$ and $H$ be groups and let $f:G \to K$ be a group homomorphism. Prove that the homomorphism $f$ is injective if and only if the kernel is trivial, that is, $\ker(f)=\{e\}$, where $e$ is the identity element of $G$.
 

 

Definitions/Hint.

We recall several relevant definitions.
• A group homomorphism $f:G\to H$ is a map such that for any $g_1, g_2 \in G$, we have
  \[f(g_1g_2)=f(g_1)f(g_2).\]
• A group homomorphism $f:G \to H$ is injective if for any $g_1, g_2 \in G$
  the equality
  \[f(g_1)=f(g_2)\] implies $g_1=g_2$.
• The kernel of a group homomorphism $f:G \to H$ is a set of all elements of $G$ that is mapped to the identity element of $H$.
  Namely,
  \[\ker(f)=\{g\in G \mid f(g)=e’\},\] where $e’$ is the identity element of $H$.

Proof.

Injective $\implies$ the kernel is trivial

Suppose the homomorphism $f: G \to H$ is injective.
Then since $f$ is a group homomorphism, the identity element $e$ of $G$ is mapped to the identity element $e’$ of $H$. Namely, we have $f(e)=e’$.
If $g \in \ker(f)$, then we have $f(g)=e’$, and thus we have
\[f(g)=f(e).\] Since $f$ is injective, we must have $g=e$. Thus we have $\ker(f)=\{e\}$.

The kernel is trivial $\implies$ injective

On the other hand, suppose that $\ker(f)=\{e\}$.
If $g_1, g_2$ are elements of $G$ such that
\[f(g_1)=f(g_2), \tag{*}\] then we have
\begin{align*}
f(g_1g_2^{-1})&=f(g_1)f(g_2^{-1}) \quad \text{ since } f \text{ is a homomorphism}\\
&=f(g_1)f(g_2)^{-1} \quad \text{ since } f \text{ is a homomorphism}\\
&=f(g_1)f(g_1)^{-1} \quad \text{ by } (*)\\
&=e’.
\end{align*}

In the second step, we used the fact $f(g_2^{-1})=f(g_2)^{-1}$, which is proved in the post “Group Homomorphism Sends the Inverse Element to the Inverse Element“.

Thus the element $g_1g_2^{-1}$ is in the kernel $\ker(f)=\{e\}$, and hence $g_1g_2^{-1}=e$.
This implies that we have $g_1=g_2$ and $f$ is injective.

The post A Group Homomorphism is Injective if and only if the Kernel is Trivial first appeared on Problems in Mathematics.