Let $A$ be a real skew-symmetric matrix, that is, $A^{\trans}=-A$.
Then prove the following statements.

(a) Each eigenvalue of the real skew-symmetric matrix $A$ is either $0$ or a purely imaginary number.

(b) The rank of $A$ is even.

Proof.

(a) Each eigenvalue of the real skew-symmetric matrix $A$ is either $0$ or a purely imaginary number.

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$. That is, we have
\[A\mathbf{x}=\lambda \mathbf{x}.\] Multiplying by $\bar{\mathbf{x}}^{\trans}$ from the left, we have
\begin{align*}
\bar{\mathbf{x}}^{\trans}A\mathbf{x}=\lambda \bar{\mathbf{x}}^{\trans} \mathbf{x}=\lambda ||\mathbf{x}||^2. \tag{*}
\end{align*}

Note that the left hand side $\bar{\mathbf{x}}^{\trans}A\mathbf{x}$ is the dot (inner) product of $\bar{\mathbf{x}}$ and $A\mathbf{x}$. Since the dot product is commutative, we have
\begin{align*}
&\text{The left hand side of (*)}\\
&=\bar{\mathbf{x}}^{\trans}A\mathbf{x}=(A\mathbf{x})^{\trans}\bar{\mathbf{x}}\\
&=x^{\trans}A^{\trans}\bar{\mathbf{x}}.
\end{align*}

Since $A$ is skew-symmetric, we have $A^{\trans}=-A$. Substituting this into the above equality, we have
\begin{align*}
&\text{The left hand side of (*)}\\
&=x^{\trans}A^{\trans}\bar{\mathbf{x}}=-\mathbf{x}^{\trans}A\bar{\mathbf{x}}
\end{align*}

Taking conjugate of $A\mathbf{x}=\lambda\mathbf{x}$ and use the fact that $A$ is real, we have
\[A\bar{\mathbf{x}}=\bar{\lambda}\bar{\mathbf{x}}.\]

Thus, we have
\begin{align*}
&\text{The left hand side of (*)}\\
&=-\mathbf{x}^{\trans}A\bar{\mathbf{x}}\\
&=-\mathbf{x}^{\trans}\bar{\lambda}\bar{\mathbf{x}}=-\bar{\lambda}||\mathbf{x}||^2.
\end{align*}

Therefore comparing the left and right hand sides of (*) yields
\[-\bar{\lambda}||\mathbf{x}||^2=\lambda ||\mathbf{x}||^2.\] Since $\mathbf{x}$ is an eigenvector, it is nonzero by definition. Thus $||\mathbf{x}||\neq 0$.

Hence we have
\[-\bar{\lambda}=\lambda,\] and this implies that $\lambda$ is either $0$ or purely imaginary number.
(To see this, let $\lambda=a+ib$, where $a, b\in \R$. Then
\[-\bar{\lambda}=-a+ib=a+ib=\lambda\] implies $a=0$, thus $\lambda=bi$.)

Remark: Another similar proof is to take conjugate of (*).

(b) The rank of $A$ is even

From part (a), we know that the eigenvalues of $A$ are $0$ or purely imaginary.

Thus if $\lambda$ is a purely imaginary eigenvalue of $A$, then its conjugate $\bar{\lambda}=-\lambda$ is also an eigenvalue of $A$ since $A$ is a real matrix.
Thus, nonzero eigenvalues come in pairs $\lambda, -\lambda$ (and their algebraic multiplicities are the same).

Let
\[\lambda_1, -\lambda_1, \lambda_2, -\lambda_2, \dots, \lambda_k, -\lambda_k\] be nonzero eigenvalues of $A$.

Since a real skew-symmetric matrix is normal, it is diagonalizable (by a unitary matrix).
Thus there exists an invertible matrix $P$ such that
\begin{align*}
&P^{-1}AP=\\
&\diag\begin{bmatrix}
\lambda_1 & -\lambda_1 & \lambda_2 & -\lambda_2 & \dots &\lambda_k &-\lambda_k & 0 &\dots 0
\end{bmatrix},
\end{align*} where $\diag[a_1,a_2,\dots, a_n]$ denotes the $n\times n$ matrix whose diagonal entries are $a_1, a_2, \dots, a_n$ and all the off diagonal entries are zero.

Since $P$ is an invertible matrix, the rank of $A$ is the same as the rank of the diagonal matrix on the right-hand side, which is easily seen to be $2k$.
Thus the rank of $A$ is $2k$, and we have proved that the rank of the real skew-symmetric matrix $A$ is even.

Related Question.

As an application of this problem, try the following problem.

For a proof, see the post↴
If $A$ is a Skew-Symmetric Matrix, then $I+A$ is Nonsingular and $(I-A)(I+A)^{-1}$ is Orthogonal.

For a proof, check out the post ↴
The Determinant of a Skew-Symmetric Matrix is Zero.

More Eigenvalue and Eigenvector Problems

Problems about eigenvalues and eigenvectors are collected on the page:

Eigenvectors and Eigenspaces

The post Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even first appeared on Problems in Mathematics.

Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.

Hint.

It follows from Sylow’s theorem that if $Q_1$ and $Q_2$ are both $p$-Sylow subgroups of a group $H$, then they are conjugate.
Namely, there exists $h\in H$ such that $h^{-1}Q_1h=Q_2$.

For more details, check out the post Sylow’s theorem (summary)

To prove the problem, let $g\in G$ be any element and try to show that both $P$ and $g^{-1}Pg$ are $p$-Sylow subgroups of $N$.
Then use the fact above with $Q_1=P$, $Q_2=g^{-1}Pg$, and $H=N$.

We use the following notations: $A < B$ means that $A$ is a subgroup of a group $B$, and $A \triangleleft B$ denotes that $A$ is a normal subgroup of $B$.

Proof.

For any $g \in G$, since $P < N$ and $N \triangleleft G$, we have \begin{align*} g^{-1}Pg < g^{-1}Ng=N. \end{align*} Thus $g^{-1}Pg$ is a $p$-Sylow subgroup in $N$. In general, any two $p$-Sylow subgroups in a group are conjugate by Sylow's theorem. Since $P$ and $g^{-1}Pg$ are both $p$-Sylow subgroups in $N$, there exists $n \in N$ such that \[n^{-1}Pn=g^{-1}Pg.\] Since $n\in N$ and $P$ is normal in $N$, we have $n^{-1}Pn=P$. Hence we obtain \[P=g^{-1}Pg.\] Since $g\in G$ is arbitrary, this implies that $P$ is a normal subgroup in $G$.

The post If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup first appeared on Problems in Mathematics.

Let $G$ be a group. We fix an element $x$ of $G$ and define a map
\[ \Psi_x: G\to G\] by mapping $g\in G$ to $xgx^{-1} \in G$.
Then prove the followings.
(a) The map $\Psi_x$ is a group homomorphism.

(b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$, where $Z(G)$ is the center of the group $G$.

(c) The map $\Psi_y=\id$ for all $y\in G$ if and only if $G$ is an abelian group.

Proof.

(a) The map $\Psi_x$ is a group homomorphism

For any elements $g, h\in G$, we have
\begin{align*}
\Psi_x(gh)=x(gh)x^{-1}\stackrel{(*)}{=} xgx^{-1}xhx^{-1}=\Psi_x(g) \Psi_x(h),
\end{align*}
where we inserted the identity element $e=x^{-1}x$ between $g$ and $h$ to obtain (*).
Hence $\Psi_x$ is a group homomorphism.

(b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$

$(\implies)$ Suppose that $\Psi_x=\id$. Then for any $g\in G$, we have
\begin{align*}
\Psi_x(g)=\id(g)
\end{align*}
and thus we have
\begin{align*}
xgx^{-1}=g.
\end{align*}
This implies that we have $xg=gx$ for all $g \in G$, and hence $x\in Z(G)$.

$(\impliedby)$ On the other hand, if $x$ is in the center $Z(G)$, then we have
\[\Psi_x(g)=xgx^{-1}=xx^{-1}g=g\] for any $g\in G$, where the second equality follows since $x \in Z(G)$.
This yields that $\Psi_x=\id$.

(c) The map $\Psi_y=\id$ for all $y\in G$ if and only if $G$ is an abelian group

$(\implies)$ Suppose that the map $\Psi_y=\id$ for all $y\in G$. Then by part (b), we have $y\in Z(G)$ for all $y\in G$. This means that we have $G=Z(G)$, and hence $G$ is an abelian group.

$(\impliedby)$ Now suppose that $G$ is an abelian group. Then for any $y\in G$ we have
\[\Psi_y(g)=ygy^{-1}=yy^{-1}g=g=\id(g)\] for any $g\in G$, where the second equality follows since $G$ is an abelian group.
Thus we have $\Psi_y=\id$ for any $y \in G$.

The post Group Homomorphism, Conjugate, Center, and Abelian group first appeared on Problems in Mathematics.

Let $G$ be a group. Assume that $H$ and $K$ are both normal subgroups of $G$ and $H \cap K=1$. Then for any elements $h \in H$ and $k\in K$, show that $hk=kh$.
 

Proof.

It suffices to show that $h^{-1}k^{-1}hk \in H \cap K$.
In fact, if this it true then we have $h^{-1}k^{-1}hk=1$, and thus $hk=kh$.

Since $h\in H$ and $H$ is a normal subgroup of $G$, we see that the conjugate $k^{-1}hk\in H$.
Thus we have
\[h^{-1}k^{-1}hk =h^{-1}(k^{-1}hk)\in H. \tag{*}\]

Also, since $k^{-1}\in K$ and $K$ is a normal subgroup of $G$, we have the conjugate $h^{-1}k^{-1}h\in K$.
Hence, we see that
\[h^{-1}k^{-1}hk =(h^{-1}k^{-1}h)k\in K. \tag{**}\]

From (*) and (**), we see that the element $h^{-1}k^{-1}hk$ is in both $H$ and $K$, hence in $H\cap K$ as claimed.

The post Two Normal Subgroups Intersecting Trivially Commute Each Other first appeared on Problems in Mathematics.

Let $G$ be a group and let $A$ be an abelian subgroup of $G$ with $A \triangleleft G$.
(That is, $A$ is a normal subgroup of $G$.)

If $B$ is any subgroup of $G$, then show that
\[A \cap B \triangleleft AB.\]

Proof.

First of all, since $A \triangleleft G$, the product $AB$ is a subgroup of $G$.
To show that $A\cap B$ is a normal subgroup of $AB$, let $x\in A \cap B$ and $ab\in AB$, where $a\in A$ and $b \in B$.
Then we have the conjugate
\begin{align*}
(ab)x(ab)^{-1}=a(bxb^{-1})a^{-1}. \tag{*}
\end{align*}

We show that the right hand side of (*) is in both $A$ and $B$.
Since $x\in A\cap B \subset A$ and $A$ is a normal subgroup of $G$, we have
\[bxb^{-1}\in A.\] Thus the right hand side of (*) is in $A$.

Also, since the elements $a, bxb^{-1}, a^{-1}$ are all in the abelian group $A$, we have
\begin{align*}
a(bxb^{-1})a^{-1}=aa^{-1}(bxb^{-1})=bxb^{-1}\in B
\end{align*}
since $x, b\in B$.

Therefore $(ab)x(ab)^{-1}$ is in both $A$ and $B$, and hence
\[(ab)x(ab)^{-1} \in A \cap B\] and the group $A \cap B$ is a normal subgroup of $AB$.

The post Abelian Normal Subgroup, Intersection, and Product of Groups first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
1 & -1\\
2& 3
\end{bmatrix}.\]

Find the eigenvalues and the eigenvectors of the matrix
\[B=A^4-3A^3+3A^2-2A+8E.\]

(Nagoya University Linear Algebra Exam Problem)
 

Hint.

Apply the Cayley-Hamilton theorem.
That is if $p_A(t)$ is the characteristic polynomial of the matrix $A$, then the matrix $p_A(A)$ is the zero matrix.

Solution.

Let us first find the characteristic polynomial $p_A(t)$ of the matrix $A$.
We have
\begin{align*}
p_A(t)&=\det(A-tI)=\begin{vmatrix}
1-t & -1\\
2& 3-t
\end{vmatrix}\\
&=(1-t)(3-t)-(-1)(2)=t^2-4t+5.
\end{align*}
Solving $t^2-4t+5=0$, we see that the matrix $A$ has the eigenvalues $2\pm i$ but it is not a good idea to use this directly to find the eigenvalues of the matrix $B$.
Instead, note that by the Cayley-Hamilton theorem, we know that
\[p_t(A)=A^2-4A+5I=O,\] where $I$ is the $2\times 2$ identity matrix and $O$ is the $2\times 2$ zero matrix.

Since we have
\[B=A^4-3A^3+3A^2-2A+8E=(A^2-4A+5I)(A^2+A+2I)+A-2I,\] we have
\[B=A-2I=\begin{bmatrix}
-1 & -1\\
2& 1
\end{bmatrix}.\] Since the eigenvalues of $A$ is $2\pm i$, the eigenvalues of $B=A-2I$ are
\[(2\pm i)-2=\pm i.\]

Next, we find eigenvectors.
Let us first find eigenvectors corresponding to the eigenvalue $i$.
We have
\begin{align*}
A-iI&=\begin{bmatrix}
-1-i & -1\\
2& 1-i
\end{bmatrix}
\xrightarrow{(-1+i)R_1}
\begin{bmatrix}
2 & 1-i\\
2 & 1-i
\end{bmatrix}\\
&
\xrightarrow{R_2-R_1}
\begin{bmatrix}
2 & 1-i\\
0 & 0
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_1}
\begin{bmatrix}
1 & (1-i)/2\\
0 & 0
\end{bmatrix}.
\end{align*}
Thus we have
\[x_1=-\frac{1-i}{2}\] and the eigenvectors associated with the eigenvalue $i$ are
\[\mathbf{x}=x_2\begin{bmatrix}
-\frac{1-i}{2} \\
1
\end{bmatrix},\] where $x_2$ is any nonzero complex number.
Or equivalently, scaling the vector by $-1+i$, the eigenvectors corresponding to the eigenvalue $i$ are
\[a\begin{bmatrix}
1 \\
-1-i
\end{bmatrix},\] where $a$ is any nonzero complex number.

Since $B$ is a real matrix and the eigenvalues $i$ and $-i$ are complex conjugate to each other, the eigenvectors of $-i$ are just the conjugates of eigenvectors of $i$. Thus the eigenvectors corresponding to the eigenvalue $-i$ are
\[b\begin{bmatrix}
1 \\
-1+i
\end{bmatrix},\] where $b$ is any nonzero complex number.

The post Find the Eigenvalues and Eigenvectors of the Matrix $A^4-3A^3+3A^2-2A+8E$. first appeared on Problems in Mathematics.

Let $G$ be a group and $H$ and $K$ be subgroups of $G$.
For $h \in H$, and $k \in K$, we define the commutator $[h, k]:=hkh^{-1}k^{-1}$.
Let $[H,K]$ be a subgroup of $G$ generated by all such commutators.

Show that if $H$ and $K$ are normal subgroups of $G$, then the subgroup $[H, K]$ is normal in $G$.

Proof.

We first prove that a conjugate of each generator is in $[H,K]$.
Let $h \in H, k \in K$. For any $g \in G$, we have by inserting $g^{-1}g=e$ inbetweens
\begin{align*}
g[h,k]g^{-1}&=ghkh^{-1}k^{-1}=(ghg^{-1})(gkg^{-1})(gh^{-1}g^{-1})(gk^{-1}g^{-1})\\
& = (ghg^{-1})(gkg^{-1})(ghg^{-1})^{-1}(gkg^{-1})^{-1}.
\end{align*}

Now note that $ghg^{-1} \in H$ since $H$ is normal in $G$, and $gkg^{-1} \in K$ since $K$ is normal in $G$. Thus $g[h,k]g^{-1} \in [H, K]$.
By taking the inverse of the above equality, we also see that $g[k,h]g^{-1} \in [H, K]$. Thus the conjugate of the inverse $[h,k]^{-1}=[k,h]$ is in $[H, K]$.

Next, note that any element $x \in [H,K]$ is a product of generators or their inverses. So let us write
\[x=[h_1, k_1]^{\pm 1}[h_2, k_2]^{\pm 1}\cdots [h_n, k_n]^{\pm 1},\] where $h_i \in H, k_i\in K$ for $i=1,\dots, n$.
Then for any $g \in G$, we have
\begin{align*}
gxg^{-1}=(g[h_1, k_1]^{\pm 1}g^{-1})(g[h_2, k_2]^{\pm 1}g^{-1})\cdots(g [h_n, k_n]^{\pm 1} g^{-1}).
\end{align*}

We saw that the conjugate of a generator, or its inverse, by $g \in G$ is in $[H,K]$.
Thus $gxg^{-1}$ is also in $[H, K]$.
This proves that the group $[H,K]$ is a normal subgroup of $G$.

Related Question.

Another problem about a commutator group is
A condition that a commutator group is a normal subgroup

The post Group Generated by Commutators of Two Normal Subgroups is a Normal Subgroup first appeared on Problems in Mathematics.

Let $G$ be a finite group and $P$ be a nontrivial Sylow subgroup of $G$.
Let $H$ be a subgroup of $G$ containing the normalizer $N_G(P)$ of $P$ in $G$.

Then show that $N_G(H)=H$.
 

Hint.

Use the conjugate part of the Sylow theorem.
See the second statement of the Sylow theorem.

Proof.

It is clear that $H \subset N_G(H)$.
So we show that $N_G(H) \subset H$.

Take any $a \in N_G(H)$. Since $P \subset N_G(P) \subset H$, we have
\[aPa^{-1} \subset aHa^{-1}=H,\] where the last step follows from $a \in N_G(H)$.

It follows that $P$ and $aPa^{-1}$ are both Sylow subgroups in $H$.
By the Sylow theorem, any two $p$-Sylow subgroups are conjugate. Thus there exists $b \in H$ such that $bPb^{-1}=aPa^{-1}$.

This implies that $(b^{-1}a)P(b^{-1}a)^{-1}=P$ and thus $b^{-1}a \in N_G(P) \subset H$. Hence we have $a \in H$ since $b \in H$.
This shows that $N_G(H) \subset H$, hence $N_G(H)=H$ as required.

Corollary (The Normalizer of the Normalizer of a Sylow subgroup)

We apply the result to the case $H=N_G(P)$, and obtain the following result.

The post If a Subgroup Contains a Sylow Subgroup, then the Normalizer is the Subgroup itself first appeared on Problems in Mathematics.

Let $X$ be a subset of a group $G$. Let $C_G(X)$ be the centralizer subgroup of $X$ in $G$.

For any $g \in G$, show that $gC_G(X)g^{-1}=C_G(gXg^{-1})$.
 

Proof.

$(\subset)$ We first show that $gC_G(X)g^{-1} \subset C_G(gXg^{-1})$.
Take any $h\in C_G(X)$. Then for any element $x \in X$, we have
\begin{align*}
ghg^{-1}(gxg^{-1}) &=ghxg^{-1}\\
&=gxhg^{-1} \,\,\,\, \text{ since } h\in C_G(X)\\
&=gxg^{-1}(ghg^{-1}).
\end{align*}

This computation shows that $ghg^{-1} \in C_G(gXg^{-1})$ for any $h \in C_G(X)$.
Thus this proves $g C_G(X) g^{-1} \subset C_G(gXg^{-1})$.

$(\supset)$ Conversely, take any element $k \in C_G(gXg^{-1})$.
Then for any $x \in X$, we have
\begin{align*}
g^{-1}kg(x)&=g^{-1}k(gxg^{-1})g\\
&=g^{-1}(gxg^{-1})kg \,\,\,\, \text{ since } k\in C_G(gXg^{-1})\\
&=x(g^{-1}kg).
\end{align*}
This implies that $g^{-1}kg \in C_G(X)$, hence $k\in gC_G(X)g^{-1}$ for any $k \in C_G(gXg^{-1})$. This proves $C_G(gXg^{-1}) \subset gC_G(X)g^{-1}$.

The post Conjugate of the Centralizer of a Set is the Centralizer of the Conjugate of the Set first appeared on Problems in Mathematics.

Let $G$ be a finite group of odd order. Assume that $x \in G$ is not the identity element.

Show that $x$ is not conjugate to $x^{-1}$.
 

Proof.

Assume the contrary, that is, assume that there exists $g \in G$ such that $gx^{-1}g^{-1}=x$.
Then we have
\[xg=gx^{-1}. \tag{*}\] Then we compute
\begin{align*}
(xg)^2&=(gx^{-1})(xg)=g^2\\
(xg)^3&=(xg)(xg)^2=(xg)(g^2)=xg^3\\
(xg)^4&=(xg)^2(xg)^2=g^2g^2=g^4.
\end{align*}

In general, we have by induction
\begin{align*}
(xg)^k=
\begin{cases}
xg^k & \text{ if } k \text{ is odd}\\
g^k & \text{ if } k \text{ is even}. \tag{**}
\end{cases}
\end{align*}

Let $m$ be the order of $g$. Since $G$ is a finite group of odd order, $m$ is odd.
We have by (**)
\[(xg)^{m-1}g=g^{m-1}g=g^m=1.\] Thus we have $g^{-1}=(xg)^{m-1}$. Multiplying by $xg$ from the right, we have
\begin{align*}
g^{-1}xg &=(xg)^m\\
&=xg^m \text{ by (**)}\\
&=x.
\end{align*}

Thus we have $gx=xg$, and combining with (*) we obtain $gx=gx^{-1}$.
This implies that $x^2=1$, and since the order of $G$ is odd, this implies $x=1$.

This contradicts our choice of $x$.
Therefore $x$ cannot be conjugate to $x^{-1}$.

The post If a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its Inverse first appeared on Problems in Mathematics.