Let $\R^{\times}=\R\setminus \{0\}$ be the multiplicative group of real numbers.
Let $\C^{\times}=\C\setminus \{0\}$ be the multiplicative group of complex numbers.
Then show that $\R^{\times}$ and $\C^{\times}$ are not isomorphic as groups.

Recall.

Let $G$ and $K$ be groups.
Recall that a map $f: G \to K$ is a group homomorphism if
\[f(ab)=f(a)f(b)\] for all $a, b \in G$.
A group isomorphism is a bijective homomorphism.
If there is a group isomorphism from $G$ to $K$, we say that $G$ and $K$ are isomorphic (as groups).
 
We give two proofs.

Proof 1.

Seeking a contradiction, assume that there is a group isomorphism $\phi: \C^{\times} \to \R^{\times}$.
Since $\phi$ is a group homomorphism, $\phi(1)=1$. Thus we have
\[1=\phi(1)=\phi((-1)(-1))=\phi(-1)\phi(-1)=\phi(-1)^2.\] Hence $\phi(-1)=\pm 1$. But since $\phi$ is injective and $\phi(1)=1$, we must have $\phi(-1)=-1$.

Now we have
\begin{align*}
-1=\phi(-1)=\phi(i^2)=\phi(i)^2.
\end{align*}

Since $\phi(i)\in \R^{\times}$, $\phi(i)^2$ must be a positive number. Thus we reached a contradiction.
Hence there is no isomorphism between $\R^{\times}$ and $\C^{\times}$.

Proof 2.

Suppose that there is a group isomorphism $\phi: \C^{\times} \to \R^{\times}$.
We want to find a contradiction.
Let $\zeta=e^{2\pi i /3}$ be a primitive third root of unity.

Since $\zeta^3=1$ and $\phi$ is a group homomorphism, we have
\begin{align*}
1=\phi(1)=\phi(\zeta^3)=\phi(\zeta)^3.
\end{align*}
Since $\phi(\zeta)$ is a real number, this implies that $\phi(\zeta)=1$.

This is a contradiction since $\phi$ is injective, but we have $\phi(1)=1=\phi(\zeta)$.
Therefore, there cannot be a group isomorphism between $\R^{\times}$ and $\C^{\times}$.

The post Multiplicative Groups of Real Numbers and Complex Numbers are not Isomorphic first appeared on Problems in Mathematics.

Answer the following questions regarding eigenvalues of a real matrix.

(a) True or False. If each entry of an $n \times n$ matrix $A$ is a real number, then the eigenvalues of $A$ are all real numbers.
(b) Find the eigenvalues of the matrix
\[B=\begin{bmatrix}
-2 & -1\\
5& 2
\end{bmatrix}.\]

(The Ohio State University, Linear Algebra Exam)

Hint.

Consider a $2\times 2$ matrix.
Then the eigenvalues are solutions of a quadratic polynomial.

Does a quadratic polynomial always have real solutions?

Solution.

(a) True or False. If each entry of an $n \times n$ matrix $A$ is a real number, then the eigenvalues of $A$ are all real numbers.

 False. In general, a real matrix can have a complex number eigenvalue. In fact, the part (b) gives an example of such a matrix.

(b) Find the eigenvalues of the matrix

 The characteristic polynomial for $B$ is
\[ \det(B-tI)=\begin{bmatrix}
-2-t & -1\\
5& 2-t
\end{bmatrix}=t^2+1.\]

The eigenvalues are the solutions of the characteristic polynomial. Thus solving $t^2+1=0$, we obtain eigenvalues $\pm i$, where $i=\sqrt{-1}$.
Thus the eigenvalue of a real matrix $B$ is pure imaginary numbers $\pm i$.

The post True or False: Eigenvalues of a Real Matrix Are Real Numbers first appeared on Problems in Mathematics.