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Problem 52 (Heisenberg Matrix Group)

Let $G$ be the set of all 3 by 3 matrices with entries in the real numbers of the form $$\begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix}.$$ Prove that $G$ is group under matrix multiplication. This group is often called the Heisenberg group and is connected to the Heisenberg uncertainty principle. See page 51 in your text for an interesting historical fact.

Solution

Let $G$ be the set of all upper-triangular matrices with 1's along the main diagonal. So $G=\{ \begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix}| a,b,c\in \mathbb{R}\}$. In order to prove that $G$ is a group we need to show that it is closed under the operation of matrix multiplication, that the operation is associative, that there exists an identity, and that every element in $G$ has an inverse. But, really, who would go through all that work when we could just use the subgroup test. So we will show that $G$ is a group under matrix multiplication by the subgroup test. Let $H=GL(3,\mathbb{R} )$. We know that $G\leq H$ and that it is a nonempty subset because $G$ contains the identity in $H$, so $I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\in G$ means $G$ is a nonempty subset of$H$ (this is what allows us to use the subgroup test rather than starting from scratch). Now we just need to show that $G$ is closed under the operation of $H$--which is matrix multiplication--and that $G$ is closed under taking inverses. We know from previous courses in linear algebra that the product of two upper-triangular matrices is another upper-triangular matrix. Therefore, $G$ is closed under matrix multiplication. Now we will show that it is closed under taking inverses. Let $a,b,c\in \mathbb{R}$. Pick $d=-a$,$e=-b+ac$,$f=-c$. We will claim the inverse of $A=\begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix}$ is the matrix $B=\begin{bmatrix}1&d&e\\0&1&f\\0&0&1\end{bmatrix}$. To show that $B$ is the inverse of $A$ we must show $AB=BA=I$We will now compute $$\begin{array}{rl} AB=\begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix} \begin{bmatrix}1&d&e\\0&1&f\\0&0&1\end{bmatrix} &=\begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix} \begin{bmatrix}1&-a&-b+ac\\0&1&-c\\0&0&1\end{bmatrix} \\ &=\begin{bmatrix}1&-a+a&-b+ac-ac+b\\0&1&-c+c\\0&0&1\end{bmatrix} \\ &=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I \end{array}$$

Now we will compute $$\begin{array}{rl} BA=\begin{bmatrix}1&d&e\\0&1&f\\0&0&1\end{bmatrix} \begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix} &=\begin{bmatrix}1&-a&-b+ac\\0&1&-c\\0&0&1\end{bmatrix} \begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix} \\ &=\begin{bmatrix}1&a-a&b-ac-b+ac\\0&1&c-c\\0&0&1\end{bmatrix} \\ &=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I \end{array}$$

Because $AB=BA=I$ we know that $B$ is the inverse of $A$. Clearly, $B\in G$ by definition. Therefore, $B$ is closed under taking inverses. Thus, $G$ is a group by the subgroup test. $\square$

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