# Set of invertible lower triangular Toeplitz-matrices form a group

Let $K$ be a field and $n \in \mathbb{N}$

First, I have to prove that the set of all invertible lower triangular matrices form a group under Matrix Multiplication. I proved this, by showing that the set is a subgroup of $GL_n(K)$. Correct?

Second question, which is really confusing me: Prove that the set of invertible lower triangular matrices, that are also Toeplitz-matrices form a group under Matrix Mulitplication.

I am not really sure how to start, since I can't fully imagine how these matrices look like.

• Why not start by writing down the definition of a lower triangular Toeplitz matrix? That might help! Then you could experiment with $n=2,3,4$. etc. Showing closure under products looks routine, Closure under inversion might be more challenging. – Derek Holt May 30 '17 at 21:56

Toeplitz matrices have the same entry in each descending diagonal.

We have to prove 2 things to show that the set of inveritble lower triangular matrices, that are also Toeplitz matrices form a group:

(1): The product of 2 Toeplitz matrices is Toeplitz. As a lower triangular Toeplitz matrix is determined by it's first column, we will write $T(a_1, \ldots, a_n) := \begin{pmatrix} a_{1} & 0 & \cdots & 0 \\ a_{2} & a_{1} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ a_{n} & a_{n-1} & \cdots & a_{1} \end{pmatrix}$.

Now $T(a_1, \ldots, a_n)_{i,j} = \begin{cases} 0 & \quad \text{if } i < j \\ a_{i-j+1} & \quad \text{if } i \geq j\\ \end{cases}$

Let two Toeplitz matrices $P :=T(a_1,\ldots,a_n)$ and $Q:=T(b_1,\ldots,b_n)$ be given. Now, for $i\geq j$ (as otherwise the entry is clearly $0$) we get:

$(P.Q)_{i,j} = \sum_{k=1}^n P_{i,k}.Q_{k,j} = \sum_{k=j}^{i}a_{i-k+1}.b_{k-j+1} = \sum_{k=j+1}^{i+1} a_{(i+1) - k + 1}.b_{k-(j+1)+1} = \sum_{k=1}^{n}P_{(i+1),k}.Q_{k, (j+1)} = (P.Q)_{i+1,j+1}$

And hence the entries on a similar diagonal in the product are equal, implying the resulting matrix is in fact Toeplitz.

(2): The inverse of a Toeplitz matrix is Toeplitz. Let again $P:= T(a_1, \ldots, a_n)$ be given. We know that $P$ is invertible (as it is an invertible lower triangular matrix by our assumption). Now the inverse is of the form $P^{-1} = \begin{pmatrix} b_{1,1} & 0 & \cdots & 0 \\ b_{2,1} & b_{2,2} & \cdots &0 \\ \vdots & \vdots & \ddots & \vdots \\ b_{n,1} & b_{n,2} & \cdots & b_{n,n} \end{pmatrix}$

We use induction on the diagonals to prove that in fact $b_{i,j} = b_{i+1,j+1}$:

$1 = (P.P^{-1})_{i,i} = \sum_{k=i}^i a_{i-k+1}b_{k,i} = a_{1}b_{i,i}$. Hence for all $i,j$, $b_{i,i} = b_{j,j}$, as $a_1 \neq 0$ (otherwise, the determinant, which is simply $(a_1)^n$, is $0$ and hence the matrix wouldn't be invertible).

Now suppose for given $l$, for all $n$ and $m\leq l$, $b_{m,1} = b_{m+n, 1+n} = : b_m$, i.e. the entries of the first $l$ diagonals have shown to not vary within each diagonal. Then for

$0 = (P.P^{-1})_{l+1,1} = \sum_{k=1}^{l+1}a_{l+1 - k +1}.b_{k,1} = \sum_{k=1}^{l}a_{l+1-k+1}.b_{k} + a_1.b_{l+1,1} =\sum_{k=2}^{l+1}a_{l+2 - k +1}b_{k-1} + a_{1}b_{l+2,2} = \sum_{k=2}^{l+2}a_{l+2 - k +1}b_{k,2} = (P.P^{-1})_{l+2, 2}$

and with $\sum_{k=1}^{l}a_{l+1-k+1}.b_{k} = \sum_{k=2}^{l+1}a_{l+2 - k +1}b_{k-1}$

and again $a_1 \neq0$

we deduce that $b_{l+1,1} = b_{l+2,2}$ and similarly for all $k$: $b_{l+1,1} = b_{l+1+k,1+k}$, so the entries of the $l+1$st diagonal in fact don't vary. Hence, by completing the induction, $P^{-1}$ is Toeplitz.

This gives us that invertible lower triangular matrices, which are of Toeplitz form, form a group.