# Show that a certain set of invertible matrices is a normal subgroup of another set of invertible matrices (triangular)

$$U_n=\left( \begin{array}{rrrr} 1 & * & \cdots & * \\ 0 & \ddots & * & \vdots \\ \vdots & 0 & \ddots & * \\ 0 & \cdots & 0 & 1 \\ \end{array}\right) T_n=\left( \begin{array}{rrrr} * & * & \cdots & * \\ 0 & \ddots & * & \vdots \\ \vdots & 0 & \ddots & * \\ 0 & \cdots & 0 & * \\ \end{array}\right)$$

I have to Show that $$U_n$$ is a normal subgroup of $$T_n$$

To this end I have to show (after I have verified that $$U_n$$ is a subgroup of $$T_n$$):

$$\forall_{x\in T_n}\forall_{a\in U_n}\exists_{b\in U_n}xa=ba$$

I have looked at the case n=2

the results are

$$\begin{pmatrix} a & b \\ 0 & c \\ \end{pmatrix}$$ $$\begin{pmatrix} 1 & d \\ 0 & 1 \\ \end{pmatrix}$$ $$=\begin{pmatrix} a & ad+b \\ 0 & c \\ \end{pmatrix}$$ $$=\begin{pmatrix} 1 & adc^{-1} \\ 0 & 1 \\ \end{pmatrix}$$ $$\begin{pmatrix} a & b \\ 0 & c \\ \end{pmatrix}$$

Now the Question is how can I choose the coefficients for my Matrix $$b$$ in Dependance to $$a$$?

I suspect that the coefficients of $$b$$ can be determined this way

(Note that I use the notation $$b_{kj}$$ which translates to: I look at the coefficient in the $$k$$-th row at the $$j$$-th column of the Matrix $$b$$)

$$b_{kj}\begin{cases}1 &\mbox{if } j=k \\ 0 & \mbox{if } j>k\\a_{(k-1)j}a_{kj}a_{k(j+1)} & \mbox{otherwise} \end{cases}$$

How can I prove this idea?

Lauds to our colleague studiosus for a concise and elegant answer; myself, I have reverted to the somewhat computationally-intensive approach which at least has the advantage of exposing some of the detailed structure of the situation with $$T_n$$ and its subgroup $$U_n$$:

Try it like this:

Each

$$M \in U_n \tag 1$$

may be written in the form

$$M = I + N_1, \tag 2$$

where $$N_1$$ is strictly upper triangular, that is, upper triangular with vanishing main diagonal: likewise, every

$$L \in T_n \tag 3$$

takes the form

$$L = D + N_2, \tag 4$$

where $$N_2$$, like $$N_1$$, is strictly upper triangular; however,

$$D = \text{diag}(d_1, d_2, \ldots, d_n) \tag 5$$

is a diagonal matrix having every diagonal entry $$d_i \ne 0$$; this assumption is necessary lest $$L$$ be singular, since here

$$\det L = \det D = \displaystyle \prod_1^n d_i; \tag 6$$

scrutinizing (4), we see that it may be written

$$L = D(I + D^{-1}N_2), \tag 7$$

and thus

$$L^{-1} = (I + D^{-1}N_2)^{-1}D^{-1}; \tag 8$$

here

$$D^{-1} = \text{diag}(d_1^{-1} , d_2^{-1} , \ldots, d_n^{-1} ) \tag 9$$

is also a diagonal matrix with $$d_i^{-1} \ne 0$$, $$1 \le i \le n$$; furthermore, the matrix $$D^{-1}N_2$$ occurring in (7)-(8) is also easily seen to be strictly upper triangular, being the product of a strictly upper triangular matrix with a diagonal matrix; as such, $$D^{-1}N_2$$ is nilpotent:

$$(D^{-1}N_2)^n = 0, \tag{10}$$

which in turn implies that the matrix $$I + D^{-1}N_2$$ is invertible, since

$$(I + D^{-1}N_2)^{-1} = \displaystyle \sum_0^{n - 1} (-D^{-1}N_2)^k$$ $$= I - D^{-1}N_2 + (-D^{-1}N_2)^2 + \ldots + (-D^{-1}N_2)^{n - 1}, \tag{11}$$

as is easily seen by using (10) to work out the product

$$(I + D^{-1}N_2) \displaystyle \sum_0^{n - 1} (-D^{-1}N_2)^k = (I - (-D^{-1}N_2)) \displaystyle \sum_0^{n - 1} (-D^{-1}N_2)^k$$ $$= I + (-D^{-1}N_2)^n = I. \tag{12}$$

We next observe that $$D^{-1} N_2$$ and all its powers are strictly upper triangular, and from thus via (11) we infer that that the diagonals of both $$I + D^{-1}N_2$$ and $$(I + D^{-1}N_2)^{-1}$$ are comprised solely of $$1$$s; thus we may in fact write

$$(I + D^{-1}N_2)^{-1} = I + N_3, \tag{13}$$

where $$N_3$$ is also strictly upper triangular, and we have, for any $$L$$, $$M$$ as in (2), (4):

$$LML^{-1} = (D + N_2)(I + N_1)(D + N_2)^{-1}$$ $$= D(I + D^{-1}N_2)(I + N_1)( (I + D^{-1}N_2)^{-1}D^{-1}$$ $$= D(I + D^{-1}N_2)(I + N_1)(I + N_3)D^{-1}; \tag{14}$$

we turn now to the product

$$(I + D^{-1}N_2)(I + N_1)(I + N_3) \tag{15}$$

occurring in (14); we have

$$(I + N_1)(I + N_3) = I + N_1 + N_3 + N_1N_3, \tag{16}$$

and each of the matrices $$N_1$$, $$N_3$$, $$N_1N_3$$ is strictly upper triangular, hence is their sum (it is easy to show the product of two such matrices is the same); thus we see that the product (16) is in $$U_n$$; and essentially the same argument gives us

$$(I + D^{-1}N_2)(I + N_1)(I + N_3) \in U_n \tag{17}$$

as well; finally, we observe that for any matrix such as $$M$$ (cf. (2)), and invertible $$D$$ as in (5),

$$D(I + N_1)D^{-1} = DID^{-1} + DN_1D^{-1} = I + DN_1D^{-1}, \tag{18}$$

where it is easy to affirm $$DN_1D^{-1}$$ is strictly upper triangular; now turning again to (15) we see that we may infer that

$$LML^{-1} \in U_n; \tag{14}$$

but $$L$$ may be any element of $$T_n$$; thus our proof that $$U_n$$ is normal in $$T_n$$ is complete.

In closing we observe that this argument largely hinges on two easy facts: first, that the product of two strictly upper triangular matrices is again strictly upper triangular; and second, the product of a strictly upper triangular matrix and a diagonal matrix is again strictly upper triangular. From these two easily established facts our demonstration flows.

Show that a certain set of invertible matrices is a normal subgroup of another set of invertible matrices (triangular)

I want to propose an alternative proof without computations. Note that $$U_{n}$$ consists exactly of those matrices in $$T_{n}$$ whose eigenvalues are all equal to $$1$$. Given $$A\in U_{n}$$ and $$X\in T_{n}$$, we immediately get that $$XAX^{-1}\in U_{n},$$ since eigenvalues are invariant under conjugation.

• Very nice answer, as opposed to my tour de grunge; endorsed; + 1!!! – Robert Lewis Jan 30 '19 at 2:25